Class 12 Commerce Maths - Vector Algebra

Match the column

$The\quad projection\quad of\quad \vec{a} \quad onto\quad \vec { b } \quad is\quad given\quad by\quad \frac { \vec { a } \cdot \vec { b } }{ |\vec { b } | } \\ So\quad the\quad magnitude\quad of\quad projection\quad of\quad \alpha i+\beta j\quad on\quad \sqrt { 3 } i+j\quad is\quad \frac { |\alpha \sqrt { 3 } +\beta | }{ 2 } =\quad \sqrt { 3 } \\ The\quad 2nd\quad equation\quad is\quad given\quad as\quad \alpha =2+\sqrt { 3 } \beta .\\ Squaring\quad the\quad first\quad equation\quad and\quad solving\quad those\quad two\quad simultaneously\quad we\quad get\quad \alpha =2,-1.\\ So\quad |\alpha |\quad =\quad 1,2.$

B) The function given has to be continuous as well as differentiable.
Applying continuity at

$x=1$ we get

$-3a-2 = b+a^2$ .
Differentiating we get

$f'(x)=-6ax\quad for\quad x<1\\ \quad \quad \quad =\quad b\quad \quad for\quad x\ge 1\\ So\quad for\quad f\quad to\quad be\quad differentiable\quad at\quad x=1\quad we\quad should\quad have\quad -6a\quad =\quad b\\ So\quad solving\quad the\quad two\quad equations\quad simulataneously\quad we\quad get\quad a=1\quad or\quad 2$

C)

$Let\quad x=3-3\omega +2{ \omega }^{ 2 }\\ we\quad then\quad observe\quad that\quad 2+3\omega -3{ \omega }^{ 2 }=x\omega \quad and\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -3+2\omega +3\omega ^{ 2 }=x\omega ^{ 2 }\\ The\quad expression\quad then\quad becomes\quad { x }^{ 4n+3 }+{ (x\omega })^{ 4n+3 }+{ (x\omega ^{ 2 }) }^{ 4n+3 }\\ ={ x }^{ 4n+3 }*(1+{ (\omega ) }^{ 4n+3 }+{ (\omega ^{ 2 }) }^{ 4n+3 })\\ Now\quad x\neq 0\quad so\quad we\quad must\quad have\quad (1+{ (\omega ) }^{ 4n+3 }+{ (\omega ^{ 2 }) }^{ 4n+3 })=0\\ For\quad this\quad n\quad must\quad not\quad be\quad a\quad multiple\quad of\quad 3$

D)

$Since\quad the\quad H.M\quad of\quad a\quad and\quad b\quad is\quad 4\quad we\quad have\quad \frac { 2ab }{ a+b } =4\\ Also\quad a,5,q,b\quad is\quad in\quad A.P\quad So\quad let\quad 5-a=d.\\ We\quad will\quad have\quad b-5=2d\quad which\quad implies\quad 2(5-a)=b-5.\\ Solving\quad the\quad simultaneous\quad equations\quad we\quad get\quad a=5/2\quad or\quad 6.\\ so\quad the\quad AP\quad is\quad either\quad 2.5,5,7.5,10\quad or\quad 6,5,4,3.\quad \\ So\quad |q-a|\quad is\quad either\quad 5\quad or\quad 2.$

A point is on the $x$ -axis. What are its $y$ -coordinate and $z$ -coordinates?

If a point is on the

$x$ -axis then only

$x$ -coordinate will have non-zero constant value and other coordinates will be zero.

Hence,

$y$ -coordinates and

$z$ -coordinates are zero.

Find the projection of the vector $\hat {i}+3\hat {j}+7\hat {k}$ on the vector $7\hat {i}-\hat {j}+8\hat {k}$ .

Let

$\vec {a}=\hat {i}+3\hat {j}+7\hat {k}$ and

$\hat {b}=7\hat {i}-\hat {j}+8\hat {k}$ .

Now, projection of vector

$\vec {a}$ on

$\vec {b}$ is given by,

$\dfrac {1}{|\vec {b}|}(\vec {a}\cdot \vec {b})=\dfrac {1}{\sqrt {7^2+(-1)^2+8^2}}\left \{1(7)+3(-1)+7(8)\right \}=\dfrac {7-3+56}{\sqrt {49+1+64}}=\dfrac {60}{\sqrt {114}}$

If either vector $\vec {a}=\vec {0}$ or $\vec {b}=\vec {0}$ , then $\vec {a}\cdot \vec {b}=0$ . But the converse need not be true. Justify your answer with an example.

Consider

$\vec {a}=2\hat {i}+4\hat {j}+3\hat {k}$ and

$\vec {b}=3\hat {i}+3\hat {j}-6\hat {k}$ .

Then,

$\vec {a}\cdot \vec {b}=2.3+4.3+3(-6)=6+12-18=0$

We now observe that:

$|\vec {a}|=\sqrt {2^2+4^2+3^2}=\sqrt {29}$

$\therefore \vec {a}\neq \vec {0}$

$|\vec {b}|=\sqrt {3^2+3^2+(-6)^2}=\sqrt {54}$

$\therefore \vec {b}\neq \vec {0}$

Hence, the converse of the given statement need not be true.

Find the sum of the vectors $\vec {a}=\hat {i}-2\hat {j}+\hat {k}, \vec {b}=-2\hat {i}+4\hat {j}+5\hat {k}$ and $\vec {c}=\hat {i}-6\hat {j}-7\hat {k}$ .

The given vectors are

$\vec {a}=\hat {i}-2\hat {j}+\hat {k}, \vec {b}=-2\hat {i}+4\hat {j}+5\hat {k}$ and

$\vec {c}=\hat {i}-6\hat {j}-7\hat {k}$

$\therefore \vec {a}+\vec {b}+\vec {c}=(1-2+1)\hat {i}+(-2+4-6)\hat {j}+(1+5-7)\hat {k}$

$=0\cdot \hat {i}-4\hat {j}-1\cdot \hat {k}$

$=-4\hat {j}-\hat {k}$

Find $|\vec {x}|$ , if for a unit vector $\vec {a}, (\vec {x}-\vec {a})\cdot (\vec {x}+\vec {a})=12$

Answer required

$(\vec {x}-\vec {a})\cdot (\vec {x}+\vec {a})=12$

$\Rightarrow \vec {x}\cdot \vec {x}+\vec {x}\cdot \vec {a}-\vec {a}\cdot \vec {x}-\vec {a}\cdot \vec {a}=12$

$\Rightarrow |\vec {x}|^2-|\vec {a}|^2=12$

$\Rightarrow |\vec {x}|^2-1=12,$ [

$\because |\vec {a}|=1$ as

$\vec {a}$ is a unit vector]

$\Rightarrow |\vec {x}|^2=13$

$\therefore |\vec {x}|=\sqrt {13}$

If $\vec {a}, \vec {b}, \vec {c}$ are unit vectors such that $\vec {a}+\vec {b}+\vec {c}=\vec {0}$ , find the value of $\vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a}$

We have

$|\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=1$

Also

$\vec {a}+\vec {b}+\vec {c}=\vec {0}$

Squaring we get,

$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a})=0$

$\Rightarrow 1+1+1+2(\vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a})=0$

$\therefore \vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a}=-\cfrac{3}{2}$

If $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ , then find the value of $|\vec{a}\cdot \vec{b}|$ .

$\vec{a}\cdot\vec{b} = (-\hat i -\hat j + \hat k)\cdot(-\hat i + \hat j - \hat k)$

$=1-1-1$

$=-1$

Hence,

$|\vec{a}\cdot\vec{b}| = |-1| = 1$

Projection of $\vec{b}$ in the direction of $\vec{a}$ ___________.

Vector projection of a vector

$\vec{a}$ in direction of

$\vec{b}$ is given by

$\left ( \cfrac{\vec{a}.\vec{b}}{|\vec{a}|} \right ) \cfrac{\vec{a}}{|\vec{a}|}$

If vectors $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ , then find the value of $\vec{a}\cdot\vec{b}$ .

$\vec{a}\cdot\vec{b} = (\hat i - \hat j + \hat k)\cdot(\hat i +\hat j + \hat k)$

$= 1-1+1$

$=1$

For any two vectors $\vec { a } ,\vec { b }$ , prove that ${ \left| \vec { a } \times \vec { b } \right| }^{ 2 }=\begin{vmatrix} \vec { a } .\vec { a } & \vec { a } .\vec { b } \\ \vec { b } .\vec { a } & \vec { b }.\vec { b } \end{vmatrix}$

$L.H.S:|\vec{a}\times \vec{b}|^2=|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta$

$R.H.S:(\vec{a}.\vec{a})(\vec{b}.\vec{b})-(\vec{a}.\vec{b})(\vec{b}.\vec{a})$

$=|\vec{a}|^2|\vec{b}|^2-(\vec{a}.\vec{b})^2(\because \vec{a}.\vec{b}=\vec{b}.\vec{a})$

$=|\vec{a}|^2|\vec{b}|^2-(|\vec{a}||\vec{b}|\cos\theta)^2=|\vec{a}|^2|\vec{b}|^2-|\vec{a}|^2|\vec{b}|^2\cos^2\theta$

$=|\vec{a}|^2|\vec{b}|^2(1-\cos^2\theta)=|\vec{a}|^2|\vec{b}|^2\sin^2\theta$

$=L.HS.$

Hence proved.

If $\bar{OA} = 3 \hat{i} + \hat{j} - 2\hat{k}$ and $\bar{OB} = 5 \hat{i} + 4 \hat{j} - \hat{k}$ then find the value of $\bar{AB}$

We have,

$\overrightarrow { OA } =3\hat { i } +\hat { j } -2\hat { k } ,\overrightarrow { OB } =5\hat { i } +4\hat { j } -\hat { k }$

$\overrightarrow { AB } =\overrightarrow { AO } +\overrightarrow { OB }$

$\overrightarrow { AO } =-3\hat { i } -\hat { j } +2\hat { k }$

$\overrightarrow { AB } =-3\hat { i } -\hat { j } +2\hat { k } +5\hat { i } +4\hat { j } -\hat { k }$

$\overrightarrow { AB } =2\hat { i } +3\hat { j } +\hat { k }$

Let $\varepsilon_1$ and $\varepsilon_2$ be the angles made by $\vec{A}$ and $-\overrightarrow{A}$ with the positive X-axis. Show that $\tan \varepsilon_1=\tan\varepsilon_2$ . Thus, given $\tan\varepsilon$ does not uniquely determine the direction of $\overrightarrow{A}$ .

Now,

$\varepsilon_{2}=\varepsilon_{1}+180^{\circ}$

$\Rightarrow \tan \varepsilon_{2}=\tan (\varepsilon_{1}+180^{\circ})$

$\Rightarrow \tan \varepsilon_{2}=\tan \varepsilon_{1}$

$\therefore \tan \varepsilon_{1}=\tan \varepsilon_{2}$

$\rightarrow$ So

$\tan$ (of angle made by the vector with

$x-$ axis ) does not uniquely determine the direction of the vector as the negative of a given vector has reverse direction with respect to given vector.

1051126_1019757_ans_b86fc77e32fe4279a03a82a00ff78539.PNG

Calculate the scalar product of the following vectors
$\displaystyle a \, = \, 3i \, + \, 2j \, + \, 2k \, and \, b \, = \, 18i \, - \, 22j \, - \, 5k$

Let

$\bar{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$ &

$\bar{b}=b_{1}\hat{i}+b_{2}\hat{j}+b_{3}\hat{k}$

Where,

$a_{1}=3,a_{2}=2,a_{3}=2$

$b_{1}=18,b_{2}=-22,b_{3}=-5$

$\bar{a}.\bar{b}=(a_{1}b_{1})+(a_{2}b_{2})+(a_{3}b_{3})$

$=3\times 18+(2\times -22)+(2\times -5)$

$=54-44-10=0$

In a $\Delta$ ABC, $\vec{AB}=6\vec{i}+3\vec{j}+3\vec{k}$ ; $\vec{AC}=3\vec{i}-3\vec{j}+6\vec{k}$ . D and D' are points of tri-sections of sides BC. Find $\vec{AD}$ and $\vec{AD'}$ .

We have

$\vec{BC}=\vec{AC}-\vec{AB}$

or,

$\vec{BC}=(3\hat{i}-3\hat{j}+6\hat{k})$

$-(6\hat{i}+3\hat{j}+3\hat{k})=-3\hat{i}-6\hat{j}+3\hat{k}$ .

Now, according to the problem,

$\vec{BD}=\vec{D'C}=\dfrac{1}{3}\vec{BC}=(-\hat{i}-2\hat{j}+\hat{k})$ .

$\therefore \vec{AD}=\vec{BD}-\vec{BA}=\vec{BD}+\vec{AB}=5\hat{i}+\hat{j}+4\hat{k}$ .

And

$\vec{AD'}=\vec{AC}-\vec{D'C}=4\hat{i}-\hat{j}+5\hat{k}$ .

Given $\vec{AB} = 3i + 2j - k \, , \vec{OB} = i + 2j + k$ . find $\vec{OA}$

From the triangle law of addition,

$\vec{OA}+\vec{AB}=\vec{OB}$

$\Rightarrow \vec{OA}=\vec{OB}-\vec{AB}$

$\vec{OA}=\hat{i}+2\hat{j}+\hat{k}-\left ( 3\hat{i}+2\hat{j}-\hat{k} \right )$

$=-2\hat{i}+2\hat{k}$

If vector $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ then, find unit vector.

$\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$

$\left| {\vec a} \right| = \sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2}$
Unit vector

$= \hat a$

$= \dfrac{{{a_1}\hat i + {a_2}\hat j + {a_3}\hat k}}{{\sqrt {{a_1}^2 + {a_2}^2 + {a_3}^2} }}$

The sides of a parallelogram are $2\hat i - 4\hat j +5\hat k$ and $\hat i - 2\hat j - 3\hat k$ . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.

Given,

$2\hat i - 4\hat j + 5\hat k$

$\hat i - 2\hat j - 3\hat k$

Consider

$\vec A = 2\hat i - 4\hat j + 5\hat k$

and

$\vec B = \hat i - 2\hat j - 3\hat k$

Then,

$\vec P = \vec A + \vec B$

$= \left( {2\hat i - 4\hat j + 5\hat k} \right) + \left( {\hat i - 2\hat j - 3\hat k} \right)$

$= 3\hat i - 6\hat j + 2\hat k$

Therefore,

$\vec P = 3\hat i - 6\hat j + 2\hat k$

So, unit vector along the diagonal

$= \dfrac{{\vec P}}{{\left| {\vec P} \right|}}$

$\begin{array}{l} =\dfrac { { 3\hat { i } -6\hat { j } +2\hat { k } } }{ { \sqrt { { 3^{ 2 } }+{ { \left( { -6 } \right) }^{ 2 } }+{ 2^{ 2 } } } } } \\ =\dfrac { { 3\hat { i } -6\hat { j } +2\hat { k } } }{ 7 } \end{array}$

Now

$\begin{array}{l} \vec { A } \times \vec { B } =\left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 2 & { -4 } & 5 \\ 1 & { -2 } & { -3 } \end{array} } \right| \\ =\hat { i } \left( { 12+10 } \right) -\hat { j } \left( { -6-5 } \right) +\hat { k } \left( { -4+4 } \right) \\ =22\hat { i } +11\hat { j } \end{array}$

Area of parallelogram

$= \left| {\vec A \times \vec B} \right|$

$\begin{array}{l} =\sqrt { { { 22 }^{ 2 } }+{ { 11 }^{ 2 } } } \\ =\sqrt { 605 } \\ =11\sqrt { 5 } \quad squareunits \end{array}$

If $r=t^2i-tj+(2t+1)k$ , find $|dr/dt|$ .

$r = {t^2}\hat i - t\hat j + (2t + 1)\hat k$

$\dfrac{dr}{dt}=(\dfrac{d}{dt}\,t^2)\hat i-(\dfrac{d}{dt}t)\hat j+(\dfrac{d}{dt}(2t+1))\hat k$

$=2t\hat i-\hat j+2\hat k$

$|\dfrac{dr}{dt}|=\sqrt{(2t)^2+(-1)^2+(2)^2}$

$=\sqrt {4t^2+5}$

If a vector $\vec{AB}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{OB}=3\hat{i}-4\hat{j}+4\hat{k}$ find the position vector $\vec{OA}$ .

$\begin{array}{l} \overrightarrow { AB } =\overrightarrow { OB } -\overrightarrow { OA } \\ \overrightarrow { OA } =\overrightarrow { OB } -\overrightarrow { AB } \\ \overrightarrow { OA } =\left( { 3\hat { i } -4\hat { j } +4\hat { k } } \right) -\left( { 2\hat { i } -\hat { j } +\hat { k } } \right) \\ \overrightarrow { OA } =\hat { i } -3\hat { j } +3\hat { k } \end{array}$

Show that:
$[\bar{a} \times \bar{b} \, \bar{b} \times \bar{c} \,\bar{c} \times \bar{a}] = [abc]^2$

1330694_1095264_ans_fa73b3a9aebf4972989fe10e88159eb8.pdf

Points X & Y are taken on the sides QR & RS respectively of a parallelogram PQRS, so that $\vec{QX}=4\vec{XR}$ & $\vec{RY}=4\vec{YS}$ . The line XY cuts the line PR at Z. Prove that $\vec{PZ}=\left(\dfrac{21}{25}\right)\vec{PR}$ .

Let P the origin

${\overrightarrow {PQ} = \bar a,\,\,\,\,\,\,\overrightarrow {PS} = \bar b}$

${\overrightarrow {PR} = \overrightarrow {PQ} + \overrightarrow {PS} = \bar a + \bar b}$

${\overrightarrow {PX} = \frac{{4(\bar a + \bar b) + 1(\bar a)}}{{4 + 1}} = \frac{{5\bar a + 4\bar b}}{5}}$

${\overrightarrow {PY} = \frac{{1(\bar a + \bar b) + 41\bar b}}{{1 + 4}} = \frac{{\bar a + 5\bar b}}{5}}$

${vector\,\,\,equ.\,\,of\,\,\overrightarrow {PR\,} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} \bar r = t(\bar a + \bar b) - - - - - - (1}$

${vector{\mkern 1mu} \,\,equ.\,\,of{\mkern 1mu} \,\,\overrightarrow {XY} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} \bar r{\mkern 1mu} = (1 - 3)\left( {\frac{{5\bar a + 4\bar b}}{5}} \right) + 3\left( {\frac{{\bar a + 5\bar b}}{5}} \right) - - - - \left( 2 \right)}$

$Let{\mkern 1mu} \,\,Z{\mkern 1mu} \,the{\mkern 1mu} \,\,POI\,\,{\mkern 1mu} of\,{\mkern 1mu} (1)\& {\mkern 1mu} (2)$

$t\,(\bar a + \bar b) = \left( {1 - s + \frac{s}{5}} \right)\bar a + \left( {(1 - s)\frac{4}{5}} \right)\overline b$

${equation{\mkern 1mu} \,\,\bar a,\bar b{\mkern 1mu} \,coeff}$

${t = 1 - \frac{{4s}}{5}}$

${t = \frac{4}{5} + \frac{s}{5}}$

${\therefore \,\,\,1 - \frac{{4s}}{5} = \frac{4}{5} + \frac{s}{5}}$

${\frac{1}{5} = s}$

${t = \frac{4}{5} + \frac{1}{{25}}}$

${t = \frac{{21}}{{25}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} in - - - - (3)}$

${\overline {PZ} = \frac{{21}}{{25}}{\mkern 1mu} \overrightarrow {PR} }$

Component of $(1,1)$ along $(3,4)$ is ?

Let

$a=(1,1)$

$b=(3,4)$

Projection of

$a$ along

$b =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|^2}.\overrightarrow{b}$

$=\dfrac{3.1+4.1}{3^2+4^2}(3,4)$

$=\dfrac{7}{25}(3,4)$

$\therefore$ Projection of

$(1,1)$ along

$(3,4)$ is

$\dfrac{7}{25}(3,4)$

Find the component of $\overline i + \overline k \,on\,\overline i - \overline j$ and its magnitude.

Let

$\overrightarrow{a}=\hat{c}+\hat{k}$

$\overrightarrow{b}=\hat{c}+\hat{j}$

Projection of

$\overrightarrow{a}$ on

$\overrightarrow{b}=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{b}|^2}.\overrightarrow{b}$

Now

$\overrightarrow{a}.\overrightarrow{b}=(\hat{i}+\hat{k}).(\hat{i}-\hat{j})$

$=1$

$|{\overrightarrow{b}}|^2=1^2+(-1)^2=2$

$\therefore$ Projection of

$\overrightarrow{a}$ on

$\overrightarrow{b}=\dfrac{1}{2}\overrightarrow{b}=\dfrac{1}{2}(\hat{i}-\hat{j})$

Magnitude

$=\sqrt{\left(\dfrac{1}{2}\right)^2+\left(\dfrac{-1}{2}\right)^2}$

$=\sqrt{\dfrac{1}{4}+\dfrac{1}{4}}$

$=\sqrt{\dfrac{2}{4}}$

$=\dfrac{1}{\sqrt{2}}$

If $\overline a = \left( {1, - 1} \right)$ and $\overline b = \left( {1,0} \right)$ then comp $\overline a \overline b = ..........$

Given,

$\overline a = (1, - 1)\,\,\overline b = (1,0)$

the component of

$\overline a \,\,\,on\,\,\,\overline b = \,\overline a \,.\,\overline b$

then,

$= \,\frac{{\overline a \,.\,\overline b }}{{\left| b \right|}}$

$= \,\frac{{(\widehat i - \widehat j)\,.(\widehat i)}}{{111}}\,\, = \,\frac{{1 - 0}}{1} = 1$

$so,comp\,\,\overline a \,.\overline b = 1$

ABCD is parallelogram. If $\vec { AB } =\vec { a } ,\vec { BC } =\vec { b }$ then show that $\vec { AB } =\vec { a } ,+\vec { b }$ and $\vec { BD } =\vec { b } ,-\vec { a }$ ?

Given ABCD is a parallalogaram and

$\begin{array}{l} \overline { AB } =\overline { a } \, \, ,\, \, \overline { BC } =\overline { b } \\ Here\, \, ,\, \overline { AC } \, \, is\, \, the\, result\, \, of\, \overline { AE } \, and\, \overline { EC } \\ so \\ \, \, \, \, \, \, \, \left| { \overline { AC } } \right| =\, \left| { \overline { AE } } \right| +\left| { \overline { EC } } \right| \\ \, \, \, \, \, \, \, =\sqrt { { { \left( { a+b\cos \theta } \right) }^{ 2 } }+{ { \left( { b\sin \theta } \right) }^{ 2 } } } \\ \, \, \, \, \, \, \, \, \sqrt { { a^{ 2 } }+{ b^{ 2 } }{ { \cos }^{ 2 } }\theta +2ab\cos \theta +{ b^{ 2 } }{ { \sin }^{ 2 } }\theta } \\ \, \, \, \, \, \, \, \, \sqrt { { a^{ 2 } }+{ b^{ 2 } }+2ab\cos \theta } \\ \, \, \, \, \, \, \, \, \, \, \left| { a+b } \right| \\ \, \, \, \, \, \, \, \, \, \, \overline { AB } =\overline { a } +\overline { b } \\ \, \, \, \, \, \, \, \, \, \, similerily\, \, \, \, \, \, \, \, \overline { BD } =\overline { b } -\overline { a } \\ \, \, \, \, \, \, \, \, \, \, \end{array}$

1129915_1109675_ans_77d4bbe92fa649ae9b515f0a3368e7e9.JPG

Find the value or values of $m$ for which $m\left( \hat { i } +\hat { j } +\hat { k } \right)$ is a unit vector.

Magnitude of vector is

$\sqrt{m^2+m^2+m^2}=\sqrt{3}m$

For the vector to be unit vector, magnitude of vector=1

$\sqrt{3}m=1$

therefore,

$m=\dfrac{1}{\sqrt{3}}$

If $\vec{a}$ and $\vec{b}$ are position vectors of point s A and B respectively, then the position vector of points of trisection of AB

Let P and Q be the point of trisection of AB, i.e. AP=PQ=QB

Since, point P divides AB in the ratio

$1:2$

Therefore the position vector of P=

$\dfrac{{1 \cdot \overrightarrow b + 2\overrightarrow a }}{{1 + 2}} = \dfrac{{2\overrightarrow a + \overrightarrow b }}{3}$

Since, point Qdivides AB in the ratio

$2:1$

Therefore the position vector of Q=

$\dfrac{{2 \cdot \overrightarrow b + 1 \cdot \overrightarrow a }}{{1 + 2}} = \dfrac{{\overrightarrow a + 2\overrightarrow b }}{3}$

Find a unit vector which is along the direction of vector $\hat{i}+\hat{j}+2\hat{k}$ .

Let

$\vec{r}=i+j+2\hat{k}$

$|\vec{r}|=\sqrt{1+1+4}=\sqrt{6}$

$\hat{r}=\dfrac{\vec{r}}{|\vec{r}|}$

$\Rightarrow \dfrac{1}{\sqrt{6}}(i+j+2k)$ .

1241164_1159243_ans_b867227e0d0449c2a3687328863343b9.jpg

If $\vec{a}||\vec{b}\times \vec{c}$ then $\left(\vec{a}\times \vec{b}\right)\cdot (\vec{a}\times \vec{c})$ is equal to?

$\begin{array}{l} \overline { a}\ || \overline { b } \times \overline { c } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, To\, find\, (\overline { a } \times \overline { b } ).(\overline { a } \times \overline { c } ) \\ \overline { b } \times \overline { c } \, \, \, is\, \, { \bot ^{ \gamma } }to\, \, \overline { b } ,\overline { c } \\ \overline { u } \, \, \, \, \, \, \, \, \, \, \overline { a } \, \, is\, \, { \bot ^{ \gamma } }\, \, \, to\, \, \overline { b } ,\overline { c } \\ \overline { a } ,\overline { b } =0,\, \, \, \, \, \overline { a } ,\overline { c } =0 \\ (\overline { a } \times \overline { b } ).(\overline { a } \times \overline { c } )=\, \overline { a } .\left[ { \overline { b } \times (\overline { a } \times \overline { c } ) } \right] \\ \overline { a } .\, \left[ { (\overline { b } .\overline { c } )\, \overline { a } \, -(\overline { b } .\overline { a } )\overline { c } } \right] \\ (\overline { b } .\overline { c } ){ \overline { a } ^{ 2 } }-(\overline { b } .\overline { a } )(\overline { a } .\overline { c } ) \\ \overline { a } .\overline { b } =0, \\ \overline { a } .\overline { c } =0 \\ (\overline { b } .\overline { c } ){ \overline { a } ^{ 2 } }-0=(\overline { b } .\overline { c } ){ \overline { a } ^{ 2 } } \end{array}$

Find the projection of $(a,b,c)$ on X-axis and its magnitude.

Let

$\vec { a } = < a,b,c >$ and

$\vec { b } = < x,0,0>$

then vector projection of

$\vec { a }$ onto

$\vec { b }$ using the following formula:

$\operatorname { proj } _ { \vec { b } } \vec { a } = \left( \dfrac { \vec { a } \cdot \vec { b } } { | \vec { b } | } \right) \dfrac { \vec { b } } { | \vec { b } | }$

$\operatorname { proj } _ { \vec { b } } \vec { a } = \dfrac { ax} { x } \cdot \dfrac { < x,0,0> } { \sqrt { x^2 } }$

$\Rightarrow < { a } , {0 } , { 0 } >$

Magnitude is

$a$

Let $a=i+2j+3k$ and $b=3i+j$ . Find the unit vector in the direction of $a+b$ .

Let

$\begin{array}{l} \overrightarrow { a } =\widehat { i } +2\widehat { j } +3\widehat { k } \\ \overrightarrow { b } =3\widehat { i } +\widehat { j } \\ \overrightarrow { a } +\overrightarrow { b } =\widehat { i } +2\widehat { j } +3\widehat { k } +3\widehat { i } +\widehat { j } \\ \overrightarrow { a } +\overrightarrow { b } =4\widehat { i } +3\widehat { j } +3\widehat { k } \\ So\, unit\, vector\, in\, the\, direction\, of \\ \overrightarrow { a } +\overrightarrow { b } \\ \left| { \overrightarrow { a } +\overrightarrow { b } } \right| =\sqrt { { 4^{ 2 } }+{ 3^{ 2 } }+{ 3^{ 2 } } } \\ =\sqrt { 16+9+9 } \\ =\sqrt { 34 } \\ so\, unit\, vector\, in\, the\, direction\, of\, \overrightarrow { a } +\overrightarrow { b } \, is \\ =\dfrac { 4 }{ { \sqrt { 34 } } } \widehat { i } +\dfrac { 3 }{ { \sqrt { 34 } } } \widehat { j } +\dfrac { 3 }{ { \sqrt { 34 } } } \widehat { k } \end{array}$

If $\vec {a}=\vec {b}+\vec {c}$ , then is true that $|\vec {a}|=|\vec {b}|+|\vec {c}|$ ? Justify your answer.

Given

$\vec a=\vec b+\vec c$

Consider,

$\vec b=\hat i+2\hat j+3\hat k$

$\vec c= 3\hat i-\hat j-2\hat k$

$\vec a=(\vec b +\vec c)$

$=4\hat i+\hat j+\hat k$

$\vec a=4\hat i+\hat j+\hat k$

$|\vec a|=\sqrt {4^2+1^2+1^2}=\sqrt {18}$

Similarly,

$|\vec b|=\sqrt {14}$

And,

$|\vec c|=\sqrt {14}$

Now,

$|\vec b+\vec c|=\sqrt {14}+\sqrt {14}=2\sqrt {14}$

whish is

$\ne \sqrt {18}$

So,

$|\vec a| \ne |\vec b|+|\vec c|$

Hence, False.

If the projections of the line segment $\underset{PQ}{\rightarrow}$ on the axes are 3,4,12 then the length of $\underset{PQ}{\rightarrow}$ is

Length of

$\vec{PQ}=\sqrt{3^2+4^2+12^2}$

$=\sqrt{5^2+12^2}$

$=\sqrt{13^2}$

$=13$

$($ since, projection of

$\vec{PQ}$ on x-axis is

$|\vec{PQ}|=\cos\alpha =3$

projection of

$\vec{PQ}$ on y-axis is

$|\vec{PQ}|\cos \beta =4$

projection of

$\vec{PQ}$ on z-axis is

$|\vec{PQ}|\cos\gamma =12$

$\cos^2\alpha +\cos^2\beta +\cos^2\gamma =1)$ .

1195554_1195998_ans_8f5d89250d9648139e2e6107dcf156c6.jpg

If $\hat { \mathbf { a } }$ is a unit vector and $( x - \hat { a } ) \cdot ( \vec { x } + \hat { a } ) = 1.5$ then find the value of $| \vec { x } |$ .

$(\overrightarrow x-\widehat{a}).(\overrightarrow x +\widehat{a})=1.5$

$|\overrightarrow x|^2-|\widehat{a}|^2=1.5$

$|\overrightarrow x|^2=1.5+1=2.5$

$|\overrightarrow x|=\sqrt{2.5}=1.58$

Find $a\times (b\times c)$ and $(a\times b)\times c$ where $a=(1, -1, -6), b=(1, -3, 4), c=(2, -5, 3)$ .

$\begin{array}{l} a=\left( { 1,-1,-6 } \right) \, \, \, \, \, \, b=\left( { 1,-3,4 } \right) \, \, \, \, c=\left( { 2,-5,3 } \right) \\ \overrightarrow { a } \times \left( { \overrightarrow { b } \times \overrightarrow { c } } \right) =\left( { \overrightarrow { a } .\overrightarrow { c } } \right) \overrightarrow { b } -\left( { \overrightarrow { a } .\overrightarrow { b } } \right) \overrightarrow { c } \\ =\left( { 2+5-\left( 8 \right) } \right) \overrightarrow { b } -\left( { 1+3-24 } \right) \overrightarrow { c } \\ =-11\overrightarrow { b } +20\overrightarrow { c } \\ =\left( { -11,33,-44 } \right) +\left( { 40,-100,60 } \right) \\ =\left( { 29,-67,16 } \right) \\ \left( { \overrightarrow { a } \times \overrightarrow { b } } \right) \times \overrightarrow { c } =-\overrightarrow { c } \times \left( { \overrightarrow { a } \times b } \right) \\ =\left[ { \left( { \overrightarrow { c } .\overrightarrow { b } } \right) \overrightarrow { a } -\left( { \overrightarrow { c } .\overrightarrow { a } } \right) \overrightarrow { b } } \right] \\ =\left( { \overrightarrow { c } .\overrightarrow { a } } \right) \overrightarrow { b } -\left( { \overrightarrow { c } .\overrightarrow { b } } \right) \overrightarrow { a } \\ =-11\overrightarrow { b } -\left( { 2+15+12 } \right) \overrightarrow { a } \\ =-11\left( { 1,-3,4 } \right) +-29\left( { 1,-1,-6 } \right) \\ \left( { \overrightarrow { a } \times \overrightarrow { b } } \right) \times \overrightarrow { c } =\left( { -40,62,130 } \right) \end{array}$

Find a unit vector in the direction of $\hat { i } + \hat { j }$ .

For the unit vector

$\overrightarrow a = \widehat i + \widehat j$

$\therefore \widehat a = \dfrac{{\widehat i + \widehat j}}{{\sqrt 2 }}$

Hence, which is the required answer.

The vector product of two vector is $\sqrt{3}$ times of their of two scalar produced. The angle between the two vector is-

Let

$\vec{a},\vec{b}$ be the two vectors.

Let

$\theta$ be the angle between the two vectors.

Then according to the problem,

$|\vec{a}||\vec{b}|\sin \theta=\sqrt{3}|\vec{a}||\vec{b}|\cos \theta$

or,

$\tan \theta=\sqrt{3}$

or,

$\theta=60^o$ .

Write the unit vector in the direction of $\overset { \rightarrow }{ A } =5\overset { \rightarrow }{ i } + \overset { \rightarrow }{ j } - 2\overset { \rightarrow }{ k }$ .

$\overset {\rightarrow}{A}$

$=5\overset {\rightarrow}{i}$

$+\overset {\rightarrow}{j}$

$-2\overset {\rightarrow}{k}$

$|A|=\sqrt{5^2+1^2+2^2}=\sqrt {30}$

Unit vector

$=(5\overset {\rightarrow}{i}$

$+\overset {\rightarrow}{j}$

$-2\overset {\rightarrow}{k})/\sqrt{30}$

Find the value of $\lambda$ , so that the vectors $\overrightarrow a =3\widehat{i}+2\widehat{j}+9\widehat{k}$ and $\overrightarrow b =\widehat{i}+\lambda \widehat{j}+3\widehat{k}$ are perpendicular to each other.

$\overrightarrow a =3\widehat{i}+2\widehat{j}+9\widehat{k}$

$\overrightarrow b = \widehat{i}+\lambda \widehat{j}+3\widehat{k}$

Since, vectors

$\overrightarrow a$ and

$\overrightarrow b$ are perpendicular to each other.

Therefore,

$\overrightarrow a .\overrightarrow b =0$

$(3\widehat{i}+2\widehat{j}+9\widehat{k}).(\widehat{i}+\lambda\widehat{j}+3\widehat{k})=0$

$3+2\lambda+27=0$

$2\lambda=-30$

$\lambda=-15$

Hence, required value of

$\lambda$ is

$-15$ .

Simplify:
(i) $\vec { D E }$ + $( - \vec { B E } )$
(ii) $\vec { AC }$ + $( - \vec { B C })$
(iii) $\vec { C D }$ + $\vec { B A }$ $( - \vec { B D } )$

$i)\vec{-BE}=\vec{EB}$

$\vec{DE}+\vec{EB}=\vec{DB}$

$ii)-\vec{BC}=\vec{CB}$

$\vec{AC}+\vec{CB}=\vec{AB}$

$iii)\vec {CD}+\vec {BA}+(\vec {-BD})=\vec {CD}+\vec {DB}+\vec {BA}=\vec {CB}+\vec {BA}=\vec {CA}$

Find the sum of the vectors :
$\overrightarrow a = \widehat i - 3\widehat k\,,\,\overrightarrow b = 2\widehat j - \widehat k\,,\,\overrightarrow c = 2\widehat i - 3\widehat j\, + 2\widehat k$

$\overrightarrow a = \widehat i - 3\widehat k\,,\,\overrightarrow b = 2\widehat j - \widehat k\,,\,\overrightarrow c = 2\widehat i - 3\widehat j\, + 2\widehat k$

$\overrightarrow a + \overrightarrow b + \overrightarrow c = 3\widehat i -\widehat j -2\widehat k$

The dot product of a vector with vectors $\hat{i}+\hat{j}-3\hat{k}$ , $\hat{ i}+ 3\hat{ j}-2\hat{k}$ , $2\hat{i} + \hat{j} + 4k$ are $0,5$ and $8$ respectively. Find the vector .

Let vertor be

$a\hat{i}+b\hat{j}+c\hat{k}$

Dot product with

$\hat{i}+\hat{j}-3\hat{k}=a+b-3c=0$ …………

$(1)$

Dot product with

$\hat{i}+3\hat{j}-2\hat{k}=a+3b-2c=5$ ……….

$(2)$

Dot product with

$2\hat{i}+\hat{j}+4\hat{k}=2a+b+4c=8$ ……………

$(3)$

Equation

$(2)$ -Equation

$(1)$

$\Rightarrow a+3b-2c-a-b+3c=5-0$

$\Rightarrow 2b+c=5$ ………..

$(4)$

Equation

$(3)$ -

$2\times$ equation

$(2)$

$\Rightarrow 2a+b+4c-2a-6b+4c=8-10$

$-5b+8c=-2$ …………..

$(5)$

Equation

$(5)$ -

$4\times$ equation

$(4)$

$\Rightarrow -5b+8c-16b-8c=-2-40$

$\Rightarrow -21b=-42$

$\therefore b=\dfrac{-42}{-21}=2$

Put

$b=2$ in

$(4)$ we get

$c=5-2b$

$c=5-2\times 2=5-4$

$\therefore c=1$

Put

$b=2$ ,

$c=1$ in

$(1)$

$a+2-3\times 1=0$

$a+2-3=0$

$\therefore a-1=0$

(or)

$a=1$

Hence the verctor is

$\hat{i}+2\hat{j}+\hat{k}$ .

1257502_1326679_ans_65e63731223841178304fcbfe136c472.PNG

Forces measuring $5,3$ and $1$ unit act in the directions $(6,2,3),(3,-2,6)$ and $(5,-1,1)$ . Find the resultant force.

$\begin{array}{l} \overrightarrow { { F_{ 1 } } } =5\frac { { \left( { 6,\, 2,\, 3 } \right) } }{ { \sqrt { 49 } } } =\frac { 5 }{ 7 } \left( { 6,\, 2,\, 3 } \right) \\ \overrightarrow { { F_{ 2 } } } =\frac { 3 }{ 7 } \left( { 3,\, -2,\, 6 } \right) \, \, \, \, \overrightarrow { { F_{ 3 } } } =\frac { { \left( { 5,\, -1,\, 1 } \right) } }{ { \sqrt { 27 } } } \\ \overrightarrow { { F_{ R } } } =\overrightarrow { { F_{ 1 } } } +\overrightarrow { { F_{ 2 } } } +\overrightarrow { { F_{ 3 } } } \\ =\left( { \left( { \frac { { 39 } }{ 7 } +\frac { 5 }{ { \sqrt { 27 } } } } \right) ,\, \left( { \frac { 4 }{ 7 } -\frac { 1 }{ { \sqrt { 27 } } } } \right) ,\, \left( { \frac { { 33 } }{ 7 } +\frac { 1 }{ { \sqrt { 27 } } } } \right) } \right) \end{array}$

$\overrightarrow { a } =3\hat { i } +\hat { j } -2\hat { k } ;\overrightarrow { b } =\hat { i } +\lambda \hat { j } -3\hat { k }$
$\overrightarrow { a } \bot \overrightarrow { b }$ , find value of $\lambda$ .

$\vec{a}\perp \vec{b}, \Rightarrow \vec{a}-\vec{b} = 0 \Rightarrow (3\hat{i}+\hat{j}-\hat{2k})(\hat{i}+\lambda \hat{j}-3\hat{k}) = 0$

$\Rightarrow 3+\lambda +6 = 0 \Rightarrow \lambda = -9$

1124066_1204033_ans_1010c03f349d4a8ca5f3476cc22f3506.jpg

Find $\vec{a}.\vec{b}$ , when $\vec{a}=\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{k}$ .

$\vec a=a_1\hat i+a_2\hat j+a_3 \hat k$ and

$\vec b=b_1\hat i+b_2\hat j+b_3\hat k$

then

$\vec a.\vec b=a_1b_1+a_2b_2+a_3b_3$

Here

$\vec a =\hat j+2\hat k$

$\vec a=2\hat i+\hat k$

$\vec a.\vec b=0.2+1.0+2.1$

$\vec a.\vec b=2.$

Find the projection of the vector $\widehat{i}-\widehat{j}$ on the vector $\widehat{i}+\widehat{j}$ .

$\textbf{Find the required projection of}\ \boldsymbol{\hat{i}-\hat{j}} \ \textbf{on}\ \boldsymbol{\hat{i}+\hat{j}}$

$\text{We have given two vectors}$

$\widehat{i}-\widehat{j}$ ,

$\widehat{i}+\widehat{j}$ ,

$\text{Let}$

$\overrightarrow a=\widehat{i}-\widehat{j}$

$\text{and}$

$\overrightarrow b = \widehat{i}+\widehat{j}$

$\text{Now, projection of vector}$

$\overrightarrow a$

$\text{on}$

$\overrightarrow b$

$\text{ is}$

$=\dfrac{1}{|\overrightarrow b|}(\overrightarrow a . \overrightarrow b) = \dfrac{1}{\sqrt{1+1}}(1-1)=0$

$\textbf{Hence the required projection of vectors is 0}$

Find the dot product of the following vectors:
$\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$ .

Given the vectors

$\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and

$\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$ .

Now,

$\vec{a}.\vec{b}$

$=(2\hat{i}+\hat{j}+3\hat{k}).$

$(3\hat{i}+5\hat{j}-2\hat{k})$

$\vec{a}.\vec{b}$

$=2.3+1.5-3.2$

$\vec{a}.\vec{b}$

$=6+5-6$

$\vec{a}.\vec{b}$

$=5$ .

Compute the magnitude of the following vector :
$\overrightarrow a = 2\widehat{i}-7\widehat{j}-3\widehat{k}$

$\overrightarrow a = 2\widehat{i}-7\widehat{j}-3\widehat{k}$

$|\overrightarrow a | = \sqrt{2^2 +(-7)^2+(-3)^2}$

$|\overrightarrow a | = \sqrt{4+49+9}=\sqrt{62}$

Find the value of x for which $x(\widehat{i}+\widehat{j}+\widehat{k})$ is a unit vector.

$x(\widehat{i}+\widehat{j}+\widehat{k})$ is a unit vector if

$|x(\widehat{i}+\widehat{j}+\widehat{k})|=1$

$\sqrt{x^2+x^2+x^2}=1$

$\sqrt{3x^2}=1$

$\sqrt{3}x=1$

$x=\pm\dfrac{1}{\sqrt{3}}$

Find the distance between the points A and B whose position vectors $\widehat{i}-\widehat{j}$ and $2\widehat{i}+\widehat{j}+2\widehat{k}$ .

1341286_1362847_ans_4514f1fc18224ca592b503f90b8af848.jpg

If $\vec a =a_1\hat i +a_2\hat j+a_3\hat k, \, \, \vec b=b_1\hat i+b_2\hat j+b_3\hat k$ and $\vec c=c_1\hat i+c_2\hat j+c_3\hat k$ , than verify that $\vec a \times (\vec b +\vec c)=\vec a \times \vec b +\vec a\times \vec c$

$\vec a \times (\vec b+\vec c)=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ b_{ 1 }+c_{ 1 } & b_{ 2 }+c_{ 2 } & b_{ 3 }+c_{ 3 } \end{vmatrix}$

$=\left\{a_2 (b_3 +c_3)-a_3 (b_2 +c_2)\right\} \hat i-\left\{a_1 (b_3 +c_3) - a_3 (b_1 +c_1)\right\} \hat j +\left\{a_1 (b_2 +c_2) -a_2 (b_1+c_1)\right\}\hat k$

$=\left\{(a_2 b_3-a_3 b_2)\hat i-(a_1 b_3 -a_3 b_1)\hat j +(a_1 b_2 -a_2 b_1) \hat k \right\}+\left\{(a_2 c_3 -a_3 c_2)\hat i - (a_1 c_3 -a_3 c_1) \hat j + (a_1 c_2 -a_2 c_1) \hat k \right\}$

$=\vec a \times \vec b +\vec a \times \vec c$

$ABCDE$ is a pentagon, then prove that $\overrightarrow{AB}+\overrightarrow{AE}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{ED}+\overrightarrow{AC} = 3\,\overrightarrow{AC}$ .

1397693_1662216_ans_82b693c585214aafb76a421bb4250c02.PNG

If $\vec a , \vec b , \vec c$ are three unit vector such that $\vec a\times \vec b=\vec c, \vec b\times \vec c=\vec a, \vec c\times \vec a=\vec b$ . Show that $\vec a , \vec b , \vec c$ from an orthogonal right handed tried of unit vectors.

$|\vec{a}\times \vec{b}|=|\vec{c} |\rightarrow |1||1|sin\theta=|1|\rightarrow \theta=90^o$

$\therefore$ angle between

$\vec a$ and

$\vec b$ is

$90^{\circ}$

$|\vec{b}\times\vec{c}|=|\vec{a}|\rightarrow |1||1|sin\theta =|1| \rightarrow \theta=90^o$

$\therefore$ angle between

$\vec b$ and

$\vec c$ is

$90^{\circ}$

$|\vec{c}\times\vec{a}|=|\vec{b}|\rightarrow |1||1|sin\theta=|1|\rightarrow \theta=90^o$

$\therefore$ angle between

$\vec c$ and

$\vec a$ is

$90^{\circ}$

Since, Angle between each pair is

$90^o$ and also magnitude of each is 1, they form an orthogonal right handed tried of unit vectors .

Answer the following as true or false:
Zero vector is unique.

Yes , zero vector is unique.

Any vector multiplied by the zero vector is the zero vector.

Find the projection of the vector $7\widehat{i}+\widehat{j}-4\widehat{k}$ and $2\widehat{i}+6\widehat{j}+3\widehat{k}$ .

Projection of

$\overrightarrow a$ on

$\overrightarrow b=\dfrac{\overrightarrow a .\overrightarrow b}{|\overrightarrow b|}$

Here,

$\overrightarrow a =7\widehat{i}+\widehat{j}-4\widehat{k}$ and

$\overrightarrow b=2\widehat{i}+6\widehat{j}+3\widehat{k}$

Therefore,

$\overrightarrow a . \overrightarrow b =(7\widehat{i}+\widehat{j}-4\widehat{k}).(2\widehat{i}+6\widehat{j}+3\widehat{k})=14+6-12=8$

$|\overrightarrow b|=\sqrt{2^2+6^2+3^2}=7$

Hence, projection of

$\overrightarrow a$ on

$\overrightarrow b=\dfrac{8}{7}$

Enter $1$ if true else $0$ .
$3\hat{i}+2\hat{j}+\hat{k}$ is the position vector of the mid-point of the vector joining the points $P(2,3,4)$ and $Q(4,1,-2).$

The position vector of the mid-point of the line segment joining the points

$P(2,3,4)$ and

$Q(4,1,-2)$ is

$\dfrac{(2\hat{i}+3\hat{j}+4\hat{k})+(4\hat{i}+\hat{j}-2\hat{k})}{2} = 3\hat{i}+2\hat{j}+\hat{k}.$

If $a\hat{i}+b\hat{j}+c\hat{k}$ is the position vector of the mid - point of the vector joining the points $P(2\hat{i}-3\hat{j}+4\hat{k})$ and $Q(4\hat{i}+\hat{j}-2\hat{k})$ then $a+b+c$ is equal to ___.

Given the two points are

$P(2\hat{i}-3\hat{j}+4\hat{k})$ and

$Q(4\hat{i}+\hat{j}-2\hat{k}).$

Now the position vector of the mid-point of the vector joining

$P$ and

$Q$ is

$\dfrac{2+4}{2}\hat{i}+\dfrac{-3+1}{2}\hat{j}+\dfrac{4-2}{2}\hat{k}=3\hat{i}-\hat{j}+\hat{k}.$

Comparing with

$a\hat{i}+b\hat{j}+c\hat{k}$
We get,

$a=3, b=-1,c=1$

$a+b+c$

$=3-1+1=3$

$\hat{a} \space and \space \hat{b}$ form the consecutive sides of a regular hexagon ABCDEF.

1655611_1783062_ans_371fbc73703645d6976ecf5257116c98.png

Find the projection of the vector $\displaystyle i-2j+k$ on the vector $\displaystyle 4i-4j+7k$

Let

$\vec{a} = \displaystyle i-2j+k$ and

$\displaystyle \vec{b} = 4i-4j+7k$
Now

$\vec{a}\cdot \vec{b} = (\hat{i}-2\hat j+\hat k )(4\hat i -4\hat j+7 \hat k)$

$\Rightarrow \vec { a } .\vec { b } =4+8+7=19$

Also,

$|\vec{a}|=\sqrt{1+4+1}=\sqrt{6}, |\vec{b}| =\sqrt{16+16+49}=\sqrt{81}=9$

Therefore projection of

$\vec{a}$ on

$\vec{b}$ is

$=\displaystyle \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \frac{19}{9}$

If $\displaystyle r=xi+yj+zk$ then show that r can also be expressed as $\displaystyle r= \left ( r.i \right )i+\left ( r.j \right )j+\left ( r.k \right )k$ .

Let

$\displaystyle \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$
Take dot product with

$\hat{i}$ both side,

$\displaystyle \therefore \vec{r}.\hat{i}=(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+0.\hat{j}+0.\hat{k})$

$= x\hat{i}.\hat{i}+0+0$

$\Rightarrow \vec{r}.\hat{i}= x.1= x$

$\Rightarrow (\vec{r}.\hat{i})= x \hat{i}$ .....(1)
Similarly,

$(\vec{r}.\hat{j})\hat j=y\hat j$ , ....(2)

$(\vec{r}.\hat{k})\hat k=z\hat j$ .....(3)

$\Rightarrow (\vec{r}.\hat{i}) \hat{i}+(\vec{r} .\hat j)\hat{j} +(\vec{r}.\hat{k} )\hat{k}=x\hat{i}+y\hat{j}+z\hat{k}$

$\displaystyle \therefore \vec{r}= \left ( r.i \right )i+\left ( r.j \right )j+\left ( r.k \right )k$

If $\displaystyle \vec{a}\:and\:\vec{b}$ are the vectors $\displaystyle \overrightarrow{AB}\:and\:\overrightarrow{BC}$ determined by the adjacent sides of a regular hexagon. What are the vectors determined by the other sides taken in order?

Given:

$\overrightarrow { AB } =\overrightarrow { a } \\ \overrightarrow { BC } =\overrightarrow { b }$

EF is parallel to BC and equal to its length

hence ,

$\overrightarrow { FE } =\overrightarrow { b }$

and ED vector is also eqaul to AB

$\overrightarrow { ED } =\overrightarrow { a }$

let the center point is O

then

$\overrightarrow { BO } =\overrightarrow { BC } +\overrightarrow { CO } =\overrightarrow { b } -\overrightarrow { a }$

now

$\overrightarrow { AF }$ is parallel to

$\overrightarrow { BO }$

hence ,

$\overrightarrow { AF } =\overrightarrow { b } -\overrightarrow { a }$

And

$\overrightarrow { CD }$ vector is also equal to

$\overrightarrow { BO }$

Hence,

$\overrightarrow { CD } =\overrightarrow { b } -\overrightarrow { a }$

If $D, \ E, \ F$ are the midpoints of the sides $BC, \ CA, \ AB,$ respectively, of a triangle $ABC$ and $O$ is any point, then

Let the position vector of

$A, B, C, D, E, F$ be

$a, b, c, d, e, f$ . So

$\displaystyle d= \frac {b+c}{2} c= \frac{c+a}{2} f= \frac{ a+b}{ 2}$

$\displaystyle OD + OE + OF = d + e + f = \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}$

$=a+b+c$
A)

$AD+BF+CF = (d-a) + (e-b)+ (f-c) =(d+e+f) -(a+b+c)=0$
B)

$OA+OB+OC=a+b+c=d+e+f=OD+OE+OF$

C)

$OE + OF+ DO= e +f-d=a= OA$

D)

$\displaystyle AD+ \frac{2}{3} BE+ \frac{1}{3} CF= (d-a) + \frac{ 2}{3} (e -b) + \frac{1}{3} (f -c)$
Substituting the values of

$d, e, f$ and simplifying we obtain it equal to

$\displaystyle \frac{1}{2} (c -a) = \frac{1}{2} AC$

Write the three times of the projection of the vector $\overset{\rightarrow}{a} = 2\hat{i} - \hat{j} + \hat{k}$ on the vector $\overset{\rightarrow}{b} = \hat{i} + 2\hat{j} + 2\hat{k}$

Given vectors are

$\overset{\rightarrow}{a} = 2\hat{i} - \hat{j} + \hat{k}$ and

$\overset{\rightarrow}{b} = \hat{i} + 2\hat{j} + 2\hat{k}$
Projection of

$\overset{\rightarrow}{a}$ on the vector

$\overset{\rightarrow}{b}$ is

$\displaystyle\frac { \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } }{ \left| \overset { \rightarrow }{ b } \right| }$
let p be the projection of

$\overset{\rightarrow}{a}$ on the vector

$\overset{\rightarrow}{b}$

$\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } \quad =\quad \left( 2\hat { i } -\hat { j } +\hat { k } \right) .\left( \hat { i } +2\hat { j } +2\hat { k } \right) \quad =\quad 2$

$\Rightarrow p = \displaystyle\frac { \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } }{ \left| \overset { \rightarrow }{ b } \right| } = \displaystyle\frac{ 2 }{ 3 }$

$\overrightarrow { p } \quad =\quad p\hat { b } \quad =\quad \displaystyle\frac { 2 }{ 9 } \left(\hat{ i }+2\hat { j } +2\hat { k } \right)$

$\therefore$

$\text{three times the projection =3p=2}$

Evaluate the product $(3\vec {a}-5\vec {b})\cdot (2\vec {a}+7\vec {b})$ .

$(3\vec {a}-5\vec {b})\cdot (2\vec {a}+7\vec {b})$

$=3\vec {a}\cdot 2\vec {a}+3\vec {a}\cdot 7\vec {b}-5\vec {b}\cdot 2\vec {a}-5\vec {b}\cdot 7\vec {b}$

$=6\vec {a}\cdot \vec {a}+21\vec {a}\cdot \vec {b}-10\vec {a}\cdot \vec {b}-35\vec {b}\cdot \vec {b}$

$=6|\vec {a}|^2+11\vec {a}\cdot \vec {b}-35|\vec {b}|^2$ $

If $\vec {a}=\vec {b}+\vec {c}$ , then is it true that $|\vec {a}|=|\vec {b}|+|\vec {c}|$ ? Justify your answer.

$\Delta ABC$ , let

$\vec {CB}=\vec {a}, \vec {CA}=\vec {b}$ , and

$\vec {AB}=\vec {c}$ (as shown in the following figure).
Now, by the triangle law of vector addition, we have

$\vec {a}=\vec {b}+\vec {c}$ .
It is clearly known that

$|\vec {a}|, |\vec {b}|$ , and

$|\vec {c}|$ represent the sides of

$\Delta ABC$ .
Also, it is known that the sum of the lengths of any two sides of a triangle is greater than the third side.

$\therefore |\vec {a}| < |\vec {b}| + |\vec {c}|$
Hence, it is not true that

$|\vec {a}| = |\vec {b}| + |\vec {c}|$ .

Find the projection of the vector $\hat {i}-\hat {j}$ on the vector $\hat {i}+\hat {j}$ .

Let

$\vec {a}=\hat {i}-\hat {j}$ and

$\vec {b}=\hat {i}+\hat {j}$ .
Now, projection of vector

$\vec {a}$ and

$\vec {b}$ is given by,

$\dfrac {1}{|\vec {b}|}(\vec {a}\cdot \vec {b})=\dfrac {1}{\sqrt {1+1}}\left \{1.1+(-1)(1)\right \}=\dfrac {1}{\sqrt {2}}(1-1)=0$
Hence, the projection of vector

$\vec {a}$ on

$\vec {b}$ is

$0$ .

If $\displaystyle \overrightarrow { a } =4\hat { i } -\hat { j } +\hat k$ and $\displaystyle \overrightarrow { b } =2\hat { i } -2\hat { j } +\hat k$ , then find a unit vector parallel to the vector $\displaystyle \overrightarrow { a } +\overrightarrow { b }$ .

$\displaystyle \overrightarrow { a } =4\hat { i } -\hat { j } +\hat k$ and

$\displaystyle \overrightarrow { b } =2\hat { i } -2\hat { j } +\hat k$

$\vec a+\vec b = 6i-3j+2k$
Unit vector parallel to

$a+b$ =

$(6i-3j+2k)/\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 }+{ 2 }^{ 2 } } =\dfrac { 1 }{ 7 } (6i-3j+2k).$

The two vectors $\hat j + \hat k$ and $3\hat i - \hat j + 4\hat k$ represent the two sides $AB$ and $AC$ , respectively of a $\triangle ABC$ . Find the length of the median through $A$

$\textbf{Step-1: Apply mid point formula and simplify}$

$\text{Let $$A$$ be the origin.}$

$\text{Thus, we get the position vectors of points $$B$$ and $$C$$ as}$

$\vec{b} = \hat{j} + \hat{k}$ and

$\vec{c} = 3\hat{i} - \hat{j} + 4\hat{k}$

$\text{Now, the midpoint of}$

$BC$

$\text{would be}$

$\dfrac{\vec{b} + \vec{c}}{2}$

$i.e.$

$\dfrac{\hat{j} + \hat{k} + 3\hat{i} - \hat{j} + 4\hat{k}}{2}$

$= \dfrac{3\hat{i} + 5\hat{k}}{2}$

$\text{Hence, the length of the median through}$

$A$

$\text{thus becomes}$

$\sqrt{\dfrac{9 + 25}{4}} = \dfrac {\sqrt{34}}{2}$

Find $'\lambda'$ when the projection of $\vec {a} = \lambda \hat{i} + \hat{j} + 4\hat{k}$ on $\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$ is 4 units.

given

$\vec{a}=\lambda\hat{i}+\hat{j}+4\hat{k}$

$\vec{b}=2\hat{i}+6\hat{j}+3\hat{k}$

$Proj_{a}b=\dfrac{\vec{a}\cdot\vec{b}}{\left | \vec{b} \right |}$

$4=\dfrac{(\lambda\hat{i}+\hat{j}+4\hat{k})\cdot(2\hat{i}+6\hat{j}+3\hat{k})}{\sqrt{2^2+6^2+3^2}}$

$4=\dfrac{2\lambda+6+12}{\sqrt{49}}$

$4=\dfrac{2\lambda+6+12}{7}$

$4\times7=2\lambda+6+12$

$28=2\lambda+18$

$\lambda=5$

A and B are two points with position vectors $2\overrightarrow{a}-3\overrightarrow{b}$ and $6\overrightarrow{b}-\overrightarrow{a}$ respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1:2.

Given: $\overrightarrow{OA} =2\overrightarrow{a}-3\overrightarrow{b}$

$\overrightarrow{OB}=6\overrightarrow{b}-\overrightarrow{a}$

Since P divides AB in 1:2 internally, so, $\overrightarrow{OP}=\frac{1.\overrightarrow{OB}+2.\overrightarrow{OA}}{1+2}=\frac{1.(6\overrightarrow{b}-\overrightarrow{a})+2.(2\overrightarrow{a}-3\overrightarrow{b})}{1+2}$

Therefore, $\overrightarrow{OP}=\frac{3\overrightarrow{a}}{3}=\overrightarrow{a}$

$P$ and $Q$ are two points with position vectors $3\overrightarrow{a}-2\overrightarrow{b}$ and $\overrightarrow{a}+\overrightarrow{b}$ respectively. Write the position vector of a point R which divides the line segment $PQ$ in the ratio $2:1$ externally.

Position vector of $P =$ $3\overrightarrow{a} -2\overrightarrow{b}$

Position vector of $Q =$ $\overrightarrow{a} +\overrightarrow{b}$

Point R divides segment PQ in ratio 2:1 externally.

Position vector of $R =$ $\dfrac{(Position \space\ of \space\ P)1-(Position \space\ of \space\ Q)2}{1-2}$

Position vector of $R =$ $\dfrac{(3\overrightarrow{a} -2\overrightarrow{b})1-(\overrightarrow{a} +\overrightarrow{b})2}{-1}=\dfrac{\overrightarrow{a}-4\overrightarrow{b}}{-1}$

Therefore, position vector of $R =$ $4\overrightarrow{b}-\overrightarrow{a}.$

L and M are two points with position vectors $2\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{a}+2\overrightarrow{b}$ respectively. Write the position vector of a point N which divides the line segment LM in the ratio 2 : 1 externally.

$\overrightarrow{OL}=2\overrightarrow{a}-\overrightarrow{b}$ and

$\overrightarrow{OM}=\overrightarrow{a}+2\overrightarrow{b}$
Since

$N$ divides

$LM$ in

$2:1$ externally, so

$\overrightarrow{ON}=\frac{1.\overrightarrow{OL}-2.\overrightarrow{OM}}{1-2}=2.(2\overrightarrow{a}-\overrightarrow{b}-1.(\overrightarrow{a}+2\overrightarrow{b}))$
Therefore,

$\overrightarrow{ON}=3\overrightarrow{a}-4\overrightarrow{b}$

Find the scalar projection of the vector $\hat {i} + 3\hat {j} + 7\hat {k}$ on the vector $2\hat {i} - 3\hat {j} + 6\hat {k}.$

The scalar projection of vector

$i+3j+7k$ on the vector

$2i-3j+6k$ is

$\dfrac { (i+3j+7k).(2i-3j+6k) }{ |2i-3j+6k| }$

$=\dfrac { 2-9+42 }{ \sqrt { 49 } } =\dfrac { 35 }{ 7 } =5$

If point $C(\overline c)$ divides the segment joining the point $A(\overline a)$ and $B(\overline b)$ internally in the ratio $m : n$ , then prove that $\overline c = \dfrac{m \overline b + n \overline a}{m + n}$

Let

$O$ be the origin.
Then

$\overrightarrow{AO} = \overrightarrow a$ and

$\overrightarrow{OB} = \overrightarrow b$
Let

$P$ be a point on

$\overrightarrow{AB}$ such that

$\dfrac{\overrightarrow{AP}}{\overrightarrow{PB}} = \dfrac{m}{n}$

$\Rightarrow n (\overrightarrow{AP}) = m (\overrightarrow{PB})$

$\Rightarrow n(\overrightarrow{OP} - \overrightarrow{OA}) = m (\overrightarrow{OB} - \overrightarrow{OP})$

$\Rightarrow (m + n) OP = m OB + nOA$

$\Rightarrow (m + n) \overrightarrow{OP} = m\overrightarrow{b} + n\overrightarrow{a}$

$\Rightarrow \overrightarrow{OP} = \overrightarrow{c} = \dfrac{m\overrightarrow{b} + n\overrightarrow{a}}{m + n}$

Find the projection of vector $\overrightarrow { a } =2\hat { i } +3\hat { j } +2\hat { k }$ on a vector $\overrightarrow { b } =\hat { i } +2\hat { j } +\hat { k }$

The vector projection of

$\vec{a}$ on

$\vec{b}$ is given by

$=\dfrac{\vec{a}.\vec{b}}{|\vec{b}|}$

$=\dfrac{(2i+3j+2k).(i+2j+k)}{\sqrt{1+2^{2}+1^{2}}}$

$=\dfrac{10}{\sqrt{6}}$

If $\overline { a } ,\overline { b } ,\overline { c }$ are the position vectors of the points $A,B,C$ respectively and $3\overline { a } +4\overline { b } -7\overline { c } =\overline { 0 }$ , find the ratio in which point $B$ divides the segment $AC$ .

We have

$3\overrightarrow { a } +4\overrightarrow { b } -7\overrightarrow { c } =\overrightarrow { 0 }$

$\Rightarrow 4\overrightarrow { b } =-3\overrightarrow { a } +7\overrightarrow { c }$

$\Rightarrow \overrightarrow { b } =\cfrac { -3\overrightarrow { a } +7\overrightarrow { c } }{ 4 }$

$\Rightarrow \overrightarrow { b } =\cfrac { -3\overrightarrow { a } +7\overrightarrow { c } }{ 7-3 }$

$\Rightarrow$

$B$ divides the line segment

$AC$ in the ratio

$7:3$ externally.

If $\vec {a}$ is a non-zero vector of modulus $a$ and $m$ is a non-zero scalar then $m\vec {a}$ is a unit vector if

$m\vec{a}$ is a unit vector so its modulus is

$1$ .
Therefore

$|m\vec{a}|=1$ or

$|m||\vec a|=1$

$|\vec a|=\dfrac{1}{|m|}$
Hence the necessary condition is

$a=\dfrac{1}{|m|}$

$AC$ and $BD$ are the diagonals of a quadrilateral $ABCD$ , prove that: $\vec{AB}+\vec{DC}=\vec{AC}+\vec{DB}$ .

Let

$ABCD$ be a rectangle,

$AC$ and

$BD$ its diagonals.
In

$\Delta ABC$ ,

$\vec{AB}+\vec{BC}=\vec{AC}$

$\vec{AB}=\vec{AC}-\vec{BC}$ ...........(i)
In

$\Delta BCD$ by vector law of addition

$\vec{DC}=\vec{DB}+\vec{BC}$ ......(ii)
by adding (i) and (ii), we get

$\vec{AB}+\vec{DC}=\vec{AC}-\vec{BC}+\vec{DB}+\vec{BC}$

$\vec{AB}+\vec{DC}=\vec{AC}+\vec{DB}$

Prove that $[\overrightarrow{a}. \overrightarrow{b}. \overrightarrow{c} + \overrightarrow{d}] = [\overrightarrow{a} . \overrightarrow{b} . \overrightarrow{c}]+[\overrightarrow{a} . \overrightarrow{b} . \overrightarrow{d}]$ .

Consider,

$[\vec{a}\cdot \vec{b}\cdot \vec{c}+\vec{d}]$

$=[\vec{a}\cdot \vec{b}\cdot (\vec{c}+\vec{d})]$

$=[\vec{a}\cdot \vec{b}\cdot \vec{c}+\vec{a}\cdot \vec{b}\cdot \vec{d}]$ ............ [Since, dot is distributive over addition]

$=[\vec{a}\cdot \vec{b}\cdot \vec{c}]+[\vec{a}\cdot \vec{b}\cdot \vec{d}]$

Hence,

$[\vec{a}\cdot \vec{b}\cdot \vec{c}+\vec{d}]$

$=[\vec{a}\cdot \vec{b}\cdot \vec{c}]+[\vec{a}\cdot \vec{b}\cdot \vec{d}]$

In a triangle $ABC, D, E, F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively then prove that, $\vec {AD} = - (\vec {BE} + \vec {CF})$ .

$\triangle CFB$

$\Rightarrow \vec{CF}+\vec{FB}+\vec{BC}=0$ ................. I

$\triangle CBE$

$\Rightarrow \vec{BE}+\vec{EC}+\vec{CB}=0$ ............... II

$\triangle ADC$

$\Rightarrow \vec{AD}+\vec{DC}+\vec{CA}=0$ .............. III

$\triangle ADB$

$\Rightarrow \vec{AD}+\vec{DB}+\vec{BA}=0$ ............... IV

Adding equation III and IV, we get

$\Rightarrow2\vec{AD}+\vec{CA} +\vec{BA}=0$

$[\because \vec{DC} and \vec{DB}$ are equal and opposite]................. V

Now,

$\vec{CA}=2\vec{CE}$ and

$\vec{BA}=2\vec{BF}$ ........... VI

Using equations V and VI, we get

$\Rightarrow \vec{AD}=-(\vec{CE}+\vec{BF})$ .................. VII

Adding equation I and II, we get

$\Rightarrow \vec{BE}+\vec{CF}=\vec{CE}+\vec{BF}$

$[\because \vec{BC}=-\vec{CB},\vec{FB}=-\vec{BF},\vec{EC}=-\vec{CE}]$ ........... VIII

Putting equation VIII in equation VII, we get

$\Rightarrow \vec{AD}=-(\vec{BE}+\vec{CF})$

Hence, proved.

If the line segments joining the points $A(a,b)$ and $B(c,d)$ subtends and angle $\theta$ at the origin, then find $\cos{\theta}$

$\vec {OA}=a \hat i+b \hat j$

$\vec {OB}=c\hat i +d \hat j$

$|\vec {OA}|=\sqrt{a^2+b^2}$

$|\vec {OB}|=\sqrt {c^2+d^2}$

$\cos \theta = \dfrac{{\overrightarrow {OA} \cdot \overrightarrow {OB} }}{{|\overrightarrow {OA} ||\overrightarrow {OB} |}} = \dfrac{{ac + bd}}{{\sqrt {{a^2} + {b^2}} \sqrt {{c^2} + {d^2}} }}$

1105951_731869_ans_649bbe81a8f945069c96e15e13606e56.PNG

Write the associative law for addition of vectors.

The associative law of vector states that the sum of the vector remains same irrespective of their order or grouping in which they are arranged.

That is, if

$\vec{A},\vec{B},\vec{C}$ are three vectors then the associative law of addition is

$\vec{A}+(\vec{B}+\vec{C})=(\vec{A}+\vec{B})+\vec{C}$

Calculate the scalar product of the following vectors:
$\displaystyle a \, = \, \left \{ 2, 4, 1 \right \} \, and\ \, b \, = \, \left \{ 3, 5, 7 \right \}$

Given vectors

$a = \left \{2, 4, 1\right \}$ and

$b = \left \{3, 5, 7\right \}$

$\vec {a} = 2\hat {i} + 4\hat {j} + \hat {k}$

and

$\vec {b} = 3\hat {i} + 5\hat {j} + 7\hat {k}$

So scalar product

$\vec {a} \cdot \vec {b} = (2\hat {i} + 4\hat {j} + \hat {k})\cdot (3\hat {i} + 5\hat {j} + 7\hat {k})$

$= 2\times3 + 4\times5 + 7\times1$ [here

$\hat {i} . \hat {i} = 1, \hat {j} . \hat {j} = 1, \hat {k} . \hat {k} = 1]$

$= 6 + 20 + 7$

$= 33$ .

Three forces $2\hat { i } -\hat { j } +3\hat { k } ,\hat { j } +\hat { k }$ and $-2\hat { i } -2\hat { j } +\hat { k }$ are applied on a body. Is the body in equilibrium?

Resultant of forces

$2\hat{i}-\hat{j}+3\hat{k}$ ,

$\hat{j}+\hat{k}$ and

$-2\hat{i}-2\hat{j}+\hat{k}$ is

$(2\hat{i}-\hat{j}+3\hat{k})+(\hat{j}+\hat{k})+(-2\hat{i}-2\hat{j}+\hat{k})$

$=2\hat{i}-\hat{j}+3\hat{k}+\hat{j}+\hat{k}-2\hat{i}-2\hat{j}+\hat{k}$

$=-2\hat{j}+5\hat{k}$

As the resultant force is not zero, the body is not in equilibrium.

Find the value $\left[ \hat { i } \quad \hat { j } \quad \hat { k } \right]$ .

Find the magnitude of each of the two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ , having the same magnitude such that the angle between them is $60^o$ and their scalar product is $\displaystyle\frac{9}{2}$ .

We Know that for any two given vectors

$\vec{a}$ and

$\vec{b}$

$\vec{a}.\vec{b} = |a||b| \cos \theta$

It is also given that

$\vec{a}$ and

$\vec{b}$ are equal in magnitude

Hence,

$|a|=|b|$

After substitution we get:

$\vec{a}.\vec{b} = |a||a| \cos 60^o$

$\Rightarrow \vec{a}.\vec{b} = |a|^2 \cos 60^o$

$\Rightarrow \dfrac{9}{2} = |a|^2 \times \dfrac{1}{2}$

$\Rightarrow |a|^2 = 9$

$\Rightarrow |a|=3$

$|a| = |b|$ , magnitude of

$\vec{a}$ and

$\vec{b}$ is equal to

$3$ .

Evaluate $\overset { \wedge }{ j } \left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right)$

$\overset { \wedge }{ j. } \left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right) \\ \overset { \wedge }{ j. } \overset { \wedge }{ i } +\overset { \wedge }{ j. } \overset { \wedge }{ k } \\ \left| \overset { \wedge }{ j } \right| \left| \overset { \wedge }{ i } \right| \cos { { 90 }^{ O } } +\left| \overset { \wedge }{ j } \right| \left| \overset { \wedge }{ k } \right| \cos { { 90 }^{ O } } \\ 1\times 1\times 0+1\times 1\times 0\\ 0+0\\ 0$

Prove that:
$\vec{P}\times \vec{Q}=(\hat{k})P_xQ_y+(-\hat{j})P_xQ_z+(-\hat{k})P_yQ_x+O+\hat{i}P_yQ_z+(\hat{j})P_zQ_x+(-\hat{i})P_zQ_y$ .

$\vec{P}\times\vec{Q}=0+\hat{k}(P_{x}Q_{y})+(-j)P_{x}Q_{z}+(-\hat{k})P_{y}Q_{x}+0+i P_{y}P_{z}+j P_{z}Q_{z}+(-i)P_{z}Q_{y}+0$

$\vec{P}\times \vec{Q}=\begin{vmatrix} i & j & k \\ { P }_{ x } & { P }_{ y } & { P }_{ z } \\ { Q }_{ x } & { Q }_{ y } & { Q }_{ z } \end{vmatrix}$

$=i(P_{y}Q_{z}-Q_{y}P_{z})-j(P_{x}Q_{z}-Q_{x}P_{z})+k(P_{x}Q_{y}-P_{y}Q_{x})$

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\hat { i } +2\hat { j } -\hat { k }$ and $-\hat { i } +\hat { j } +\hat { k }$ respectively in the ratio $2:1$
(i) Internally (ii) externally

$\vec { a } =\hat { i } +2\hat { j } -\hat { k } \quad \quad \quad \vec { b } =-\hat { i } +\hat { j } +\hat { k }$
Position vector of a point for internal division

$\cfrac { m\vec { b } +n\vec { a } }{ m+n } =\cfrac { 2\left( -\hat { i } +\hat { j } +\hat { k } \right) +1\left( \hat { i } +2\hat { j } -\hat { k } \right) }{ 2+1 } =\cfrac { -\hat { i } +4\hat { j } +\hat { k } }{ 3 }$
Position vector of a point for external division

$\cfrac { m\vec { b } +n\vec { a } }{ m-n } =\cfrac { -3\hat { i } +3\hat { k } }{ 2-1 } =-3\hat { i } +3\hat { k }$

$\displaystyle \overrightarrow{AC}.$

$\vec{OA}=3\hat{i}+2\hat{j}-3\hat{k}$

$\vec{OB}=5\hat{i}+1\hat{j}-1\hat{k}$

$\vec{OC}=1\hat{i}-2\hat{j}+1\hat{k}$

$\vec{AB}=\vec{OB}-\vec{OA}=(5\hat{i}+1\hat{j}-1\hat{k})-(3\hat{i}+2\hat{j}-3\hat{k})$

$\vec{AB}=2\hat{i}-1\vec{j}+2\hat{k}$

$\hat{AB}=\dfrac{2\hat{i}-1\vec{j}+2\hat{k}}{3}\Rightarrow direction of \vec{AB}$

$\vec{AC}=\vec{OC}-\vec{OA}=(1\hat{i}-2\hat{j}+1\hat{k})-(3\hat{i}+2\hat{j}-3\hat{k})$

$\vec(AC)=-2\hat{i}-4\hat{j}+4\hat{k}$

$|\vec{AC}|=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{36}=6$

$new vector = |\vec{AC}|\hat{AB}$

$= 6\times{\dfrac{2\hat{i}-1\vec{j}+2\hat{k}}{3}}$

$=4\hat{i}-2\hat{j}+4\hat{k}$

Calculate the scalar product of the following vectors. $\displaystyle \overrightarrow{CB}, \overrightarrow{CA}, \, and \, \overrightarrow{AB}$ are a, b, and c respectively. Find the scalar product of the vectors $\displaystyle \overrightarrow{CA} \, and \, \overrightarrow{CD},$ where [CD] is the median of the triangle ABC.

The point D is given by

$\dfrac{1}{2}\left(A+B\right)$ and the median CD is given by the vector

$CD=\dfrac{1}{2}\left(A+B\right)-C$

$=\dfrac{1}{2}\left(A-C+B-C\right)$

$=\dfrac{1}{2}\left(\vec{CA}+\vec{CB}\right)$

$=\dfrac{1}{2}\left(b+a\right)$

$\vec{CA}.\vec{CD}=b.\dfrac{1}{2}\left(a+b\right)$

$=\frac{1}{2}\left(a.b+|b|^2\right)$

Find the projection of the vector $\hat {i} + 3\hat {j} + \hat {k}$ on the vector $7\hat {i} - \hat {j} + 8\hat {k}$ .

Let

$\vec {a} = \hat {i} + 3\hat {j} + 7\hat {k}$ and

$\vec {b} = 7\hat {i} - \hat {j} + 8\hat {k}$
Projection of

$\vec {a}$ on

$\vec {b} = \dfrac {\vec {a} . \vec {b}}{|\vec {b}|}$

$= \dfrac {(\hat {i} + 3\hat {j} + 7\hat {k})\cdot (7\hat {i} - \hat {j} + 8\hat {k})}{\sqrt {49 + 1 + 64}} = \dfrac {7 - 3 + 56}{\sqrt {114}}$

$= \dfrac {60}{\sqrt {114}}$ .

Let $\vec a$ & $\vec b$ be two non-zero perpendicular vectors. If a vector $\vec x$ satisfying the equation $\vec x$ x $\vec b$ = $\vec a$ is $\vec x$ = $\beta \vec b - {1 \over {{{\left| {\vec b} \right|}^2}}}$ $\vec a$ x $\vec b$ then $\beta$ can be

$\bar { a } \times \bar { b } =\left| \bar { a } \right| \left| \bar { b } \right| \sin { 90°=\left| \bar { a } \right| \left| \bar { b } \right| } \longrightarrow (1)\\ \bar { x } =\beta \bar { b } -\frac { 1 }{ \left| \bar { b } \right| } \left( \bar { a } \times \bar { b } \right) \\ \bar { x } \times \bar { b } =\beta \bar { b } \times \bar { b } -\frac { 1 }{ { \left| \bar { b } \right| }^{ 2 } } \left( \bar { a } \times \bar { b } \right) \times \bar { b } \\ \bar { x } \times \bar { b } =-\frac { 1 }{ { \left| \bar { b } \right| }^{ 2 } } \left| \bar { a } \times \bar { b } \right| \left| \bar { b } \right| \\ \bar { x } \times \bar { b } =-\frac { 1 }{ { \left| \bar { b } \right| }^{ 2 } } \left| \bar { a } \right| \left| \bar { b } \right| \left| \bar { b } \right| =-\left| \bar { a } \right| \hat { a } =-\bar { a } \\ \bar { x } \times \bar { b } =-\bar { a }$

Since

$\beta$ is not used and

$\bar { x } \times \bar { b } =-\bar { a }$

So,

$\beta$ can be any real number.

If $\vec{r}.\hat{i} = \vec{r}.\hat{j} = \vec{r}.\hat{k}$ and $|\vec{r}| = 3$ , then find vector $\vec{r}$ .

Let

$\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}$ .

Now according to the problem we have,

$a=b=c$ [As

$\vec{r}.\hat{i}=a$ ,

$\vec{r}.\hat{j}=b$

$\vec{r}.\hat{k}=c$ ]....(1)

Also given,

$|\vec{r}|=3$

or,

$a^2+b^2+c^2=9$

or,

$3a^2=9$ [Using (1)]

or,

$a=\pm\sqrt{3}$ .

So,

$\vec{r}=\pm \sqrt{3}(\hat{i}+\hat{j}+\hat{k})$ .

Find all $\lambda \in R$ such that $\left( {x,y,z} \right) \ne \left( {0,0,0} \right)$ and $\left( {\hat i + \hat j + \hat k} \right)x + \left( {3\hat i - 3\hat j + \hat k} \right)y + \left( { - 4\hat i + 5\hat j} \right)z = \lambda \left( {x\hat i + y\hat j + z\hat k} \right)$

$\overset { \wedge }{ i } (x+3y-4z)+\overset { \wedge }{ j } (x-3y+5z)+\overset { \wedge }{ k } (3x+y)=\overset { \wedge }{ i } (\lambda x)+\overset { \wedge }{ j } (\lambda y)+\overset { \wedge }{ k } (\lambda z)$

$comparing\quad both\quad sides$

$\implies\quad x+3y-4z=\lambda x\quad -(1)$

$\implies\quad x-3y+5z=\lambda y\quad -(2)$

$\implies\quad 3x+y=\lambda z\quad -(3)$

$\implies\quad Rearranging(1),(2)\& (3)$

$\implies\quad x(1-\lambda )+y(3)+2(-4)=0\quad -(4)$

$x(1)+y(-3-\lambda )+2(5)=0\quad -(5)$

$x(3)+y(1)+2(-\lambda )=0\quad -(6)$

$We\quad know\quad if\quad a1x+a2y+a3z=0$

$b1x+b2y+b3z=0$

$c1x+c2y+c3z=0$

$\implies\quad \left| \begin{matrix} a1 & a2 & a3 \\ b1 & b2 & b3 \\ c1 & c2 & c3 \end{matrix} \right| =0\\ \therefore Here,\quad \left| \begin{matrix} 1-\lambda & 3 & -4 \\ 1 & -3-\lambda & 5 \\ 3 & 1 & -\lambda \end{matrix} \right| =0$

$\implies\quad (1-\lambda )[(3\lambda -{ \lambda }^{ 2 })-5]-3(-\lambda -15)-4(1+9+3\lambda )=0$

$\implies\quad (1-\lambda )({ \lambda }^{ 2 }-3\lambda -5)+3\lambda +45-40-12\lambda =0$

$\implies\quad { \lambda }^{ 2 }+3\lambda -5-{ \lambda }^{ 3 }-3{ \lambda }^{ 2 }+5\lambda -9\lambda +5=0$

$\implies\quad -{ \lambda }^{ 3 }-2{ \lambda }^{ 2 }-\lambda =0$

$\implies\quad -\lambda ({ \lambda }^{ 2 }+2\lambda +1)=0$

$\implies\quad \lambda =0\quad or\quad { \lambda }^{ 2 }+2\lambda +1=0$

$\implies\quad { (\lambda +1) }^{ 2 }=0$

$\implies\quad \lambda =-1$

$\therefore \lambda =0\quad or\quad \lambda =-1$

Show that magnitude of vector product of two vectors is numerically equal to the area of a parallelogram formed by two vectors.

$Formula\quad for\quad area\quad of\quad parallelogram$

$=base\times height$

$=\left| \overline { a } \right| .h$

$h=\left| \overline { b } \right| \sin { \theta }$

$\therefore Area=\left| \overline { a } \right| .\left| \overline { b } \right| \sin { \theta }$

$We\quad know\quad \left| \overline { a } \times \overline { b } \right| =\left| \overline { a } \right| .\left| \overline { b } \right| \sin { \theta }$

$Hence\quad proved.$

1022270_1025499_ans_c432956839e14c3bb65bd1e0e1dfe0e4.png

If ${\vec x}$ satisfying the conditions ${\vec b}$ . ${\vec x}$ = $\beta$ & ${\vec b}$ x ${\vec x}$ = ${\vec a}$ is $\vec x = {{\left( {{\beta ^2} - 12} \right)\vec b} \over {{{\left| {\vec b} \right|}^2}}} + {{\vec a \times \vec b} \over {{{\left| {\vec b} \right|}^2}}}$ then $\beta$ can be

$\vec b.\vec x=\beta$

$\vec x=\dfrac{(\beta^2-12)\vec b}{|\vec b|^2}+\dfrac{\vec a\times \vec b}{|\vec b|^2}$

$\vec b.\vec x=\beta^2-12=\beta\implies \beta^2-\beta-12=0\implies \beta=4,-3$

Establish the following vector inequalities geometrically or
otherwise:
(a) $\left| {a + b} \right| \le \left| a \right| + \left| b \right|$
(b) $\left| {a + b} \right| \ge \left| {\left| a \right| - \left| b \right|} \right|$

$From\quad triangle\quad law\quad of\quad vectors:$

$\overline { c } =\overline { a } +\overline { b }$

$From\quad triangle\quad property:$

$we\quad know\quad sum\quad of\quad two\quad sides\ge third\quad side.$

$\implies\quad \left| \overline { a } \right| +\left| \overline { b } \right| \ge \left| \overline { c } \right|$

$\implies\quad \left| \overline { a } \right| +\left| \overline { b } \right| \ge \left| \overline { a } +\overline { b } \right|$

$\implies\quad \left| \overline { a } +\overline { b } \right| \le \left| \overline { a } \right| +\left| \overline { b } \right|$

$\therefore a)is\quad established.$

$Similarily,we\quad know\quad from\quad triangle\quad property:$

$Difference\quad of\quad two\quad sides\le third\quad side.$

$\implies\quad \left| \left| \overline { a } \right| -\left| \overline { b } \right| \right| \le \left| \overline { c } \right|$

$\implies\quad \left| \left| \overline { a } \right| -\left| \overline { b } \right| \right| \le \left| \overline { a } +\overline { b } \right|$

$\implies\quad \left| \overline { a } +\overline { b } \right| \ge \left| \left| \overline { a } \right| -\left| \overline { b } \right| \right|$

$\therefore b)is\quad established.$

(a) Find the unit vector which is parallel to the vector $\overrightarrow {\rm{A}} {\rm{ = 2}}\widehat {\rm{i}}{\rm{ + 2}}\widehat {\rm{j}}{\rm{ - 2}}\widehat {\rm{k}}$
(b) Find the unit vector which is perpendicular to both of the
vector $\overrightarrow {\rm{A}} {\rm{ = 2}}\widehat {\rm{i}}\;{\rm{and}}\;\overrightarrow B {\rm{ = 3}}\widehat {\rm{i}}{\rm{ + 4}}\widehat {\rm{j}}{\rm{ + 12}}\widehat {\rm{k}}$

$a)\left| \overline { A } \right| =\sqrt { { 2 }^{ 2 }+{ 2 }^{ 2 }+{ (-2) }^{ 2 } } =\sqrt { 12 }$

$\overset { \wedge }{ A } =\cfrac { \overline { A } }{ \left| \overline { A } \right| } =\cfrac { 2\overset { \wedge }{ i } +2\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { 12 } }$

$=\cfrac { 2(\overset { \wedge }{ i } +\overset { \wedge }{ j } -\overset { \wedge }{ k } ) }{ 2\sqrt { 3 } }$

$=\cfrac { \overset { \wedge }{ i } +\overset { \wedge }{ j } -\overset { \wedge }{ k } }{ \sqrt { 3 } }$

$b)Vector\quad perpendicular\quad to\quad \overline { A } and\overline { B } .$

$\overline { A } \times \overline { B } =\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 2 & 0 & 0 \\ 3 & 4 & 12 \end{matrix} \right|$

$\implies\quad \overset { \wedge }{ i } (0)-\overset { \wedge }{ j } (24)+\overset { \wedge }{ k } (8)$

$\implies\quad -24\overset { \wedge }{ j } +8\overset { \wedge }{ k }$

$\implies\quad 8(-3\overset { \wedge }{ j } +\overset { \wedge }{ k } )$

$unit\quad vector:\quad \cfrac { \overline { A } \times \overline { B } }{ \left| \overline { A } \times \overline { B } \right| }$

$\implies\quad \cfrac { 8(-3\overset { \wedge }{ j } +\overset { \wedge }{ k } ) }{ 8\sqrt { 9+1 } }$

$\implies\quad \cfrac { -3\overset { \wedge }{ j } +\overset { \wedge }{ k } }{ \sqrt { 10 } }$

Suppose the vectors $\overrightarrow a ,\overrightarrow {\rm{b}} ,\overrightarrow {\rm{c}}$ on a plane satisfy the condition that $|\overrightarrow a | = |\overrightarrow {\rm{b}} | = |\overrightarrow {\rm{c}} | = |\overrightarrow {\rm{a}} + \overrightarrow b | = 1{\rm{ }};{\rm{ }}\overrightarrow {\rm{c}}$ is perpendicular to $\overrightarrow {\rm{a}}$ and $\overrightarrow {\rm{b}} {\rm{.}}\overrightarrow {\rm{c}} {\rm{ > 0}}$ , then

${ \left| \overline { a } +\overline { c } \right| }^{ 2 }={ \left| \overline { a } \right| }^{ 2 }+{ \left| \overline { c } \right| }^{ 2 }+2\overline { a } .\overline { c }$

$\because \overline { a } \quad is\quad perpendicular\quad to\quad \overline { c } .$

$\therefore \overline { a } -\overline { c } =0\quad -(i)$

$\implies\quad { \left| \overline { a } +\overline { c } \right| }^{ 2 }={ \left| \overline { a } \right| }^{ 2 }+{ \left| \overline { a } \right| }^{ 2 }$

$\implies\quad { \left| \overline { a } +\overline { b } \right| }^{ 2 }={ \left| \overline { a } \right| }^{ 2 }+{ \left| \overline { b } \right| }^{ 2 }+2\overline { a } .\overline { b }$

$\because \left| \overline { a } \right| =\left| \overline { b } \right| =\left| \overline { a } +\overline { b } \right| =1$

$\implies\quad 1=1+1+2\overline { a } .\overline { b }$

$\implies\quad \overline { a } .\overline { b } =\cfrac { -1 }{ 2 }$

$\implies\quad 2\overline { a } .\overline { b } =-1$

$\overline { a } .\overline { b } =\cfrac { -1 }{ 2 }$

$\implies\quad \left| \overline { a } \right| .\left| \overline { b } \right| \cos { \theta } =\cfrac { -1 }{ 2 }$

$\implies\quad \cos { \theta } =\cfrac { -1 }{ 2 }$

$\implies\quad \theta =\pi -\cfrac { \pi }{ 3 } =\cfrac { 2\pi }{ 3 }$

$Angle\quad between\quad \overline { a } \quad and\quad \overline { b } =\cfrac { 2\pi }{ 3 }$

$\overline { a } .\overline { c } =0\quad -from(i)$

$\implies\quad \left| \overline { a } \right| .\left| \overline { c } \right| \cos { \alpha } =0$

$\implies\quad \cos { \alpha } =0\\ implies\quad \alpha =\cfrac { \pi }{ 2 }$

$Angle\quad between\quad \overline { a } \quad and\quad \overline { c } =\cfrac { \pi }{ 2 }$

Let $\overrightarrow a = 2\widehat i + 3\widehat i + 2\widehat k$ and $\overrightarrow b = \widehat i + 2\widehat i + \widehat k$ .
Find the projection of
1. $\overrightarrow a$ on $\overrightarrow b$
2. $\overrightarrow b$ on $\overrightarrow a$

Consider given the given vectors, $\overrightarrow{a}=2\widehat{i}+3\widehat{j}+2\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}+\widehat{k}$

Now, projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ $=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{b} \right|}=\dfrac{\left( 2\widehat{i}+3\widehat{j}+2\widehat{k} \right).\left( \widehat{i}+2\widehat{j}+\widehat{k} \right)}{\left| \widehat{i}+2\widehat{j}+\widehat{k} \right|}$

$= \dfrac{\left( 2\widehat{i}+3\widehat{j}+2\widehat{k} \right).\left( \widehat{i}+2\widehat{j}+\widehat{k} \right)}{\left| \widehat{i}+2\widehat{j}+\widehat{k} \right|}=\dfrac{2.1+3.2+2.1}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}}$

$=\dfrac{10}{\sqrt{6}}$

Projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ ,

$=\dfrac{\left( \widehat{i}+2\widehat{j}+\widehat{k} \right).\left(2\widehat{i}+3\widehat{j}+2\widehat{k} \right)}{\left| 2\widehat{i}+3\widehat{j}+2\widehat{k} \right|}=\dfrac{1.2+2.3+1.2}{\sqrt{{{2}^{2}}+{{3}^{2}}+{{2}^{2}}}}=\dfrac{10}{\sqrt{17}}$

Hence, this is the answer.

Find the projection of the vector $\vec {a}=4\hat {i}-2\hat {j}+\hat {k}$ on the vector $\vec {b}=3\hat {i}+6\hat {j}+2\hat {k}$ . Also find component of $\vec {a}$ along $\vec {b}$ and perpendicular to $\vec {b}$ .

$\vec a = 4\hat i - 2\hat j + \hat k$

$\vec b = 3\hat i + 6\hat j + 2\hat k$

$\left| {\vec b} \right| = \sqrt {9 + 36 + 4} = 7$

$\hat b = \dfrac{{\vec b}}{{|\vec b|}} = \dfrac{3}{7}\hat i + \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k$

Projection of

$\vec a$ on

$\vec b=\vec a.\hat b = \left( {4\hat i - 2\hat j + \hat k} \right).\left( {\dfrac{3}{7}\hat i + \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k} \right) = \dfrac{2}{7}$

Component of vector

$\vec a$ on

$\vec b=\dfrac{2}{7} \hat b=\dfrac{2}{7}\left( {\dfrac{3}{7}\hat i + \dfrac{6}{7}\hat j + \dfrac{2}{7}\hat k} \right)={\dfrac{3}{{49}}\hat i + \dfrac{6}{{49}}\hat j + \dfrac{2}{{49}}\hat k}$

Component of

$\vec a$ along perpendicular to

$\vec b=\vec a-\dfrac{2}{7}\hat b$

$={\dfrac{{190}}{{49}}\hat i - \dfrac{{110}}{{49}}\hat j + \dfrac{{45}}{{49}}\hat k}$

If $\overrightarrow a$ is a unit vector and $\left( {\overrightarrow x - \overrightarrow a } \right)$ , $\left( {\overrightarrow x - \overrightarrow b } \right)$ =35, then the $\left| {\overline x } \right|$ .
Solve the equation ${x^4} - 8{x^3} + 24{x^2} - 32x + 20 = 0$ if 3+1 is a root.

1165764_1051493_ans_d8c114e81050459dbc508d82b0afd5c5.jpg

For any two vectors $\vec{a}$ and $\vec{b}$ , If $(\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0.$
Then $| \vec{a}|-|\vec{b}|=?$ .

$(\overline { a } +\overline { b } ).(\overline { a } -\overline { b } )=0$

$\implies\quad { \left| \overline { a } \right| }^{ 2 }-{ \left| \overline { b } \right| }^{ 2 }=0$

$\implies\quad \left| \overline { a } \right| =\left| \overline { b } \right|$

$(\left| \overline { a } \right| =-\left| \overline { b } \right| \quad$ Not possible because both positive).

$\implies\quad \left| \overline { a } \right| -\left| \overline { b } \right| =0$

Find the value of $\lambda$ such that points $P(1, 2), Q(-2, 3)$ and $R(\lambda +1, \lambda)$ are not forming a triangle?

$P-\quad \overset { \wedge }{ i } +2\overset { \wedge }{ j }$

$Q-\quad -2\overset { \wedge }{ i } +3\overset { \wedge }{ j }$

$R-\quad (\lambda +1)\overset { \wedge }{ i } +\lambda \overset { \wedge }{ j }$

$\overline { PQ } =(-2-1)\overset { \wedge }{ i } +(3-2)\overset { \wedge }{ j }$

$=-3\overset { \wedge }{ i } +\overset { \wedge }{ j }$

$\overline { QR } =(\lambda +1+2)\overset { \wedge }{ i } +(\lambda -3)\overset { \wedge }{ j }$

$=(\lambda +3)\overset { \wedge }{ i } +(\lambda -3)\overset { \wedge }{ j }$

$PQR$ form a triangle,

$\overline { PQ } +\overline { QR } =\overline { PR }$

$\overline { PR } =(\lambda +1-1)\overset { \wedge }{ i } +(\lambda -2)\overset { \wedge }{ j }$

$=\lambda \overset { \wedge }{ i } +(\lambda -2)\overset { \wedge }{ j }$

They don't form a triangle.

$Hence,\overline { PQ } +\overline { QR } \neq \overline { PR }$

$\implies\quad \overset { \wedge }{ i } (\lambda +3-3)+\overset { \wedge }{ j } (\lambda -3+1)\neq \overset { \wedge }{ i } (\lambda )+\overset { \wedge }{ j } (\lambda -2)$

$\implies\quad \overset { \wedge }{ i } (\lambda )+\overset { \wedge }{ j } (\lambda -3+1)\neq \overset { \wedge }{ i } (\lambda )+\overset { \wedge }{ j } (\lambda -2)$

$\implies\quad \overset { \wedge }{ i } (\lambda )+\overset { \wedge }{ j } (\lambda -2)\neq \overset { \wedge }{ i } (\lambda )+\overset { \wedge }{ j } (\lambda -2)\quad$

$\because$ L.H.S and R.H.S can't be equal,

$\therefore$ Both have to be zero,then

$\overline { PR } =\overline { 0 } ,$

hence no side and hence,no triangle will be possible.$$

$\implies\quad \lambda =0,2.$

But sine

$\lambda$ can't be equal to

$0$ and

$2$ at the same time.

Hence,there is no value of

$\lambda$ for which the given points can't form a triangle.

Prove that $\left[ {\vec a\,\,\vec b\,\,\vec c\,\, + \,\,\vec d} \right] = \left[ {\vec a\,\,\vec b\,\,\vec c\,} \right] + \left[ {\vec a\,\,\vec b\,\,\vec d\,} \right]$

$\left[\vec{a}\vec{b}\vec{c}+\vec{d}\right]=\left[\vec{a}\vec{b}\vec{c}\right]+\left[\vec{a}\vec{b}\vec{d}\right]$

$\vec{a}=x_{1}\hat{i}+y_{1}\hat{j}+z_{1}\hat{k}$

$\vec{b}=x_{2}\hat{i}+y_{2}\hat{j}+z_{2}\hat{k}$

$\vec{c}=x_{3}\hat{i}+y_{3}\hat{j}+z_{3}\hat{k}$

$\vec{d}=x_{4}\hat{i}+y_{4}\hat{j}+z_{4}\hat{k}$

$\left[\vec{a}\vec{b}.\vec{c}+\vec{d}\right]=\begin{vmatrix} { x }_{ 1 } & { y }_{ 1 } & { z }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { z }_{ 2 } \\ { x }_{ 3 }+{ x }_{ 4 } & { y }_{ 3 }+{ y }_{ 4 } & { z }_{ 3 }+{ z }_{ 4 } \end{vmatrix}$

$=\begin{vmatrix} { x }_{ 1 } & { y }_{ 1 } & { z }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { z }_{ 2 } \\ { x }_{ 3 } & { y }_{ 3 } & { z }_{ 3 } \end{vmatrix}+\begin{vmatrix} { x }_{ 1 } & { y }_{ 1 } & { z }_{ 1 } \\ { x }_{ 2 } & { y }_{ 2 } & { z }_{ 2 } \\ { x }_{ 4 } & { y }_{ 4 } & { z }_{ 4 } \end{vmatrix}$ (properties of determinants)

$=\left[\vec{a}\vec{b}\vec{c}\right]+\left[\vec{a}\vec{b}\vec{d}\right]$

Show that vectors $\left| {\overline a } \right|\left| {\overline a + } \right|\overline a \left| {\overline b } \right.$ and $\left| b \right|\overline a - \left| a \right.\left| {\overline b } \right.$ are orthogonal.

Given that, $\left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a}$ and $\left| \overrightarrow{a} \right|\overrightarrow{b}-\left| \overrightarrow{b} \right|\overrightarrow{a}$

Now ,

$\left( \left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a} \right).$ $\left( \left| \overrightarrow{a} \right|\overrightarrow{b}-\left| \overrightarrow{b} \right|\overrightarrow{a} \right)$ = $\left( \left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a} \right).$ $\left( \left| \overrightarrow{a} \right|\overrightarrow{b}+\left| \overrightarrow{b} \right|\overrightarrow{a} \right).$

$={{\left| \overrightarrow{a} \right|}^{2}}{{\overrightarrow{b}}^{2}}-\left| \overrightarrow{a} \right|\overrightarrow{b}\left| \overrightarrow{b} \right|\overrightarrow{a}+\left| \overrightarrow{b} \right|\overrightarrow{a}\left| \overrightarrow{a} \right|\overrightarrow{b}-{{\left| \overrightarrow{b} \right|}^{2}}{{\overrightarrow{a}}^{2}}$

$\because \,\,\overrightarrow{x}={{\left| \overrightarrow{x} \right|}^{2}}$

$\therefore \,\,{{\left| \overrightarrow{a} \right|}^{2}}{{\left| \overrightarrow{b} \right|}^{2}}-\left| \overrightarrow{a} \right|\overrightarrow{b}\left| \overrightarrow{b} \right|\overrightarrow{a}+\left| \overrightarrow{b} \right|\overrightarrow{a}\left| \overrightarrow{a} \right|\overrightarrow{b}-{{\left| \overrightarrow{b} \right|}^{2}}{{\left| \overrightarrow{a} \right|}^{2}}$

$={{\left| \overrightarrow{a} \right|}^{2}}{{\left| \overrightarrow{b} \right|}^{2}}-\left| \overrightarrow{a} \right|\overrightarrow{b}\left| \overrightarrow{b} \right|\overrightarrow{a}+\left| \overrightarrow{b} \right|\overrightarrow{a}\left| \overrightarrow{a} \right|\overrightarrow{b}-{{\left| \overrightarrow{a} \right|}^{2}}{{\left| \overrightarrow{b} \right|}^{2}}$

$=-\left| \overrightarrow{a} \right|\overrightarrow{b}\left| \overrightarrow{b} \right|\overrightarrow{a}+\left| \overrightarrow{b} \right|\overrightarrow{a}\left| \overrightarrow{a} \right|\overrightarrow{b}$

$=-\left| \overrightarrow{a} \right|\overrightarrow{b}\left| \overrightarrow{b} \right|\overrightarrow{a}+\left| \overrightarrow{b} \right|\left| \overrightarrow{a} \right|\overrightarrow{a}\overrightarrow{b}$

$=0$

hence, both given vector are orthogonal

Two vertices of a triangle are at $-\hat{i}+3\hat{j}$ and $2\hat{i}+5\hat{j}$ and its orthocentre is at $\hat{i}+2\hat{j}$ . Find the position vector of thrid vertex.

We know that

$AB \perp CD$ and

$AC \perp BF$

$BC \perp AE$ and

$m_{1},m_{2}=-1$ for two lines which are

$\perp$

$\therefore AB \perp CH \Rightarrow \dfrac{(5-3)}{(2-(-1))}\left(\dfrac{y-2}{x-1}\right)=-1$

$2y-4=3-3x$

$3x+2y-7=0$ ______(1)

$BH \perp AC\Rightarrow \dfrac{(5-2)}{(2-1)}\left(\dfrac{y-3}{x+1}\right)=-1$

$3y-9=-x-1$

$x+3y-8=0$ _____ (2)

solving equations (1) and (2)

$(1)-3\times (2)$

$3x+2y-7-3x-9y+24=0$

$-7y+17=0$

$y=\dfrac{17}{7}$

$\therefore x=8-\dfrac{3\times 7}{7}=\dfrac{5}{7}$

$\therefore C\equiv \dfrac{5}{7}\hat{i}+\dfrac{17}{7}\hat{j}$

1417464_1088152_ans_e003add345b94b00a393234987008204.png

Angle between $\overrightarrow{a}$ & $\overrightarrow{b}$ . Find a vector along the angle bisector of $\overrightarrow{a}$ & $\overrightarrow{b}$ .

1340165_1082751_ans_293d7cfd0f0e47b6b7cd4cba2732ed4d.jpg

Let $\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k }$ and $\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ k }$ .Find the component of $\overset { \rightarrow }{ b }$ on $\overset { \rightarrow }{ a }$ .

Given,

$\begin{array}{l} \overrightarrow { a } =2\widehat { i } +3\widehat { j } +2\widehat { k } \\ \overrightarrow { b } =\widehat { i } +2\widehat { j } +\widehat { k } \end{array}$

Vector projection of

$\overrightarrow b$ on to

$\overrightarrow a$

$\begin{array}{l} =\left( { \dfrac { { \overrightarrow { a } .\overrightarrow { b } } }{ { \left| { \overrightarrow { a } } \right| } } } \right) .\dfrac { { \overrightarrow { a } } }{ { \left| { \overrightarrow { a } } \right| } } \\ =\left( { \dfrac { { 2+6+2 } }{ { \sqrt { 4+9+4 } } } } \right) \dfrac { { \left( { 2\widehat { i } +3\widehat { j } +2\widehat { k } } \right) } }{ { \sqrt { 4+9+4 } } } \\ =\dfrac { 10 }{ 17 } \left( { 2\widehat { i } +3\widehat { j } +2\widehat { k } } \right) \end{array}$

Find the projection of $3\hat i + 4\hat j + \hat k$ on $\hat i + \hat j - \hat k$ and its magnitude.

Let

$\vec { a } = < 3,4,1 >$ and

$\vec { b } = < 1,1,-1>$

then vector projection of

$\vec { a }$ onto

$\vec { b }$ using the following formula:

$\operatorname { proj } _ { \vec { b } } \vec { a } = \left( \dfrac { \vec { a } \cdot \vec { b } } { | \vec { b } | } \right) \dfrac { \vec { b } } { | \vec { b } | }$

$\operatorname { proj } _ { \vec { b } } \vec { a } = \dfrac { 6} { \sqrt { 3 } } \cdot \dfrac { < 1 ,1,-1 > } { \sqrt { 3 } }$

$\Rightarrow < { 2 } , {2 } , { -2 } >$

Magnitude is

$2\sqrt { 3 }$

If $\hat { a }$ and $\hat { b }$ are unit vector and $\theta$ is the angle between them, show that $\cos { \left( \dfrac { \pi }{ 4 } -\dfrac { \theta }{ 2 } \right) } =\dfrac { 1 }{ 2\sqrt { 2 } } \left( \left| \hat { a } -\hat { b } \right| +\left| \hat { a } +\hat { b } \right| \right)$

1111458_1114899_ans_069afea2415a44c5aab9a3236719c6af.jpeg

If $\ \overline r = \left( {x + y + 2} \right)\overline i + \left( {2x - y + 3} \right)\overline j + \left( {x + 2y + 7} \right)\overline k$ where $\overline r .\overline i = 3,\overline r .\overline j = 5$ then $\overline r .\overline k =$

$\bar{r}=(x+y+2)i+(2x-y+3)j+(x+2y+7)k$

$\because \bar{r}.i=3$

$\Rightarrow r+y+2+=3$

$\Rightarrow x+y=1\ ----- \left(1\right)$

$\because \bar{r}.\bar{j}=5$

$\Rightarrow 2x-y+3=5$

$\Rightarrow 2x-y=2\ -----\left(2\right)$

solving

$\left(1\right)$ and

$\left(2\right)$

$\ \ x+y\ =1\\ 2x-y=2\\ \_ \_ \_ \_ \_ \_ \_ \_ \_\\ 3x\quad=\ 3 \Rightarrow x=1\ and\ y=0$

$\therefore .\bar{r}\widehat{k}=x+2y+7$

$=1+2 \times0+7=8$

A line passes through the points whose position vectors are $\hat { i } + \hat { j } - 2 \hat { k }$ and $\hat { i } - 3 \hat { j } + \hat { k }$ . The position vector of a point on it at unit distance from the first point is

$\begin{array}{l} A\left( { 1,1,-2 } \right) B\left( { 1,-3,1 } \right) \\ Equation\, of\, line\, through\, A\, and\, B \\ \dfrac { { x-1 } }{ 0 } =\dfrac { { y-1 } }{ 4 } =\dfrac { { z+2 } }{ { -3 } } \\ P\left( { 1,4r+1,-3r-2 } \right) \\ PA=1 \\ \sqrt { { { \left( { 4r } \right) }^{ 2 } }+{ { \left( { -3r } \right) }^{ 2 } } } =1 \\ 5\left| r \right| =1 \\ r=\pm \dfrac { 1 }{ 5 } \\ r=\dfrac { 1 }{ 5 } ,P\left( { 1,\dfrac { 9 }{ 5 } ,-\dfrac { { 13 } }{ 5 } } \right) \\ r=-\dfrac { 1 }{ 5 } ,P\left( { 1,\dfrac { 1 }{ 5 } ,-\dfrac { 7 }{ 5 } } \right) \end{array}$

If $C$ is the midpoint of $AB$ and $P$ is any point outside $AB$ , then $\overrightarrow{PA}+\overrightarrow{PB}=$

$\overrightarrow{PA}=\overrightarrow{PC}+\overrightarrow{CA}$

$\overrightarrow{PB}=\overrightarrow{PC}+\overrightarrow{CB}$

$\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)$

$=2\overrightarrow{PC}$

$\left(\because \overrightarrow{CA}=-\overrightarrow{CB}\right)$
Hence

$\overrightarrow{CA}+\overrightarrow{CB}=0$
and

$\overrightarrow{PA}+\overrightarrow{PB}=2\overrightarrow{PC}$

Let $\vec {a}=\hat {i}+4\hat {j}+2\hat {k},\,\vec b=3\hat i-2\hat j+7\hat k$ and $\vec {c}=2\hat {i}-\hat j+4\hat {k}$ . Find a vector $\vec d$ which is perpendicular to both $\vec {a}$ and $\vec {b}$ , and $\vec {c}\cdot \vec {d}=15$ .

Consider the problem

Given,

$\begin{array}{l} \vec { a } =\hat { i } +4\hat { j } +2\hat { k } \\ \\ \vec { b } =3\hat { i } -2\hat { j } +7\hat { k } \\ \\ \vec { c } =2\hat { i } -\hat { j } +4\hat { k } \end{array}$

Since

$d$ is perpendicular to both

$\vec a$ and

$\vec b$ , it is paralle to

$\vec a \times \vec b$ .

Suppose,

$d=\lambda (\vec a \times \vec b)$ for some scalar $$

$\lambda$ .

then,

$\begin{array}{l} d=\lambda \left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & 4 & 2 \\ 3 & { -2 } & 7 \end{array} } \right| \\ \\ =\lambda \left[ { \left( { 28+4 } \right) \hat { i } -\left( { 7-6 } \right) \hat { j } +\left( { -2-12 } \right) \hat { k } } \right] \\ \\ =\lambda\left[ { 32\hat { i } -\hat { j } -14\hat { k } } \right] \, \, \, \, \, \, -----\, \, \, \, \left( 1 \right) \end{array}$

And,

$\vec c .\vec d=15$

$\begin{array}{l} \left( { 2\hat { i } -\hat { j } +4\hat { k } } \right) .\lambda \left( { 32\hat { i } -\hat { j } -14\hat { k } } \right) =15 \\ \\ \lambda \left( { 64+1-56 } \right) =15 \\ \\ \lambda =\frac { 5 }{ 3 } \end{array}$

From

$(1)$

$\vec d = \frac{5}{3}\left( {32\hat i - \hat j - 14\hat k} \right)$

Find the position vector of the midpoint of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$ .

Let the midpoint of

$PQ$ be

$R$

Position vector of

$P=(2-0)\hat i+(3-0)\hat j+(4-0)\hat k$

$\vec{OP}=2\hat i+3\hat j+4\hat k$

Position vector of

$Q=(4-0)\hat i+(1-0)\hat j+(-2-0)\hat k$

$\vec{OQ}=4\hat i+\hat j-2\hat k$

Now,

Position vector of

$R=\dfrac{\vec {OQ}+\vec {OP}}{2}$

$\vec {OR}=\dfrac{(4\hat i+\hat j -2\hat k)+(2\hat i+3\hat j+4\hat k)}{2}$

$\vec {OR}=\dfrac{(4+2)\hat i+(1+3)\hat j+(-2+4)\hat k}{2}$

$\vec {OR}=\dfrac{6\hat i+4\hat j+2\hat k}{2}$

$\vec {OR}=\dfrac{2(3\hat i+2\hat j+\hat k)}{2}$

$\vec {OR}=3\hat i+2\hat j+\hat k$

Therefore, position vector of midpoint of

$PQ$ is

$3\hat i+2\hat j+\hat k$

1029564_1144170_ans_8a11a782d64b4de1b0ad9f8d275a132e.png

If $A, B, C$ and $D$ are any four points E and F are the middle points of $AC$ and $BD$ , respectively then prove that $\vec{CB}+\vec{CD}+\vec{AD}+\vec{AB}=4\vec{EF}$ .

$E$ is midpoint of

$AC$

$F$ i midpoint of

$BD$

$AB+AD=2AF----(1)$

$CB+CD=2CF---(2)$

$\vec {AB}+\vec {AD}+\vec {CB}+\vec {CD} =2(\vec {AF}+\vec {CF})$

$\vec {AB}+\vec {AD}+\vec {CB}+\vec {CD} =-2(\vec {FA}+\vec {FC})$

on adding

$(1)$ %

$(2)$ we get

$\vec {AB}+\vec {AD}+\vec {CB}+\vec {CD}=-2 (2\vec {FE})$

$\Rightarrow \ \vec {AB}+\vec {AD}+\vec {CB}+\vec {CD}=4\vec {EF}$

Hence proved

if $\overline{a},\overline{b},\overline{c}$ are co planar
Prove that :
$[\overline {a}\ \ \ \ \overline {b}+\overline {c}\ \ \ \ \overline {a}+\overline {b}+\overline {c}]=0$

$[\bar{a}\bar{b}+\bar{c}\, \bar{a}+\bar{b}+\bar{c}]$

$(\bar{a}+\bar{b}+\bar{c}). (\bar{a}\times \bar{b}+\bar{a}\times \bar{c})$

Since

$\bar{a},\bar{b}\bar{c}$ are co planar

$\bar{a}+\bar{b}+\bar{c}$ and

$\bar{a}\times \bar{b}+\bar{a}\times\bar{c}$ are

$\perp$ to each other

$(\bar{a}+\bar{b}+\bar{c}).(\bar{a}\times \bar{b}+\bar{a}\times \bar{c}) = 0$

1239447_1144551_ans_f7863c5a8bf94f708978cf28305d0f3c.jpg

If $\vec{a} +\vec{b}+\vec{c}=\vec{0}$ , $|\vec{a}|=3 .|\vec{b}|=\vec|c|=5$ find $\vec {a}.\vec{b}+\vec{b}.\vec{c}+\vec{c}.\vec{a}$

It is given that,

$\vec a+\vec b+\vec c=0$

Squaring both sides,

$(\vec a+\vec b+\vec c)^2=0$

$|\vec a|^2+|\vec b|^2+|\vec c|^2+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=0$

$3^2+4^2+5^2+2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=0$

$2(\vec a.\vec b+\vec b.\vec c+\vec c.\vec a)=-50$

Hence,

$\vec a.\vec b+\vec b.\vec c+\vec c.\vec a=-25$

If $\vec{a}=\vec{i}+\vec{j}+2\vec{k}, \vec{b}=3\vec{i}+2\vec{j}-\vec{k}$ find $(\vec{a}+3\vec{b}).(2\vec{a}-\vec{b}).(\vec{a}+3\vec{b})$ .

$\vec{a}=\hat{i}+\hat{j}+2\hat{k}$

$\vec{b}=3\hat{i}+2\hat{j}-\hat{k}$

$(\vec{a}+3\vec{b}).(2\vec{a}-\vec{b}).(\vec{a}+3\vec{b})\Rightarrow$

$(\vec{a}+3\vec{b})=(\hat{i}+\hat{j}+2\hat{k}+9\hat{i}+6\hat{j}-3\hat{k})=10\hat{i}+7\hat{j}-\hat{k}$

$(2\vec{a}-\vec{b})=(2\hat{i}+2\hat{j}+4\hat{k}-3\hat{i}-2\hat{j}+\hat{k})=-\hat{j}+5\hat{k}$

$\Rightarrow (-10\hat{i}+7\hat{j}-\hat{k})(-\hat{j}+5\hat{k}).(10\hat{i}+7\hat{j}-\hat{k})$

$\Rightarrow (-10+5)(10\hat{i}+7\hat{j}-\hat{k})=-5(10\hat{j}+7\hat{j}-\hat{k})$

For any vector $\vec { r }$ prove that $\vec { r }=$ $\left(\vec { r }.\vec { i }\right) \vec { r }+\left(\vec { r } .\vec { j }\right) \vec { j }+\left(\vec { r }.\vec { k }\right)\vec { k }$

To prove :

$\vec{r} = (\vec{r}.\vec{c})\vec{r}+(\vec{r}.\vec{j})\vec{j}+(\vec{r}.\vec{k})\vec{k}$

Solving R.H.S

$\vec{r} = (x\hat{i}+y\hat{j}+z\hat{k})$

So,

$(\vec{r},\vec{i}) = [(x\hat{i}+y\hat{j}+z\hat{k}).\hat{i}]\hat{i}+[(x\hat{i}+y\hat{j}+z\hat{k}).\hat{j}] \hat{j}+[(x\hat{i}+y\hat{j}+z\hat{k}).\hat{k}]\hat{k}$

$=[x\hat{i},\hat{i}+y\hat{j}.\hat{i}+z\hat{k}.\hat{i}]\hat{i}+[x\hat{i}.\hat{j}+y\hat{j}.\hat{j+2\hat{k}.\hat{j}}]\hat{i}+[x\hat{i}.\hat{k}+y\hat{j}.\hat{k}+2\hat{k}.\hat{k}]\hat{k}$

$= [\because \hat{i}.\hat{i}=1=\hat{j}.\hat{j} = \hat{k}.\hat{k}$ &

$\hat{i}.\hat{j}= \hat{j},\hat{k} = \hat{k}.\hat{i} = 0]$

$= x\hat{i} +y\hat{j}+z\hat{k}$

$= r = R.H.S$

$\Rightarrow L.H.S = R.H.S$

Hence proved.

1148989_1144475_ans_41260b93e12d46e784a4418cd5de1305.jpeg

If $\bar{x},\bar{y},\bar{z}$ are mutually perpendicular vectors of eqaul magnitude, then find the measure of an angle that $\bar{x}+\bar{y}+\bar{z}$ makes with any of the three vectors.

Assuming,

$\begin{array}{l} \overline { x } \bot \overline { y } \bot \overline { z } \, \, \, \, \, \, \, and\, \, equal\, magnitude\, is\, \, \left| { \overline { x } } \right| =\, \left| { \overline { y } } \right| =\left| { \overline { z } } \right| =k \\ Lets, \\ \left( { \overline { x } +\overline { y } +\overline { z } } \right) \, \, .\, \, \overline { x } =\left| { \overline { x } +\overline { y } +\overline { z } } \right| \cdot \left| { \overline { x } } \right| .\cos \theta \\ simplify\, of, \\ \left| { \overline { x } } \right| \, .\, \overline { x } +\overline { x } .\overline { y } +\overline { x } .\, \overline { z } =\left| { \overline { x } +\overline { y } +\overline { z } } \right| \cdot \left| { \overline { x } } \right| .\cos \theta \\ { \left| { \overline { x } } \right| ^{ 2 } }+0+0=\left| { \overline { x } +\overline { y } +\overline { z } } \right| \cdot \left| { \overline { x } } \right| .\cos \theta \\ we\, know\, that,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ { k^{ 2 } }=\left| { \overline { x } +\overline { y } +\overline { z } } \right| \cdot k.\cos \theta \\ \cos \theta =\frac { k }{ { \left| { \overline { x } +\overline { y } +\overline { z } } \right| } } \\ lets\, \, squring\, of\, \left| { \overline { x } +\overline { y } +\overline { z } } \right| \end{array}$

$\begin{array}{l} { \left| { \overline { x } +\overline { y } +\overline { z } } \right| ^{ 2 } }={ \left| { \overline { x } } \right| ^{ 2 } }+{ \left| { \overline { y } } \right| ^{ 2 } }+{ \left| { \overline { z } } \right| ^{ 2 } }+2\left| { \overline { x } } \right| \left| { \overline { y } } \right| +2\left| { \overline { y } } \right| \left| { \overline { z } } \right| +2\left| { \overline { x } } \right| \left| { \overline { z } } \right| \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ={ k^{ 2 } }+{ k^{ 2 } }+{ k^{ 2 } }+0+0+0 \\ { \left| { \overline { x } +\overline { y } +\overline { z } } \right| ^{ 2 } }=\, \, \, 3{ k^{ 2 } } \\ \therefore \, \, \left| { \overline { x } +\overline { y } +\overline { z } } \right| =\sqrt { 3 } k \\ we\, \, have, \\ \cos \theta =\frac { k }{ { \left| { \overline { x } +\overline { y } +\overline { z } } \right| } } \\ \cos \theta =\frac { k }{ { \sqrt { 3 } k } } =\frac { 1 }{ { \sqrt { 3 } } } \\ \therefore \, \, \, \theta =\, { \cos ^{ -1 } }\frac { 1 }{ { \sqrt { 3 } } } \end{array}$

Let $\overrightarrow{A}=2\hat{i}+\hat{k},\overrightarrow{B}=\hat{i}+\hat{j}+\hat{k}$ and $\overrightarrow{C}=4\hat{i}-3\hat{j}+7\hat{k}$ .Determine a vector $\overrightarrow{R}$ satisfying $\overrightarrow{R}.\overrightarrow{A}=0$ and $\overrightarrow{R}\times \overrightarrow{B}=\overrightarrow{C}\times \overrightarrow{B}$

$\overrightarrow{R}\times \overrightarrow{B}=\overrightarrow{C}\times \overrightarrow{B} \Rightarrow \left(\overrightarrow{R}-\overrightarrow{C}\right)\times \overrightarrow{B}=0$

$\Rightarrow \overrightarrow{R}=\overrightarrow{C}+\lambda \overrightarrow{B}$

$\overrightarrow{R}.\overrightarrow{A} \Rightarrow \overrightarrow{C}.\overrightarrow{A}+\lambda \overrightarrow{B}.\overrightarrow{A}=0$

$\Rightarrow \left(4\hat{i}-3\hat{j}+7\hat{k}\right)\left(2\hat{i}+\hat{k}\right)+\lambda \left(\hat{i}+\hat{j}+\hat{k}\right)\left(2\hat{i}+\hat{k}\right)=0$

$\Rightarrow \left(8+7\right)+\lambda\left(2+1\right)=0$

$\Rightarrow 15+3\lambda=0$

$\Rightarrow \lambda=-5$

$\therefore \overrightarrow{R}=\overrightarrow{C}-5\overrightarrow{B}$

$=\left(4\hat{i}-3\hat{j}+7\hat{k}\right)-5\left(\hat{i}+\hat{j}+\hat{k}\right)$

$=4\hat{i}-3\hat{j}+7\hat{k}-5\hat{i}-5\hat{j}-5\hat{k}$

$=-\hat{i}-8\hat{j}+2\hat{k}$

Let $\overrightarrow{b}=4\hat{i}+3\hat{j}$ and $\overrightarrow{c}$ be two vectors perpendicular to each other in the $xy-$ plane.Find all vectors in the same plane having projections $1$ and $2$ along $\overrightarrow{b}$ and $\overrightarrow{c}$ respectively.

$\overrightarrow{c}\bot \overrightarrow{b} \Rightarrow \overrightarrow{c}=3\hat{i}-4\hat{j}$

Let

$\overrightarrow{a}=\alpha\overrightarrow{b}+\beta\overrightarrow{c}$ ---------

$\left(1\right)$

$\overrightarrow{a}.\dfrac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\pm 1$

$\Rightarrow 5\alpha=\pm 1$ or

$\alpha=\pm \dfrac{1}{5}$

$\overrightarrow{a}.\dfrac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|}=\pm 2 \Rightarrow 5\beta=\pm 2$ or

$\beta=\pm \dfrac{2}{5}$

Now

$\left(1\right) \Rightarrow \overrightarrow{a}=\pm \left(2\hat{i}-\hat{j}\right), \pm \dfrac{\left(2\hat{i}-11\hat{j}\right)} {5}$

Working:

$\overrightarrow{a}=\alpha\overrightarrow{b}+\beta\overrightarrow{c}$

$=\pm \dfrac{1}{5}\left(4\hat{i}+3\hat{j}\right)\pm \dfrac{2}{5}\left(3\hat{i}-4\hat{j}\right)$

$=\dfrac{1}{5}\left(4\hat{i}+3\hat{j}+6\hat{i}-8\hat{j}\right), \dfrac{-1}{5}\left(4\hat{i}+3\hat{j}-6\hat{i}+8\hat{j}\right)$

$=\dfrac{1}{5}\left(10\hat{i}-5\hat{j}\right), \dfrac{-1}{5}\left(-2\hat{i}+11\hat{j}\right)$

$=\left(2\hat{i}-\hat{j}\right), \left(\dfrac{2\hat{i}-11\hat{j}}{5}\right)$

if the sum of two vectors is a unit vector , prove that the magnitude of their difference $\sqrt 3$ .

Given, if the sum of teo unit vector is a unit vector

Let

$\vec{a}$ and

$\vec{b}$ are unit vectors

i.e,

$\vec{a}+\vec{b}=\vec{c}$

$\left|\vec{a}+\vec{b}\right|^{2}=\left|\vec{c}\right|^{2}=\vec{a}^{2}+2\vec{a}\vec{b}+\vec{b}^{2}$

$=1+2.\left(1\right). \left(1\right).\cos\theta+1=1$

$\cos\theta=-\dfrac{1}{2}$

Now

$\left|\vec{a}-\vec{b}\right|^{2}=\vec{a}^{2}+2\vec{a}\vec{b}+\vec{b}^{2}$

$=1-2\left|{a}\right|.\left|{b}\right|\cos\theta+1$

$=1-2.\left(\dfrac{-1}{2}\right)+1$

$=1+1+1=3$

$\vec{a}-\vec{b}=\sqrt{3}$

Find the position vector of the midpoint of the vector joining the points $P(2,2,4)$ and $Q(4,1,-2)$ .

Let

$\begin{array}{l} \overrightarrow { OP } =2\widehat { i } +2\widehat { j } +4\widehat { k } \\ \overrightarrow { OQ } =4\widehat { i } +\widehat { j } -2\widehat { k } \end{array}$

Since it is given that

$\overrightarrow R$ divides

$\overrightarrow {PQ}$ into two equal halves, we can use mid point formula here.

Mid point formula

$\begin{array}{l} \overrightarrow { OR } =\frac { { \overrightarrow { OP } +\overrightarrow { OQ } } }{ { 1+1 } } \\ =\frac { { \left( { 2\hat { i } +2\hat { j } +4\hat { k } } \right) \left( { 4\hat { i } +\hat { j } -2\hat { k } } \right) } }{ 2 } \\ =\frac { { \left( { 6\hat { i } +3\hat { j } +2\hat { k } } \right) } }{ 2 } \\ \overrightarrow { OR } =\frac { 6 }{ 2 } \hat { i } +\frac { 3 }{ 2 } \hat { j } +\frac { 2 }{ 2 } \hat { k } \end{array}$

Find the projection of $\vec a = \hat i - 3\hat k$ on $\vec b = 3\hat i + \hat j - 4\hat k$

Projection of

$\vec {a}= \hat {i}-3 \hat {k}$

$\vec {b}= 3\hat {i}+ \hat {j}-4 \hat {k}$

Projection

$=\dfrac{\vec {a}.\vec {b}}{|\vec {b}|^{2}} \vec {b}$

$=\dfrac{3+12}{(\sqrt{26})^{2}} (3 \hat {i}+ \hat {j} -4\hat {k})$

$=\dfrac{15}{26} (3\hat {i}+ \hat {j}-4\hat {k})$

$=\dfrac{45}{26} \hat {i}+ \dfrac{15\hat {i}}{26}-\dfrac{60 \hat {k}}{26}$

Find vector $\vec { c }$ such that $\vec { c } \cdot \hat { i } = \vec { c } \cdot \hat { j } = \vec { c } \cdot \hat { k }$ and $| \vec { c } | = 100$

$\begin{array}{l} \overrightarrow { C } =\overrightarrow { C } .\widehat { i } =\overrightarrow { C } .\widehat { j } =\overrightarrow { C } .\widehat { k } \, \, \, and\, \, \left| { \overrightarrow { C } } \right| =100 \\ Let\, \overrightarrow { C } =x\widehat { i } +y\widehat { j } +z\widehat { k } \\ { x^{ 2 } }+{ y^{ 2 } }+{ z^{ 2 } }=100\, \, \, \, \, \, \, \, ---\left( i \right) \\ Now, \\ \overrightarrow { C } \widehat { .i } =\left( { x\widehat { i } +y\widehat { j } +z\widehat { k } } \right) .\widehat { i } =x \\ \overrightarrow { C } .\widehat { j } =y \\ \overrightarrow { C } .\widehat { k } =z \\ \therefore x=y=z \\ Putting\, these\, value\, in\, eq{ u^{ n } }\left( i \right) \, we\, get \\ 3{ x^{ 2 } }=100 \\ { x^{ 2 } }=\frac { { 100 } }{ 3 } \\ \therefore x=\pm \frac { { 10 } }{ { \sqrt { 3 } } } \\ { { Re } }quired\, vector=\pm \frac { { 10 } }{ { \sqrt { 3 } } } \left( { \widehat { i } +\widehat { j } +\widehat { k } } \right) \end{array}$

Which is the required answer.

Find the projection of vector $2\hat i + \hat j$ on the vector $\hat i + 2\hat j$ .

Projection of

$\vec{a}=2\hat{i}+\hat{j}$ on vector

$(\hat{i}+2\hat{j})$

projection =

$\dfrac{\vec{a}.\vec{b}}{|\vec{b}|^{2}} \vec{b}$

$=\dfrac{2+2}{(\sqrt{5})}(\hat{i}+2\hat{j})$

$=\dfrac{4}{5}(\hat{i}+2\hat{j})$

$\dfrac{4\hat{i}}{5}+\dfrac{8\hat{i}}{5}$

in $\triangle ABC$ if $A=2\hat {i}+4\hat {j}-\hat {k},B=4\hat {i}+\hat {j}+\hat {k}$ and $C=3\hat {i}+6\hat {j}-3\hat {k}$ and $D$ is mid point of the side $BC$ ,then the length of $AD$ is $\sqrt m$ /2 then m=

undefined

1983963_1160069_ans_35e928c627844f8c97e4bd1724de0b40.jpg

Find unit vector in the direction of vector $a=2i+3j+k$ .

unit vector

$\hat{a} = \frac{\vec{a}}{\left | \vec{a} \right |}$

Given

$\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$

$\left | \vec{a} \right | = \sqrt{2^{2}+3^{2}+1}$

$= \sqrt{14}$

unit vector

$\hat{a} = \frac{2\hat{i}+3\hat{j}+\hat{k}}{\sqrt{14}}$

$= \frac{2}{\sqrt{14}}\hat{i}+\frac{3}{\sqrt{14}}\hat{j}+\frac{1}{\sqrt{14}}\hat{k}$

zero vector

$\vec{0} = 0\hat{i}+0\hat{j}+0\hat{k}$

1164881_1183203_ans_947a14ba01f942f6b456775f5c66ca7b.jpg

$\vec{a}=4\hat{i}-\hat{j}+\hat{k} \quad and \quad \vec{b}=p\hat{i}+2\hat{j}+3\hat{k}$ are mutually perpendicular , then find the value of p.

$\vec{a}=4\hat{i}-\hat{j}+\hat{k}$ and

$\vec{b}=p\hat{i}+2\hat{j}+3\hat{k}$ are initially perpendicular

$\Rightarrow \vec{a}\cdot \vec{b}=0$

$\Rightarrow 4p-2+3=0$

$\Rightarrow 4p+1=0$

$\Rightarrow p=-\dfrac{1}{4}$ .

1202100_1196408_ans_d64bd3c06ca34fe58e47ee6ee9ebb8b2.jpg

If $\left| \overline { a } \right| =2,\left| \overline { b } \right| =7$ and $\overline { a } \times \overline { b } =3\overline { i } +2\overline { j } +6\overline { k }$ then $\overline { a }.\overline { b }=$

$\overrightarrow { a } \times \overrightarrow { b } =\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| sin\theta$

$\left| \overrightarrow { a } \times \overrightarrow { b } = \right| \left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| \left| sin\theta \right|$

$\left| 3 \hat i+2 ]hat j+ 6 \hat K \right|=2\times7 \left| sin \theta \right|$

$\sqrt { 3^2+2^2+6^2 } =14\left| sin\ theta \right|$

$\sqrt { 49 }=14 \left| sin\ theta \right|$

$\frac { 7 }{ 14 } =sin\theta$

$\frac { 1}{ 2} =sin\theta =sin\theta$

$\theta =30^o$

$\overrightarrow { a } .\overrightarrow { b } =\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| cos\theta |$

$=2\times7\times\times cos30$

$=2\times7\times\frac { \sqrt { 3 } }{ 2 }$

$=7\sqrt { 3 }$

$=\sqrt {49\times3}$

$=\sqrt {147}$

In a $\Delta A B O , E$ is the midpoint of $\overline { O B }$ and $D$ is a point on $AB$ such that $A D : D B = 2:1$ if $\overline { O D }$ and $\overline { A E }$ intersect at $P$ then $\overline { O P } : \overline { P D } =$

With

$O$ as origin let

$a$ and

$b$ the position vectors of

$A$ and

$B$ respectively.

Then the positoion vector of

$E$ , the mid-point of

$OB$ , is

$\dfrac{b}{2}$

Again, since

$AD:DB=2:1$ , the position vector of

$D$ is

$\dfrac {1.a+2b}{1+2}=\dfrac{a+2b}{3}$

Equation of

$OD$ and

$AE$ are

$r=t\dfrac{a+2b}{3}...(1)$ and

$r=a+s\left(\dfrac{b}{2}-a\right)$ or

$r=(1-s)a+s\dfrac{b}{2}...(2)$

If the intersect at

$P$ , then we will have identical values of

$r$ .

Hence companing the coefficient of

$a$ and

$b$ , we get

$\dfrac{t}{3}=1-s,\dfrac{2t}{3}=\dfrac{s}{2}\therefore t=\dfrac{3}{5}$ or

$s=\dfrac{4}{5}$ .

Putting for

$t$ in

$(1)$ or for

$S$ in

$(2)$ , we get the position vector of position of intersection

$p$ as

$\dfrac{a+2b}{5}..(3)$

now let

$p$ divide

$OD$ in the ratio

$\lambda : 1$ .

Hence by ratio formula the

$P.V$ of

$P$ is

$\dfrac { \dfrac { \lambda \left( a+2b \right) }{ 3 } +1.0 }{ \lambda +1 } =\dfrac { \lambda }{ 3\left( \lambda +1 \right) } \left( a+2b \right) ... (4)$

Comparing

$(3)$ and

$(4)$ , we get

$\dfrac { \lambda }{ 3\left( \lambda +1 \right) } =\dfrac { 1 }{ 5 } \Rightarrow 5\lambda =3\lambda +3\Rightarrow 2\lambda =3\Rightarrow \lambda =\dfrac { 3 }{ 2 }$

$\therefore OP:PD=3:2$

If the vector $- \hat i + \hat j - \hat k$ bisects the angle between the vector c and the vector $3\hat i + 4\hat j$ ,then the unit vector in the direction of c,is

$\begin{array}{l} \overrightarrow { a } =3\hat { i } +4\hat { j } \\ \overrightarrow { c } \, \, { { bisect } } \\ \overrightarrow { b } =-\hat { i } +\hat { j } -\hat { k } \end{array}$

So, we can write

$\begin{array}{l} \overrightarrow { a } +\overrightarrow { c } =\lambda \overrightarrow { b } & \\ \hat { a } +\hat { c } =\lambda \hat { b } & \left[ { \hat { a } =\frac { { 3\hat { i } +4\hat { j } } }{ { \sqrt { 9+16 } } } =\frac { { 3\hat { i } +4\hat { j } } }{ { \sqrt { 15 } } } =\frac { { 3\hat { i } +4\hat { j } } }{ 5 } } \right] \\ \frac { { 3i } }{ 5 } +\frac { { 4j } }{ 5 } +\hat { c } =\lambda \left( { \frac { { -\hat { i } +\hat { j } -\hat { k } } }{ { \sqrt { 3 } } } } \right) \to \left( i \right) & \end{array}$

Let the angle between a & b is

$\theta$

$\begin{array}{l} \overrightarrow { a } .\overrightarrow { b } =\left| { \overrightarrow { a } } \right| \left| { \overrightarrow { b } } \right| \cos \theta \Rightarrow \cos \theta =\frac { { \overrightarrow { a } .\overrightarrow { b } } }{ { \left| { \overrightarrow { a } } \right| \left| { \overrightarrow { b } } \right| } } =\frac { { -3+4 } }{ { 5\sqrt { 3 } } } =\frac { 1 }{ { 5\sqrt { 3 } } } \\ \overrightarrow { a } +\overrightarrow { c } =\lambda \overrightarrow { b } \\ \overrightarrow { a } .\overrightarrow { b } +\overrightarrow { c } .\overrightarrow { b } =\lambda \left( { { { \left| { \overrightarrow { b } } \right| }^{ 2 } } } \right) \\ \left| \sigma \right| \left| { \overrightarrow { a } } \right| \cos \theta +\left| { \overrightarrow { c } } \right| \left| { \overrightarrow { b } } \right| \cos \theta =3\lambda \\ 1\times \sqrt { 3 } \times \frac { 1 }{ { 5\sqrt { 3 } } } +1.\sqrt { 3 } \times \frac { 1 }{ { 5\sqrt { 3 } } } =3\lambda \\ \frac { 1 }{ 5 } +\frac { 1 }{ 5 } =3\lambda \Rightarrow \frac { 2 }{ 5 } =3\lambda \Rightarrow \frac { 2 }{ { 15 } } =\lambda \\ \frac { { 3\hat { i } } }{ 5 } +\frac { { 4\hat { j } } }{ 5 } +\hat { c } =\frac { 2 }{ { 15 } } \left( { -\hat { i } +\hat { j } -\hat { k } } \right) \\ \overrightarrow { c } =\frac { { -2 } }{ { 15 } } \hat { i } +\frac { 2 }{ { 15 } } \hat { j } -\frac { 2 }{ { 15 } } \hat { k } -\frac { { 3\hat { i } } }{ 5 } -\frac { { 4\hat { j } } }{ 5 } \\ \overrightarrow { c } =\frac { { -11 } }{ { 15 } } \hat { i } -\frac { { 10 } }{ { 15 } } \hat { j } -\frac { 2 }{ { 15 } } \hat { k } \\ \overrightarrow { c } =\frac { { -1 } }{ { 15 } } \left( { 11\hat { i } +10\hat { j } +2\hat { k } } \right) \end{array}$

$A ( 1 , - 1 , - 3 ) , B ( 2,1 , - 2 ) \quad \& \quad C ( - 5,2 , - 6 )$ are the position vectors of the vertices of a triangle $ABC$ , then the length of the bisector of its internal angle $A$ is

$\begin{array}{l} Given,\, we\, have\, three\, vector\, position: \\ A(1,-1,-3),B(2,1,-2)\& C(-5,2,-6) \\ distance\, of\, AB=\sqrt { { { (1-2) }^{ 2 } }+{ { (-1-1) }^{ 2 } }+{ { (-3-(-2)) }^{ 2 } } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { distance\, formula:\, \sqrt { { { ({ x_{ 2 } }-{ x_{ 1 } }) }^{ 2 } }+{ { ({ y_{ 2 } }-{ y_{ 1 } }) }^{ 2 } }+{ { ({ z_{ 2 } }-{ z_{ 1 } }) }^{ 2 } } } } \right. \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt { 1+4+1 } =\sqrt { 6 } \, \, \, \, \, \, \, \, \, \, \, \\ And, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, AC=\sqrt { { { (1-(-5)) }^{ 2 } }+{ { (-1-2) }^{ 2 } }+{ { (-3-(-6)) }^{ 2 } } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt { 36+9+9 } =\sqrt { 54 } =3\sqrt { 6 } \\ Now, \\ \Rightarrow \frac { { AB } }{ { AC } } =\frac { { BD } }{ { DC } } =\frac { { \sqrt { 6 } \, } }{ { 3\sqrt { 6 } } } =\frac { 1 }{ 3 } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \\ \Rightarrow \frac { { BD } }{ { DC } } =\frac { 1 }{ 3 } \, \, =\frac { m }{ n } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, (where\, \, m:1\, ,\, \, n:3) \\ Now,\, u{ { sing } }\, sector\, formula\, :E=\frac { { m{ x_{ 2 } }+n{ x_{ 1 } } } }{ { m+n } } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E=\frac { { (1\times (-5)+(3\times 2)) } }{ { 1+3 } } ,\frac { { (1\times 2+3\times 1) } }{ { 1+3 } } ,\frac { { (1\times -6+3\times -2) } }{ { 1+3 } } \\ \, \, \, \, \, \, \, \, \, \, \therefore \, \, \, \, \, E=\frac { 1 }{ 4 } ,\frac { 5 }{ 4 } ,-3 \\ \, \, \, \, We know,\, \, co-ordinates\, \, \, of\, A(1,-1,-3)\, \\ find\, distance\, of\, AE: \\ AE=\sqrt { { { (1-\frac { 1 }{ 4 } ) }^{ 2 } }+{ { (-1-\frac { 5 }{ 4 } ) }^{ 2 } }+{ { (-3-(-3)) }^{ 2 } } } \\ \, \, \, \, \, \, \, \, \, \, =\sqrt { \frac { 9 }{ { 16 } } +\frac { { 81 } }{ { 16 } } +0 } =\sqrt { \frac { { 90 } }{ { 16 } } } =\frac { { 3\sqrt { 10 } } }{ 4 } \\ So,\, length\, of\, { { int } }ernal\, angle\, \, of\, A\, \, is\, AE=\frac { { 3\sqrt { 10 } } }{ 4 } \, .\, \, \, Answer \end{array}$

A girl walks $4\ km$ towards west, then she walks $3\ k$ in the direction of $30$ east of north and stops. Determine the girl's displacement from her initial point of departure.

$\begin{array}{l} BAO={ 90^{ 0 } }-{ 30^{ 0 } }={ 60^{ 0 } } & \\ \sin { 60^{ 0 } } =\frac { { BD } }{ { 3kn } } & \\ \frac { { \sqrt { 3 } } }{ 2 } =\frac { { BD } }{ 3 } & \\ BD=\frac { { 3\sqrt { 3 } } }{ 2 } & \\ \cos { 60^{ 0 } } =\frac { { AD } }{ 3 } & OD=OA-AD \\ AD=\frac { 3 }{ 2 } & =4\frac { { -3 } }{ 2 } =\frac { 5 }{ 2 } \\ O{ B^{ 2 } }=B{ D^{ 2 } }+O{ D^{ 2 } } & \\ ={ \left( { \frac { { 3\sqrt { 3 } } }{ 2 } } \right) ^{ 2 } }+{ \left( { \frac { 5 }{ 2 } } \right) ^{ 2 } } & \\ =\frac { { 9\times 3 } }{ 4 } +\frac { { 25 } }{ 4 } =\frac { { 52 } }{ 4 } =13 & \\ OB=\sqrt { 13 } & \end{array}$
1219678_1214012_ans_062dc602d5934ad2b1d42cc71adaf682.PNG

1219678_1214012_ans_062dc602d5934ad2b1d42cc71adaf682.PNG

If A , B , C and D are any four points E and F are the middle points of AC and BD, respectively then prove that $\underset{CB}{\rightarrow}+\underset{CD}{\rightarrow}+\underset{AD}{\rightarrow}+\underset{AB}{\rightarrow}=4 \underset{EF}{\rightarrow}$

$\vec {AB}+\vec {AD}+\vec {CB}=4\vec {CF}$

$AB+AD=2AF----(1)$

$CB+CD=2CF---(2)$

$AB+AD+CB+CD=(AF+CF)$

$AB+AD+CB+CD=2(\vec {FA}+\vec {CF})$

$AB+AD+CB+CD=-2(2FC)$

$AB+AD+CB+CD=4EF$

1353985_1204248_ans_15bc1f1291b14300bb85100870d1a368.png

Let $\vec{a}$ be an unit vector and $m$ being a scalar then find the value of $m$ from the equation $|m\vec{a}|=2$ .

Given,

$|m\vec{a}|=2$ .

or,

$|m|.|\vec{a}|=2$

or,

$|m|.1=2$ [ Since

$\vec{a}$ is given to be unit vector]

or,

$m=\pm 2$ .

Find by vector method the perimeter of the triangle whose vectors are $( - 1 , - 1,9 ) , ( 3,1,5 )$ and $( 0 , - 5,1 ) .$

$AB=(1-,1-,9)-(3,1,5)=(-4,-2,4)\\ \left| AB \right| =\sqrt { { 4 }^{ 2 }+{ 2 }^{ 2 }+{ 4 }^{ 2 } } =6\\ BC=(3,1,5)-(0,-5,1)=(3,6,4)\\ \left| BC \right| =\sqrt { 9+36+16 } =\sqrt { 61 } \\ CA=(0,-5,1)-(-1,-1,9)=(1,-4,-8)\\ \left| CA \right| =\sqrt { { 1 }^{ 2 }+{ 4 }^{ 2 }+{ 8 }^{ 2 } } =\sqrt { 64+16+1 } =9$

Perimeter

$=6+\sqrt { 61 } +9\\ =15+7.81\\ =22.81$

1186589_1222466_ans_c9373b8d6ccb4ad0afdda857edbf2246.png

1.Find the projection of the vector i-j on the vector i+j.
2.Find the projection of the vector $2\widehat i + 3\widehat j + 2\widehat k$ on the vector
$\widehat i + 2\widehat j + \widehat k.$ .
3.Find the projection of $\overrightarrow a = \widehat i + 3\widehat j + \widehat k\,along\,\overrightarrow b = 2\widehat i - 3\widehat j + 6\widehat k.$

$1)$ Projection of vector

$\overrightarrow { a } \quad on\quad \overrightarrow { b }$ is given by

$\cfrac { \overrightarrow { a } .\overrightarrow { b } }{ \left| \overrightarrow { b } \right| }$

$\cfrac { \left( \hat { i } -\hat { j } \right) .\left( \hat { i } -\hat { j } \right) }{ \left| \hat { i } +\hat { j } \right| } =\cfrac { 1-1 }{ \sqrt { 2 } } =0$

$2) \cfrac { \left( 2\hat { i } -3\hat { j } +2\hat { k } \right) .\left( \hat { i } +2\hat { j } +\hat { k } \right) }{ \left| \hat { i } +2\hat { j } +\hat { k } \right| } =\cfrac { 2+6+2 }{ \sqrt { 1+4+1 } } =\cfrac { 10 }{ \sqrt { 16 } }$

$3)\cfrac { \overrightarrow { a } .\overrightarrow { b } }{ \left| \overrightarrow { b } \right| } =\cfrac { \left( \hat { i } +3\hat { j } +2\hat { k } \right) .\left( 2\hat { i } -3\hat { j } +6\hat { k } \right) }{ \left| 2\hat { i } -3\hat { j } +\hat { 6k } \right| } =\cfrac { 2-9+6 }{ \sqrt { 4+9+36 } } =\cfrac { -1 }{ \sqrt { 49 } } =\cfrac { -1 }{ 7 }$

Find the position vectors of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $2\vec{a}+3\vec{b}$ and $\vec{a}-3\vec{b}$ respectively, externally in the ratio $1:2$ . Also, show that $P$ is the midpoint of the line segment $R$ .

R.E.F image

$\vec{R} = \dfrac{2\times \vec{P}-1\times \vec{Q}}{2-1}$

$= 4\vec{a}+6\vec{b}-\vec{a}+3\vec{b} = 3\vec{a}+9\vec{b}$

Mid-point of

$Q\vec{R} = \dfrac{\vec{Q}+\vec{R}}{2} = \dfrac{(\vec{a}-3\vec{b}+(3\vec{a}+9\vec{b}))}{2}$

$= 2\vec{a}+3\vec{b} = \vec{P}$

For what value of $\lambda ,$ the vector $i - \lambda j + 2k$ and $8i + 6j - k$ are at right angles?

$i - \lambda j + 2k$ and

$8i + 6j - k$ are right angles

so dot product should be zero

$1\times 8 + (-\lambda) \times 6 + 2 (-1) = 0$

$8 - 6\lambda - 2 = 0$

$\lambda = 1$ .

Using vector method, if $Q$ is the point of concurrence of the medians of the triangle $ABC$ ,then prove that $\vec {QA} + \vec{QB}+\vec{QC} = \vec { 0 }$

Let

$\vec{a},\vec{b},\vec{c}$ be the position vector of vertices

A,B.C respectively. then position vector of

centroid Q is

$\dfrac{\vec{a}+\vec{b}+\vec{c}}{3}$

LHS

$= \overrightarrow{QA}+\overrightarrow{QB}+\overrightarrow{QC}$

$=(\overrightarrow{QA}-\overrightarrow{OQ})+(\overrightarrow{OB}-\overrightarrow{OQ})+(\overrightarrow{OC}-\overrightarrow{OQ})$

$= \overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}-3\overrightarrow{OQ}$

$= \vec{a}+\vec{b}+\vec{c}-3 \left(\dfrac{\vec{a}+\vec{b}+\vec{c}}{3}\right)$

$= \vec{a}+\vec{b}+\vec{c}-\vec{a}-\vec{b}-\vec{c} = \vec{0}$

If $\vec { DA } =\vec { a }$ , $\vec { AB } =\vec { b }$ , $\vec { CB } =\vec { ka }$ where $k>0$ X, Y are the mid points of DB and AC respectively, such that $\left| \vec { a } \right| =17$ and $\left| \vec { XY } \right| =4$ , then k equal to ?

Let D be origin

$\vec{DA}=\vec{A}-\vec{D}=\vec{a}$

$\Rightarrow \vec{A}=\vec{a}$

$\vec{AB}=\vec{B}-\vec{A}=\vec{b}$

$\Rightarrow \vec{B}=\vec{a}+\vec{b}$

$\vec{CB}=\vec{B}-\vec{C}=k\vec{a}$

$\Rightarrow \vec{C}=\vec{a}+\vec{b}-k\vec{a}$

$\vec{x}=\dfrac{\vec{a}+\vec{b}}{2}$ ,

$\vec{y}=\dfrac{2\vec{a}+\vec{b}-k\vec{a}}{2}$

$\vec{xy}=\dfrac{\vec{a}-k\vec{a}}{2}=\left(\dfrac{1-k}{2}\right)\vec{a}$

$(\vec{xy})=\left(\dfrac{1-k}{2}\right)|\vec{a}|$

$\Rightarrow 4=\left(\dfrac{1-k}{2}\right)17$

$\Rightarrow 1-k=\dfrac{8}{17}$

$\Rightarrow k=\dfrac{9}{17}$ .

1134149_1242218_ans_a330f66910cc409e997e446c2ec04c10.jpg

Get the area of $\triangle ABC$ for $A(1,1,2),B(2,3,5),C(1,3,4)$ by use of vectors.

$\begin{array}{l} Area\, of\, \Delta ABC=\frac { 1 }{ 2 } \left| { \overrightarrow { AB } \times \overrightarrow { AC } } \right| \\ \overrightarrow { AB } =\left( { 1,2,3 } \right) \\ \overrightarrow { AB } \times \overrightarrow { AC } \left| { \begin{array} { *{ 20 }{ c } }{ \hat { i } } & { \hat { j } } & { \hat { k } } \\ 1 & 2 & 3 \\ 0 & 2 & 2 \end{array} } \right| =\hat { i } \left( { -2 } \right) -\hat { j } \left( 2 \right) +\hat { k } \left( { 2-0 } \right) \\ =-2\hat { i } -2\hat { j } +2\hat { k } \\ Area\, =\frac { 1 }{ 2 } \sqrt { 4+4+4 } \\ =\frac { { \sqrt { 12 } } }{ 2 } \\ =\frac { { \sqrt { 12 } } }{ 2 } \\ =\frac { { 2\sqrt { 3 } } }{ 2 } =\sqrt { 3 } . \end{array}$

Let $\vec {a},\vec {b},\vec {c}$ be three non-zero vectors such that $\vec {c}$ is a unit vectors perpendicular is both $\vec {a}$ and $\vec {b}$ . If the angle between $\vec {a}$ and $\vec {b}$ is $\dfrac {\pi}{6}$ , prove that $[ \vec {a}\ \vec {b}\ \vec {c} ]^{2}=\dfrac {1}{4}|\vec {a}|^{2}|\vec {b}|^{2}$ .

Since,

$\vec { c } =\lambda \left( { \vec { a } \times \vec { b } } \right) \\ \left| { \vec { c } } \right| =1\, \, \, \, \, \,$

Now,

$\\ { \left[ { \vec { a } \, \, \, \vec { b } \, \, \, \vec { c } } \right] ^{ 2 } }={ \left| { \vec { c } .\left( { \vec { a } \times \vec { b } } \right) } \right| ^{ 2 } } \\ ={ \left[ { \left| { \vec { c } } \right| \left| { \vec { a } } \right| \left| { \vec { b } } \right| \sin \dfrac { \pi }{ 6 } \cos { 0^{ \circ } } } \right] ^{ 2 } } \\ =\dfrac { { { { \left| { \vec { a } } \right| }^{ 2 } }{ { \left| { \vec { b } } \right| }^{ 2 } } } }{ 4 } \\$

$Hence,\, proved.$

A man walks $2a-b\ km$ due North from a fixed point $O$ and then walks a distance of $3a+2b\ km$ due South. What is his final position with regard to $O$ .

R.E.F image

On walking north, he reaches

$(2a-b)$ units due north to

$A$ .

He moves

$3a + 2b$ due south from

$A$ .

So, subtract

$3a + 2b$ from

$2a -b$ .

$\therefore 2a - b - 3a - 2b = -a - 3b$

$\therefore$ He is

$(a+3b)$ distance due south of O.

1209910_1284377_ans_57155c87948a4fce8dbf8be53bb0ac6b.jpg

Prove that
$\vec ({ a } \vec { b })\times (\vec { c } \vec { d }) = [\vec ({ a } \vec { b } \vec { d }] \vec [{ a } \vec { b } \vec { c }] \vec { d }$

$(\vec{a}\times\vec{b})\times (\vec{c}\times \vec{d})=[(\vec{a}\times \vec{b})\cdot \vec{d}]\vec{c}-[(\vec{a}\times \vec{b})\cdot \vec{c}]\vec{d}=[\vec{a}\vec{b}\vec{d}]\vec{c}-[\vec{a}\vec{b}\vec{c}]\vec{d}$ .
1181982_1289720_ans_6404a2fe258c493ea630386d158cc191.jpg

1181982_1289720_ans_6404a2fe258c493ea630386d158cc191.jpg

If A, B, C, D are the points with position vectors $\hat{i} - \hat{j} + \hat{k}, 2 \hat{i} - \hat{j} + 3 \hat{k}, 2 \hat{i} - 3 \hat{k}, 3 \hat{i} - 2 hat{j} + \hat{k}$ , respectively, find the projection of $\overline{AB}$ along $\overline{CD}$

$\textbf{Step 1: Projection of a vector on some other vector.}$

$\text{Here, }$

$\overline {OA}=\widehat i+\widehat j-\widehat k,\ \overline {OB}=2\widehat i-\widehat j+3\widehat k,\ \overline {OC}=2\widehat i-3\widehat k$ and

$\overline {OD}=3 i-2\widehat i+\widehat k$

$[\textbf{Given}]$

$\therefore \overline {AB}=\overline {OB}-\overline {OA}=(2-1) \widehat i+(-1-1)\widehat j+(3+1)\widehat k$

$=\widehat i-2\widehat j+4\widehat k$

$....(1)$

$\text{And }$

$\overline {CD}=\overline {OD}-\overline {OC}=(3-2) \widehat i+(-2-0)\widehat j+(1+3)\widehat k$

$=\widehat i-2\widehat j+4\widehat k$

$....(2)$

$\text{ So, the projection of }$

$\overline {AB}$

$\text{ along }$

$\overline {CD}=\overline {AB}\cdot \dfrac{\overline {CD}}{|\overline {CD}|}$

$=\dfrac{(\widehat i-2\widehat j+4\widehat k)(\widehat i-2\widehat j+4\widehat k)}{\sqrt{1^2+2^2+4^2}}$

$=\dfrac{1+4+16}{\sqrt{21}}=\dfrac{21}{\sqrt{21}}$

$=\sqrt{21}$ units.

$\textbf{Hence, Projection of }\boldsymbol{\overline{AB}} \textbf{ along }\boldsymbol{\overline{CD} \textbf{ is }\boldsymbol{\sqrt{21}}.}$

If $\vec {a}=\hat {i}+\hat {j}+2\hat {k}$ and $\vec {b}=3\hat {i}+2\hat {j}-\hat {k}$ , find the value of $(\vec {a}+3\vec {b}).(2\vec {a}-\vec {b})$

$\vec{a} = \hat{i} + \hat{j} + 2 \hat{k}$

$\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$

$2 \vec{a} = 2 \hat{i} + 2 \hat{j} + 4 \hat{k}$

$2 \vec{a} - \vec{b} = 2 \hat{i} + 2 \hat{j} + 4 \hat{k} - \left( 3 \hat{i} + 2 \hat{j} - \hat{k} \right) = - \hat{i} + 0 \hat{j} + 5 \hat{k}$

$3b = 9 \hat{i} + 6 \hat{j} - 3 \hat{k}$

$\vec{a} + 3 \vec{b} = \hat{i} + \hat{j} + 2 \hat{k} + 9 \hat{i} + 6 \hat{j} - 3 \hat{k} = 10 \hat{i} + 7 \hat{j} -\hat{k}$

Therefore,

$\left( \vec{a} + 3 \vec{b} \right)\left( 2 \vec{a} - \vec{b} \right)$

$= \left( - \hat{i} + 0 \hat{j} + 5 \hat{k} \right) \cdot \left( 10 \hat{i} + 7 \hat{j} - \hat{k} \right)$

$= -10 + 0 - 5 = -15$

If P and Q are the midpoints of the diagonals AC and BD respectively of a quadrilateral ABCD, prove that $\overline { AB } +\overline { AD } +\overline { CB } +\overline { CD } =4\overline { PQ }$ .

We have,

P and Q are the midpoints of diagonal AC and BD of quadrilateral.

Then,

Prove that,

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{PQ}$

Proof:-

Since Q is the mid point of diagonals AC

Then,

$\overrightarrow{AB}+\overrightarrow{AD}=2\overrightarrow{AQ}\,\,......\,\,\left( 1 \right)$

Similarly

$\overrightarrow{CB}+\overrightarrow{CD}=2\overrightarrow{CQ}\,\,......\,\,\left( 2 \right)$

On adding (1) and (2) to, we get,

$\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=2\overrightarrow{AQ}+2\overrightarrow{CQ}$

$\Rightarrow \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=2\left( \overrightarrow{AQ}+\overrightarrow{CQ} \right)$

$\Rightarrow \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=-2\left( \overrightarrow{QA}+\overrightarrow{QC} \right)$

$\Rightarrow \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=-2\left( 2\overrightarrow{QP} \right)$

$\Rightarrow \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=-2\left( -2\overrightarrow{PQ} \right)$

$\Rightarrow \overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{PQ}$

Hence proved.

Hence, this is the answer.

Show that vectors $2\hat{i}-\hat{k},\hat{i}-3\hat{j}-5\hat{k}$ and $3\hat{i}-4\hat{j}-4\hat{k}$ form the vertices of the triangle.

We have,

Given three vectors

Let,

$\overrightarrow{OA}=2\widehat{i}-\widehat{k}$

$\overrightarrow{OB}=\widehat{i}-3\widehat{j}-5\widehat{k}$

$\overrightarrow{OC}=3\widehat{i}-4\widehat{j}-4\widehat{k}$

So,

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

$\overrightarrow{AB}=\left( \widehat{i}-3\widehat{j}-5\widehat{k} \right)-\left( 2\widehat{i}-\widehat{k} \right)$

$\overrightarrow{AB}=-\widehat{i}-3\widehat{j}-4\widehat{k}$

$\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}$

$\overrightarrow{BC}=\left( 3\widehat{i}-4\widehat{j}-4\widehat{k} \right)-\left( \widehat{i}-3\widehat{j}-5\widehat{k} \right)$

$\overrightarrow{BC}=2\widehat{i}-\widehat{j}+\widehat{k}$

$\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}$

$\overrightarrow{CA}=\left( 2\widehat{i}-\widehat{k} \right)-\left( 3\widehat{i}-4\widehat{j}-4\widehat{k} \right)$

$\overrightarrow{CA}=-\widehat{i}+4\widehat{j}+3\widehat{k}$

Now,

$\left| \overrightarrow{AB} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{\left( -4 \right)}^{2}}}$

$\left| \overrightarrow{AB} \right|=\sqrt{26}$

$\left| \overrightarrow{BC} \right|=\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}=\sqrt{6}$

$$\begin{align}

$\left| \overrightarrow{CA} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{4}^{2}}+{{3}^{2}}}$

$\left| \overrightarrow{CA} \right|=\sqrt{1+16+9}=\sqrt{26}$

Now,

$AB=CA$

Hence, it is a isosceles triangle.

Let $u=(2,4,-5)$ and $v=(1,-6,9)$ then find $3u-5v$ .

Given,

$u=(2,4,-5)$ and

$v=(1,-6,9)$ .

Now,

$3u-5v$

$=3(2,4,-5)-5(1,-6,9)$

$=(1,42,-60)$ .

Show that the points $A(2, 1, -1)$ , $B(0, -1, 0)$ , $C(4, 0, 4)$ and $D(2, 0, 1)$ are coplanar.

$\vec{AB}=(-2,-2,1)$

$\vec{BC}=(4,1,4)$

$\vec{CD}=(-2,0,-3)$

$\implies \triangle =\begin{vmatrix}-2&-2&1\\4&1&4\\-2&0&-3\end{vmatrix}=0$

$\implies A,B,C,D$ are co-planar.

The two adjacent sides of a parallelogram, $\vec{a}=2i+4j-5k$ and $\vec{b}=i+2j+3k$ , then find the unit vector parallel to its diagonal.

Let

$\vec{a}$ and

$\vec{b}$ are adjacent side of a parallelogram, where

$\vec{a}=2\hat{i}+4\hat{j}−5\hat{k}$

and

$\vec{b}=\hat{i}+2\hat{j}+3\hat{k}$

Let diagonal be

$\vec{c}$

Hence

$\vec{c}=\vec{a}+\vec{b}$

$=2\hat{i}+4\hat{j}−5\hat{k}+\hat{i}+2\hat{j}+3\hat{k}=3\hat{i}+6\hat{j}-2\hat{k}$

magnitude of

$\vec{c}=\sqrt{{3}^{2}+{6}^{2}+{2}^{2}}=\sqrt{9+36+4}=7$

Unit vector in direction of

$\vec{c}=\dfrac{\vec{c}}{\left|\vec{c}\right|}=\dfrac{1}{7}\left(3\hat{i}+6\hat{j}-2\hat{k}\right)$

Let $u=(1,-2,3)$ and $v=(4,5,-1)$ find $u.v$ .

Given,

$u=(1,-2,3)$ and

$v=(4,5,-1)$ .

Now,

$u.v$

$=4-10-3$

$=-9$ >

Find the angle between two vectors $\overrightarrow a$ and $\overrightarrow b$ with magnitude $\sqrt{3}$ and $2$ , respectively having $\overrightarrow a . \overrightarrow b=\sqrt{6}$ .

$\overrightarrow a . \overrightarrow b = |a||b| \cos \theta$

$\sqrt{6}=\sqrt{3}\times 2 \times \cos \theta$

$\cos \theta = \dfrac{1}{\sqrt{2}}$

$\theta = \dfrac{\pi}{4}$

Using the vector equation of the straight line passing through two points, prove that the points whose vectors are $a, b$ and $(3a-2b)$ are collinear.

Let

$A,B$ and

$C$ are the points whose position vectors are

$\vec{a}$ ,

$\vec{b}$ and

$\vec{c}$

$\vec{AB}=\vec{b}-\vec{a}$

$\vec{BC}=3\vec{a}-2\vec{b}-\vec{a}=-2\left(\vec{b}-\vec{a}\right)$

$\Rightarrow \vec{AC}=-2\vec{AB}$

Hence

$\vec{a}$ ,

$\vec{b}$ and

$\vec{c}$ are collinear.

If vector $\vec {AB}=2\hat {i}-\hat {j}+\hat {k}$ and $\vec {OB}=3\hat {i}-4\hat {j}+4\hat {k}$ find the position vector $\vec {OA}$

$\vec{AB}=2\hat{i}-\hat{j}+\hat{k}=\vec{b}-\vec{a}$ ….

$(1)$ ;

$\vec{OA}=3\hat{i}-4\hat{j}+4\hat{k}=\vec{a}-\vec{0}$ …………

$(2)$

Equation

$(2)$ - Equation

$(1)$

$\vec{a}-\vec{0}=\vec{OA}=\hat{i}-3\hat{j}+3\hat{k}$ .

1214471_1396112_ans_7d999d658a50437fa4e878969fc7bc0b.jpg

If vector $\vec{AB}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{OB}=3\hat{i}-4\hat{j}+4\hat{k}$ , find the position vector $\vec{OA}$

Let position vector of

$O, A, B$ be

$\vec{O}, \vec {a}, \vec{b}$

$\Rightarrow \overrightarrow{AB} - \vec{b} - \vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ ...(1)

$\overrightarrow{OB} = \vec{b} - \vec{O} = 3\hat{j}-4\hat{j} + 4\hat{k}$ ...(2)

$\overrightarrow{OA} = \vec{a} - \vec{O}$

Eqn. (2) - eqn. (1)

$overline{OA} = \hat{i} - 3\hat{j} + 3\hat{k}$

1227001_1398656_ans_a31f5b22aefe497f8dce880860ce50fd.jpg

Write the position vector of the point which divides the join of points with position vectors $3 \vec { a } - 2 \vec { b }$ and $2 \vec { a } + 3 \vec { b }$ in the ratio $2 : 1.$

Given position vectors :

$\vec{3a} - \vec{2b}$ and

$\vec{2a} + \vec{3b}$

Given ratio :

$2 : 1$

Since, when a point R divides a line segment having end points P and Q in the ratio

$m : n$ and the position vector of P is

$\vec{OP}$ and the position vector of

$Q$ is

$\vec{OQ}$ , then the position vector of

$R$ is given by,

$\vec{OR} = \dfrac{m(\vec{OQ}) + n(\vec{OP)}}{m + n}$

Here,

$\vec{OP} = \vec{3a} - \vec{2b}$

$\vec{OQ} = \vec{2a} + \vec{3b}$

$m = 2 \ \& \ n = 1$

$\Rightarrow \vec{OR} = \dfrac{2 (\vec{2a} + \vec{3b}) + 1 (\vec{3a} - \vec{2b})}{2+1} = \dfrac{4\vec{a} + 6 \vec{b} + 3 \vec{a} - 2 \vec{b}}{3}$

$\Rightarrow \boxed{\vec{OR} = \dfrac{7\vec{a} + 4 \vec{b}}{3}}$

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat { i } +2\hat { j } -\hat { k }$ and $\hat { -i } +\hat { j } +\hat { k }$ respectively, in the ratio 2 :externally

Let position vectors of points

$P,Q$ and

$R$ from origin

$O$ are

$\vec {OP},\vec {OQ}$ and

$\vec {OR}$ respectively.
According to question,

$\vec {PO}=\hat { i } +2\hat { j } -\hat { k }$
and

$\vec{OQ}=\hat { -i } +\hat { j } +\hat { k }$
Wehn point

$R$ divides the line externally

$\vec{OR}=\dfrac{m\vec{OQ}-n\vec{OP}}{m-n}$

$=\dfrac{2 (- \hat i+ \hat j+ \hat k) -1(\hat i+ 2 \hat j - \hat k)}{2-1}$

$={(-2-1)\hat i+(2-2)\hat j+(2+1) \hat k}$

$=-3 \hat i+3 \hat k$
Hence Position vector of

$R=-3 \hat i+3 \hat k$

Dot product of a vector with $\hat{i}+\hat{j}-3\hat{k},\hat{i}+3\hat{j}-2\hat{k}$ and $2\hat{i}+\hat{j}+4\hat{k}$ are $0,5$ and $8$ respectively. Find the vector.

Let

$\vec a=\hat i+\hat j-3\hat k, \vec b=\hat i-3\hat j-2\hat k, \vec c=2\hat i+\hat j+4\hat k$

let

$\vec x$ be the required vector and

$\vec x=x_1\hat i+x_2\hat j+x_3\hat k$

$\vec {x}.\vec {a}=0\Rightarrow x_1+x_2-3x_3=0.....(1)$

$\vec {x}.\vec {b}=5\Rightarrow x_1+3x_2-2x_3=5.....(2)$

$\vec {x}.\vec {c}=8\Rightarrow 2x_1+x_2-4x_3=0.....(3)$

$(2)-(1)$ we get

$2x_2+x_3=5\Rightarrow x_3=5-2x_2$

$(3)-2\times (2)$ we get

$-5x_2+8x_3=-2$

$-5x_2+8(5-2x_2)=-2$

$-5x_2+40-16x_2=-2\Rightarrow -21x_2=-12$

$x_2=2$

$x_3=5-2\times 2=1$

$x_1+x_2-3x_3=0\Rightarrow x_1+2-3=0$

$x_1=1, x_2=2, x_3=1$

$\vec {x}=\hat i+2\hat j+\hat k$ .

Find $\vec{a}.\vec{b}$ , when $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=4\hat{i}-4\hat{j}+7\hat{k}$ .

$\vec a=a_1\hat i+a_2\hat j+a_3 \hat k$ and

$\vec b=b_1\hat i+b_2\hat j+b_3\hat k$

then

$\vec a.\vec b=a_1b_1+a_2b_2+a_3b_3$

Here

$\vec a =\hat i-2\hat j+\hat k$

$\vec b=4\hat i-4\hat j-7\hat k$

$\vec a.\vec b=4+(-2)\cdot (-4)+1\cdot 7$

$=4+8+7=19$ .

If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ , then prove that
$tan \dfrac{\theta }{2}=\dfrac{\left | \hat{a}-\hat{b} \right |}{\left | \hat{a}+\hat{b} \right |}$

1395687_1662212_ans_54af5cd4a7514eefbc11af292bfb084b.jpg

If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ , then prove that

$cos \dfrac{\theta }{2}=\dfrac{1}{2}\left | \hat{a}+\hat{b} \right |$

Since

$\hat a$ and

$\hat b$ are unit vectors we have

$|\hat a|=1$

$|\hat b|=1$

Also

$|\hat a+\hat b|^2=|\hat a|^2+|\hat b|^2+2|\hat a||\hat b|\cos \theta$ ,

$\theta$ is angle between

$\hat a$ and

$\hat b$

$|\hat a+\hat b|^2=1+1+2\cos \theta$

$=2(1+\cos \theta)\Rightarrow |\hat a+\hat b|^2=4\cos^2\theta/2$

$|\hat a+\hat b|=2\cos \theta/2$

$\cos \theta/2=\dfrac{1}{2}|\hat a+\hat b|$ .

Find $|\overrightarrow a |$ and $|\overrightarrow b|$ , if $(\overrightarrow a -\overrightarrow b).(\overrightarrow a +\overrightarrow b)=27$ and $|\overrightarrow a |=2|\overrightarrow b|$ .

$(\overrightarrow a -\overrightarrow b).(\overrightarrow a +\overrightarrow b)=27,|\vec a|=2|\vec b|$

$\Rightarrow |\overrightarrow a|^2-|\overrightarrow b|^2=27$

$\Rightarrow 4|\overrightarrow b|^2-|\overrightarrow b|^2=27$

$\Rightarrow 3|\overrightarrow b|^2=27$

$\Rightarrow |\overrightarrow b|^2=9$

$\Rightarrow |\overrightarrow b|=3$

Therefore,

$|\overrightarrow a |=2\times 3=6$

Thus,

$|\overrightarrow a|=6$ and

$|\overrightarrow b|=3$ .

Show that the four points A,B,C,D with position vectors $\vec{a},\vec{b}, \vec{c}, \vec{d}$ respectively such that $3\vec{a}-2\vec{b}+5\vec{c}-6\vec{d} = \vec{0},$ are coplanar. Also find the position vector of the point of intersection of the line segments AC and BD.

A, B, C, D has position vectors

$\vec{a}, \vec{b}, \vec{c}, \vec{d}$

$3\vec{a}-2\vec{b}+5\vec{c}-6\vec{d}=0$

$\Rightarrow 3\vec{a}+5\vec{c}=2\vec{b}+\vec{d}$

$\Rightarrow 3\vec{OA}+5\vec{OC}=2\vec{OB}+6\vec{OD}$

$\Rightarrow \dfrac{3\vec{OA}+5\vec{OC}}{8}=\dfrac{2\vec{OB}+6\vec{OD}}{8}$

$\vec{OQ}$ be the point which divides AC in the ratio

$5:3$

$\therefore \vec{OQ}=\dfrac{3\vec{OA}+5\vec{OC}}{8}=\dfrac{2\vec{OB}+6\vec{OD}}{8}$

so, Q also divides BD in the ratio

$6:2$

$\therefore$ Q also lies on Bd

$\therefore$ Q is the point of intersection of AC and BD

$\therefore$ Point of intersection

$=\dfrac{3\vec{a}+5\vec{c}}{8}$ (or)

$\dfrac{2\vec{b}+6\vec{c}}{8}$ .

1404517_1662250_ans_bed57bf29def43c48342498fed1eeb85.png

$ABCDE$ is a pentagon, then prove that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EA} = \vec{0}$ .

1431621_1662214_ans_a565c91d40d84927b34fb36f450dff9e.png

If O is a point in space, ABC is a triangle and D,E,F are the mid-points of the sides BC,CA and AB respectively of the triangle,prove that $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC} = \overrightarrow{OD}+\overrightarrow{OE}+\overrightarrow{OF}.$

1397690_1662262_ans_471eb27b362b45fb8e73cec94d4e23bb.jpg

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ with position vectors $\vec{OP} = 2\vec{a}+b$ and $\vec{OQ}=\vec{a}-2\vec{b},$ respectively in the ratio $1:2$ internally and externally.

Internally

$1:2$

$OP, OQ$

$R = \dfrac{2\vec{OP}+ 10\vec{Q}}{3}$

$R = \dfrac{3\vec{a} + 2 \vec{b}+\vec{a}-2\vec{b}}{3}$

$= \dfrac{5\vec{a}}{3}$

Externalls

$-1 : 2$

$OP, OQ$

$R = \dfrac{20\vec{P} - \vec{OQ}}{2-1}$

$= \dfrac{4a + 2b - a + 2b}{1}$

$= 3\vec{a} + 4\vec{b}$

If P is a point and ABCD is a quadrilateral and $\overrightarrow{AP}+\overrightarrow{PB}+\overrightarrow{PD} = \overrightarrow{PC},$ show that ABCD is a paralleogram.

$ABCD$ is quadrilateral

$P$ is a point

$\overline {AP}+\overline {PB}+\overline {PD}=\overline {PC}....(i)$

$\Rightarrow \ \overline {AP}+\overline {PB}=\overline {PC}-\overline {PD}=\overline {DC}$

$\Rightarrow \ \overline {AB}=\overline {DC}$

Also from

$(i)$

$\overline {AD} +\overline {PB}+\overline {PD}=\overline {PC}$

$\overline {AP}+\overline {PD}=\overline {PC}-\overline {PB}=\overline {BC}$

$\Rightarrow \ \overline {AD}=\overline {BC}$

In a quadrilateral

$ABCD$

$\overline {AB}=\overline {DC}$ and

$\overline {AD}=\overline {BC}$

$\therefore \$ Quadrilateral

$ABCD$ is parallelogram

The projection of the line segment joining the points $(1, -1, 3)$ and $(2, -4, 11)$ on the line joining the points $(-1, 2, 3)$ and $(3, -2, 10)$ is _________.

Given points.

$A(1, -1, 3) \, B(2, -4, 11)$

$\overline{AB} = (2 - 1, -4 + 1, 11 - 3)$

$\equiv (1, -3 , 8)$

$C (-1, 2, 3) \, D(3, -2, 10)$

$\overline{CD} = (3 + 1 , -2-2, 10 - 3)$

$\equiv(4, -4 , 7)$ -Direction ratios.

Projection of

$\overline{AB}$ on

$\overline{CD}$

Ref. image

$= \dfrac{\overline{AB} . \overline{CD}}{|\overline{CD}|}$

$= \dfrac{(1 \times 4) + (-3 \times - 4) +(8 \times 7)}{\sqrt{4^2 + 4^2 + 7^2}}$

$= \dfrac{4 + 12 + 56}{\sqrt{81}}$

$= \dfrac{72}{9}$

$= 8$

1477760_1697521_ans_fcf4d200e5b74326a4214700bc0ceae7.png

Find the projection of $\vec{b}+\vec{c}$ on $\vec{a}$ , where $\vec{a}=2\hat{i}-2\hat{j}+\hat{k}, \vec{b}=\hat{i}+2\hat{j}-2\hat{k}$ and $\vec{c}=2\hat{i}-\hat{j}+4\hat{k}$ .

Let

$\vec{d}=\vec{b}+\vec{c}=\hat{i}+2\hat{j}-2\hat{k}+2\hat{i}-\hat{j}+4\hat{k}=3\hat{i}+\hat{j}+2\hat{k}$

Projection of

$\vec{d}$ on

$\vec{a}$ is

$=\dfrac{1}{\left|\vec{a}\right|}\left(\vec{d}.\vec{a}\right)$

$=\dfrac{1}{\left|\sqrt{{2}^{2}+{\left(-2\right)}^{2}+{1}^{2}}\right|}\left(\left(3\hat{i}+\hat{j}+2\hat{k}\right).\left(2\hat{i}-2\hat{j}+\hat{k}\right)\right)$

$=\dfrac{1}{\left|\sqrt{4+4+1}\right|}\left(6-2+2\right)$

$=\dfrac{1}{\left|\sqrt{9}\right|}\left(6\right)$

$=\dfrac{6}{3}=2$

The two vectors $\hat{j}+\hat{i}$ and $3\hat{i}-\hat{j}+4\hat{k}$ represent the sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$ respectively of triangle $ABC$ . Find the length of the median through $A$ .

$\vec{AB}=\hat{i}+\hat{j}$

$\vec{AC}=3\hat{i}-\hat{j}+4\hat{k}$

Median through A is

$\vec{AD}$

$=\dfrac{\vec{AB}+\vec{AC}}{2}$

$=\dfrac{(\hat{i}+\hat{j})+(3\hat{i}-\hat{j}+4\hat{k})}{2}$

$=\dfrac{(4\hat{i}+4\hat{k})}{2}$

$\vec{AD}=2\hat{i}+2\hat{k}$

$\therefore$ Length of Median

$=|\vec{AD}|=\sqrt{4^2+4^2}=\sqrt{8}$

$=2\sqrt{2}$ .

1404498_1662907_ans_1c18d84da4af454ea2b30cfc6fc4ae4a.png

Find the vector sum of $N$ coplanar forces, each of magnitude $F$ , When each force makes an angle $2\pi /N$ with that preceding it.

Consider a regular polygon of N sides. Each side of the polygon is same. C is the centre of polygon (Fig. 2.60).

$\theta = \frac{2\pi}{N} \rightarrow \theta$ is the angle subtended at the centre of polygon by any one side.
We can prove that

$\alpha = \theta$ as

$\angle ABC + \angle CBD + \alpha = 180^0$ .......(i)
But

$\angle ABC= \angle CBD = 90^0 - \frac{\theta}{2}$ ........(ii)
From Eqs. (i) and (ii), we get

$\alpha = \theta$

$\Rightarrow \alpha = \frac{2\pi}{N}$
So if we have N vectors of equal magnitude and they are arrangged in such a way that each vector makes an angle

$\frac{2\pi}{N} (=\alpha)$ with its preceding vector, we find that the head of the last vector will coincide with the tail of the first vector. Hence, the resultant of all these vectors becomes zero.
1570406_1733193_ans_b22bfbaa18954d6491823bdba67e2a5d.PNG

1570406_1733193_ans_b22bfbaa18954d6491823bdba67e2a5d.PNG

The vector equation of a line which passes through the points $(3, 4 , -7)$ and $(1,-1, 6)$ is ______.

Given points are

$A(3, 4, -7), B(1,-1, 6)$

$\vec{A} = 3 \hat{i} + 4\hat{j} - 7\hat{k}$ ,

$\vec{B} = \hat{i} - \hat{j} + 6\hat{k}$

$\therefore$ vector equation

$\vec{r} = (3\hat{i} + 4\hat{j} - 7\hat{k}) + \lambda [\hat{i} - \hat{j} + 6 \hat{k} - (3\hat{i} + 4\hat{j} - 7\hat{k})]$

$= 3\hat{i} + 4\hat{j} - 7\hat{k} + \lambda [ -2\hat{i} - 5 \hat{j} + 13\hat{k}]$

Find the sum of the vectors $\vec{a}=(\hat{i}-3\hat{k})$ , $\vec{b}=(2\hat{j}-\hat{k})$ and $\vec{c}=(2\hat{i}-3\hat{j}+2\hat{k})$ .

1540353_1707429_ans_91b474e3fa2f4fdb94d4e92db58efbed.jpeg

Find the sum of the vectors $\vec{a}=(\hat{i}-2\hat{j}), \vec{b}=(2\hat{i}-3\hat{j})$ and $\vec{c}=(2\hat{i}+3\hat{k})$ .

1540354_1707430_ans_ae895d08d40d4c94a1a416b88b0c4fe5.jpeg

If $\left| \overrightarrow { a } \right| =\sqrt { 3 } ,\left| \overrightarrow { b } \right| =2$ and angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is $60^o$ , Then find $\overrightarrow{a}.\overrightarrow{b}.$

Given ,

$| \overrightarrow {a} | = \sqrt {3}, | \overrightarrow {b} | = 2$ and

$\theta =60^0$

$\overrightarrow {a} . \overrightarrow {b} = | \overrightarrow {a} | | \overrightarrow {b} | cos \theta$

$= \sqrt {3}.2. cos 60^0 = \sqrt {3} \quad \quad \quad [ \because cos 60^0 = \dfrac {1}{2} ]$

Find the projection of $\overrightarrow{a}$ on $\overrightarrow{b},$ if $\over$ and $\overrightarrow{a}.\overrightarrow{b}.=8$ and $\overrightarrow{b}=2\hat{i} +6 \hat{j} + 3\hat{k}$ .

Given ,

$\overrightarrow {a} . \overrightarrow {b} = 8$ and

$\overrightarrow {b} = 2 \hat {i} + 6 \hat {j} +3 \hat {k}$

We know projection of

$\overrightarrow {a}$ on

$\overrightarrow {b}$ =

$\dfrac { \overrightarrow {a} . \overrightarrow {b} }{ | \overrightarrow {b} | }$ =

$\dfrac {8}{ \sqrt { 4 +36+ 9} }$ =

$\dfrac {8}{7}$

If $\overrightarrow {a} , \overrightarrow {b} , \overrightarrow {c}$ are three vectors such that $| \overrightarrow {a} | = 5 , | \overrightarrow {b} |= 12$ and $| \overrightarrow {c} | = 13$ and $\overrightarrow {a} + \overrightarrow {b} +\overrightarrow {c} = \overrightarrow {0}$ then find the value of $\overrightarrow {a}.\overrightarrow {b} + \overrightarrow {b}.\overrightarrow {c} + \overrightarrow {c} . \overrightarrow {a}.$

${\textbf{Step-1: Find }}$

${\mathbf{\bar a . \bar b + \bar b. \bar c+\bar c.\bar a.}}$

${\text{Given,}}$

$\bar a+\bar b+\bar c=0$

${\text{and}}$

$∣\bar a∣=5,∣\bar b∣=12,∣\bar c∣=13$

${\text{Now}}$

$∣\bar a+\bar b+\bar c∣^2=(\bar a+\bar b+\bar c).(\bar a+\bar b+\bar c) ∵[∣k∣^2=k. k]$

$⇒0=∣\bar a∣^2+∣\bar b∣^2+∣\bar c∣^2+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$

$⇒0=5^2+{12}^2+{13}^2+2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$

$⇒25+144+169=2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$

$⇒-338=2(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)$

$⇒(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)=-169$

${\textbf{Hence , }}$

${\mathbf{(\bar a.\bar b+\bar b.\bar c+\bar c.\bar a)=-169.}}$

Find the sum of the vectors.
$\overrightarrow{a}=\hat{i} - 2\hat{j} + \hat{k}=\overrightarrow{b}-2\hat{i}+4\hat{j} + 5k$ and $\overrightarrow{c} = \hat{i}-6\hat{j}-7\hat{k}$ .

$\overrightarrow {a} + \overrightarrow {b} + \overrightarrow {c} =( 1-2 + 1) \hat {i} + ( - 2 + 4 -6 ) \hat {j} + ( 1 + 5 - 7) \hat {k}$

$= -4 \hat {j} - \hat {k}$

Find the value of $\overrightarrow{a}. \overrightarrow{b}$ if $\left| \overrightarrow { a } \right| =10,\left| \overrightarrow { b } \right| =2$ and $\left| \overrightarrow { a } \times \overrightarrow { b } \right| =16$

$\because | \overrightarrow {a} \times \overrightarrow {b} | = 16 \quad \quad | \overrightarrow {a}| | \overrightarrow {b}| sin \theta = 16$

$\Rightarrow 10 \times 2 sin \theta = 16 \quad \quad \quad sin \theta = \dfrac {16}{20} = \dfrac {4}{5}$

$\Rightarrow cos \theta = \sqrt {1 sin^2 \theta } = \sqrt { 1 -\dfrac {16}{25} } = \pm \dfrac {3}{5}$

$\therefore \overrightarrow {a}. \overrightarrow {b} = | \overrightarrow {a}| | \overrightarrow {b} | cos \theta = \pm 10 \times 2 \times \dfrac {3}{5} = \pm 2$

Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ , whose position vectors are $(2 \overrightarrow{a}+\overrightarrow{b})$ and $( \overrightarrow{a}-3\overrightarrow{b})$ respectively, externally in the ratio $1:2$ Also, show that $P$ is the mid point of the segment $RQ$ .

1698705_1783836_ans_1325961f8b17400db28aebeabeee9046.jpg

If $\overrightarrow{a}, \overrightarrow{b},\overrightarrow{c}$ are three vectors such that $\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{a}.\overrightarrow{c}$ and $\overrightarrow{a} \times \overrightarrow{b}=\overrightarrow{a} \times \overrightarrow{c},\overrightarrow{a} \neq 0$ , then show that $\overrightarrow{b}=\overrightarrow{c}$ .

Given

$\overrightarrow {a} . \overrightarrow {b} = \overrightarrow {a} . \overrightarrow {c}$

$\Rightarrow \overrightarrow {a}. \overrightarrow {b} - \overrightarrow {a} . \overrightarrow {c} = 0 \quad \Rightarrow \quad \overrightarrow {a} .( \overrightarrow {b} - \overrightarrow {c} =0$

$\Rightarrow either \overrightarrow {b} = \overrightarrow {c} \quad or \quad \overrightarrow {a} \bot ( \overrightarrow {b} - \overrightarrow {c} )$

Also given

$\overrightarrow {a} \times \overrightarrow {b} = \overrightarrow {a} \times \overrightarrow {c} \Rightarrow \overrightarrow {a} \times \overrightarrow {b} - \overrightarrow {a} \times \overrightarrow {c} = 0$

$\Rightarrow \overrightarrow {a} \times ( \overrightarrow {b} - \overrightarrow {c} ) = 0 \quad \quad \Rightarrow \quad \overrightarrow {a} || \overrightarrow {b} - \overrightarrow {c} \quad or \quad \overrightarrow {b} = \overrightarrow {c}$

But

$\overrightarrow {a}$ cannot be both parallel and perpendicular to

$( \overrightarrow {b} - \overrightarrow {c}) .$ hence

$\overrightarrow {b} = \overrightarrow {c}$

If $\vec{A} \space and \space \vec{B}$ be two vectors and k be any scalar quantity greater than zero, then prove that $|\vec{A} + \vec{B}|^2 \leq (1 + k) |\vec{A}|^2 + \left( 1 + \dfrac{1}{k}\right) |\vec{B}|^2$ .

1771301_1783420_ans_b56d2a80988b4e27b371f281484ba5f2.jpeg

If $\vec{a}$ and $\vec{b}$ are any two unit vectors, then find the greatest positive integer in the range of $\dfrac{3|\vec{a}+\vec{b}|}{2}+2|\vec{a}-\vec{b}|$

Let angle between

$\bar{a}$ and

$\vec{b}$ be

$\theta$
We have

$|\vec{a}|=|\vec{b}|=1$
Now

$|\vec{a}+\vec{b}|=2 \cos \dfrac{\theta}{2}$ and

$|\vec{a}-\vec{b}|=2 \sin \dfrac{\theta}{2}$

Consider

$F(\theta)=\dfrac{3}{2}\left(2 \cos \dfrac{\theta}{2}\right)+2\left(2 \sin \dfrac{\theta}{2}\right)$

$\therefore F(\theta)=3 \cos \dfrac{\theta}{2}+4 \sin \dfrac{\theta}{2}, \theta \in[0, \pi]$

Let $\vec{u}$ be a vector on rectangular coordinate system with sloping angle $60^{\circ}$ . Suppose that $|\vec{u}-\hat{i} |$ is geometric mean of $|\vec{u}|$ and $|\vec{u}-2 \hat{i}|,$ where $\hat{i}$ is the unit vector along $x$ -axis. Then find the value of $(\sqrt{2}+1)|\vec{u}|$

since angle between

$\vec{u}$ and

$\hat{i}$ is

$60^{\circ}$

$\vec{u} \cdot i=|\vec{u}||\hat{i}| \cos 60^{\circ}=\dfrac{|\vec{u}|}{2}$

Given that

$|\vec{u}-\hat{i}|,|\vec{u}|,|\vec{u}-2 \hat{i}|$ are in G.P., so

$|\vec{u}-\hat{i}|^{2}=|\vec{u}||\vec{u}-2 \hat{i}|$

Squaring both sides,

$\left[\left.| \vec{u}\right|^{2}+|\hat{i}|^{2}-2 \vec{u} \cdot \hat{i}\right]^{2}=|\vec{u}|^{2}\left[|\vec{u}|^{2}+4|\hat{i}|^{2}-4 \vec{u} \cdot \hat{i}\right]$

$\left[|\vec{u}|^{2}+1-\dfrac{2|\vec{u}|}{2}\right]^{2}=|\vec{u}|^{2}\left[|\vec{u}|^{2}+4-4 \dfrac{|\vec{u}|}{2}\right] \Rightarrow|\vec{u}|^{2}+2|\vec{u}|-1=0 \Rightarrow|\vec{u}|=-\dfrac{2 \pm 2 \sqrt{2}}{2} \Rightarrow|\vec{u}|=\sqrt{2}-1$

$P_{1}$ and $P_{2}$ are planes passing through origin. $L_{1}$ and $L_{2}$ are two lines on $P_{1}$ and $P_{2}$ , respectively, such that their intersection is the origin. Show that there exist points $A, B$ and $C,$ whose permutation $A^{\prime}, B^{\prime}$ and $C^{\prime},$ respectively, can be chosen such that
(i) $\mathrm{A}$ is on $L_{1}, B$ on $P_{1}$ but not on $L_{1}$ and $C$ not on $P_{1}$
(ii) $A^{\prime}$ is on $L_{2}, B^{\prime}$ on $P_{2}$ but not on $L_{2}$ and $C^{\prime}$ not on $P_{2}$

The following figure shows the possible situation for planes

$P_{1}$ and

$P_{2}$ and the lines

$L_{1}$ and

$L_{2}$

Now if we choose points

$A, B$ and

$C$ as

$A$ on

$L_{1}, B$ on the line of intersection of

$P_{1}$ and

$P_{2}$ but other than the origin and

$C$ on

$L_{2}$ again other than the origin, then we can consider

$A$ corresponds to one of

$A^{\prime}, B^{\prime}, C^{\prime}$

$B$ corresponds to one of the remaining of

$A^{\prime}, B^{\prime}$ and

$C^{\prime}$

$C$ corresponds to third of

$A^{\prime}, B^{\prime}$ and

$C^{\prime},$ e.g.,

$A^{\prime} \equiv C ; B^{\prime} \equiv B ; C^{\prime} \equiv A$

Hence one permutation of

$[A B C]$ is

$[C B A] .$ Hence proved.

1656619_1785393_ans_c02c5b0574a3426ca0464c165aada535.PNG

Let $\vec{A}, \vec{B}$ and $\vec{C}$ be vectors of length, 3,4 and 5, respectively. Let $\vec{A}$ be perpendicular to $\vec{B}+\vec{C}, \vec{B}$ to $\vec{C}+\vec{A}$ and $\vec{C}$ to $\vec{A}+\vec{B}$ . Then the length of vector $\vec{A}+\vec{B}+\vec{C}$ is______________

1857128_1785404_ans_e864d9e0525c4816a86155e0242b21e3.jpg

If $c$ be a given non-zero scalar, and $\vec{A}$ and $\vec{B}$ be given non-zero vectors such that $\vec{A} \perp \vec{B}$ , find the vector $\vec{X}$ which satisfies the equations $\vec{A} \cdot \vec{X}=c$ and $\vec{A} \times \vec{X}=\vec{B}$ .

$\vec{A} \times \vec{X}=\vec{B}\\\Rightarrow(\vec{A} \times \vec{X}) \times \vec{A}=\vec{B} \times \vec{A}\\\Rightarrow(\vec{A} \cdot \vec{A}) \vec{X}-(\vec{X} \cdot \vec{A}) \vec{A}=\vec{B} \times \vec{A}$

$\Rightarrow(\vec{A} \cdot \vec{A}) \vec{X}-c \vec{A}=\vec{B} \times \vec{A}$

$\Rightarrow \vec{X}=\dfrac{\vec{B} \times \vec{A}+c \vec{A}}{(\vec{A} \cdot \vec{A})}$

Prove that : $| \overrightarrow {a} \times \overrightarrow {b}|^2 = \begin{vmatrix} \overrightarrow {a}. \overrightarrow {a} & \overrightarrow {a}. \overrightarrow {b} \\ \overrightarrow {a}. \overrightarrow {b} & \overrightarrow {b} . \overrightarrow {b} \end{vmatrix}$

Let

$\theta$ be the angle between

$\overrightarrow {a}$ and

$\overrightarrow {b}$ then

$LHS\ \ \ \ | \overrightarrow {a} \times \overrightarrow {b}|^2 = ( \overrightarrow {a} \times \overrightarrow {b}) .( \overrightarrow {a} \times \overrightarrow {b})$

$=( ab\ sin\ \theta ) \hat {n} . (ab\ sin\ \theta) \hat {n} - ( a^2 b^2\ sin^2\ \theta)( \hat {n} \hat {n}) = a^2b^2\ sin^2\ \theta$

$= a^2b^2 ( 1-cos^2\ \theta) =a^2b^2 -( ab\ cos\ \theta )^2$

$( \overrightarrow {a}. \overrightarrow {a} )( \overrightarrow {b} . \overrightarrow {b})- ( \overrightarrow {a}. \overrightarrow {b})^2$ ...........(ii)

$(i) \quad and \quad (ii) \quad RHS = LHS$ [hence proved].

Find three-dimensional vectors $\vec{v}_{1}, \vec{v}_{2}$ and $\vec{v}_{3}$ satisfying $\vec{v}_{1} \cdot \vec{v}_{1}=4, \vec{v}_{1} \cdot \vec{v}_{2}=-2, \vec{v}_{1} \cdot \vec{v}_{3}=6, \vec{v}_{2} \cdot \vec{v}_{2}=2,$ $\vec{v}_{2} \cdot \vec{v}_{3}=-5, \vec{v}_{3} \cdot \vec{v}_{3}=29$

Given data are insufficient to uniquely determine the three vectors as there are only six equations involving nine variables.

Therefore, we can obtain infinite number of sets of three vectors,

$\vec{v}_{1}, \vec{v}_{2}$ and

$\vec{v}_{3},$ satisfying these conditions.

From the given data, we get

$\vec{v}_{1} \cdot \vec{v}_{1}=4 \Rightarrow\left|\vec{v}_{1}\right|=2$

$\vec{v}_{2} \cdot \vec{v}_{2}=2 \Rightarrow\left|\vec{v}_{2}\right|=\sqrt{2}$

$\overrightarrow{v_{3}} \cdot \overrightarrow{v_{3}}=29 \Rightarrow|\overrightarrow{v_{3}}|=\sqrt{29}$

Also

$\vec{v}_{1} \cdot \vec{v}_{2}=-2$

$\Rightarrow\left|v_{\mathrm{t}}\right|\left|v_{2}\right| \cos \theta=-2 \quad$ (where

$\theta$ is the angle between

$\vec{v}_{1}$ and

$\vec{v}_{2}$ )

$\Rightarrow \cos \theta=\dfrac{-1}{\sqrt{2}}$

$\Rightarrow \theta=135^{\circ}$

since any two vectors are always coplanar, let us suppose that

$\vec{v}_{1}$ and

$\vec{v}_{2}$ are in the

$x-y$ plane. Let

$\vec{v}_{1}$ be along the positive direction of the

$x$ -axis. Then

$\vec{v}_{1}=2 \hat{i}, \quad\left(\because\left|\vec{v}_{1}\right|=2\right)$

$\vec{v}_{2}$ makes an angle

$135^{\circ}$ with

$\vec{v}_{1}$ and lies in the

$x-y$ plane, also keeping in mind

$\left|\vec{v}_{2}\right|=\sqrt{2},$ we obtain

$\vec{v}_{2}=-\hat{i} \pm \hat{j}$

Again let

$\vec{v}_{3}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

$\because \vec{v}_{3} \cdot \vec{v}_{1}=6 \Rightarrow 2 \alpha=6 \Rightarrow \alpha=3$

and

$\vec{v}_{3} \cdot \vec{v}_{2}=-5 \Rightarrow-\alpha \pm \beta=-5 \Rightarrow \beta=\pm 2$

Also

$\left|\vec{v}_{3}\right|=\sqrt{29} \Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}=29$

$\Rightarrow \gamma=\pm 4$

Hence

$\vec{v}_{3}=3 \hat{i} \pm 2 \hat{j} \pm 4 \hat{k}$

Prove by vector method that the diagonals of a parallelogram bisect each other.

We draw a parallelogram with one diagonal coincident to x-axis and the intersect of two diagonals is on origin. One diagonal into divided into

$\overrightarrow{a}$
and

$m\overrightarrow{a}$ , the other is

$\overrightarrow{b}$ and

$n\overrightarrow{b}$ .
Now,

$\Rightarrow \overrightarrow{a}$ +

$\overrightarrow{l}$ =

$\overrightarrow{b}$ and

$\overrightarrow{l}$ +

$m\overrightarrow{a}$ =

$n\overrightarrow{b}$
(where m and n are scalars)

$\Rightarrow \overrightarrow{a}$ -

$\overrightarrow{b}$ =

$m\overrightarrow{a}$ -

$n\overrightarrow{b}$

$\Rightarrow (m-1)\overrightarrow{a}$ =

$(n-1)\overrightarrow{b}$
So,

$\overrightarrow{a}$ =

$\begin{pmatrix} |a|\\ 0 \end{pmatrix}$

$\overrightarrow{b}$ =

$\begin{pmatrix} |b|cos\ \theta\\ |b|sin\ \theta \end{pmatrix}$

y direction:

$0=(n−1)|b|sin\ \theta$
In the parallelogram,

$\theta>0$ and

$|b|>0$

$\therefore n=1$ and

$\overrightarrow{b} =n\overrightarrow{b}$
Hence, diagonal

$b$ is bisected.
and since,

$\overrightarrow {a}=m\overrightarrow {a}$
diagonal

$a$ is bisected.
1652848_1785174_ans_4e36cccd92c0469092ea9aeb35f57f08.JPG

Let $\vec{b}=4 \hat{i}+3 \hat{j}$ and $\vec{c}$ be two vectors perpendicular to each other in the $x y$ -plane. All vectors in the same plane having projections 1 and 2 along $\vec{b}$ and $\vec{c},$ respectively, are given by______________

1856883_1785411_ans_a8804c5b9ac74f5ba006da90310fc76d.jpg

Find a vector a magnitude $11$ in the direction opposite to that of $\bar{PQ}$ , where $P$ and $Q$ are the points $(1,3,2)$ and $(-1, 0, 8)$ , respectively.

The vector with initial point

$P(1,3,2)$ and terminal point

$Q(-1, 0, 8)$
is given by

$\bar{PQ}=(-1-1)\hat{i}+(0-3)\hat{j}+(8-2)\hat{k}=-2\hat{i}-3\hat{j}+6\hat{k}$

Thus,

$\bar{QP}=-\bar{PQ}=2\hat{i}+3\hat{j}-6\hat{k}$

$\Rightarrow |\bar{QP}|=\sqrt{2^{2}+3^{2}+(-6)^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$

Therefore, unit vector in the direction of

$\bar{QP}$ is given by

$\bar{QP}=\dfrac{\bar{QP}}{|\bar{QP}|}=\dfrac{2\hat{i}+3\hat{j}-6\hat{k}}{7}$

Hence the required vector of magnitude

$11$ in direction of

$\bar{PQ}$ is

$11\bar{QP}=11\left(\dfrac{2\hat{i}+3\hat{j}-6\hat{k}}{7}\right)=\dfrac{22}{7}\hat{i}+\dfrac{33}{7}\hat{j}-\dfrac{66}{7}\hat{k}$ .

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ with position vectors $\bar{OP}=2\bar{a}+\bar{b}$ and $\bar{OQ}=\bar{a}-2\bar{b}$ , respectively, in the ratio $1:2$ externally.

The position vector of the point

$R$ dividing the join of

$P$ and

$Q$ in the ratio

$1:2$ externally is given by

$\bar{OR}=\dfrac{2(2\vec{a}+\vec{b})-1(\vec{a}-2\vec{b})}{2-1}=3\vec{a}+4\vec{b}$ .
Since, the position vector of a point

$R$ divides the line segment joining the points

$P$ and

$Q$ , whose position vectors are

$\vec p$ and

$\vec q$ in the ration

$m:n$ internally, is given by

$\dfrac{m\vec q +n\vec p}{m+n}$

If $\vec a$ is any non zero vector, then $(\vec a, \hat i)\hat i+(\vec a, \hat j)\hat j +(\vec a, \hat k)\hat k$ equals ______ .

Let

$\vec a=a_1\hat i+a_2\hat j+a_3\hat k$

$\therefore \vec a.\hat i=a_1. \vec a. \hat j=a_2$ and

$\vec a. \hat k=a_3$

$\therefore (\vec a. \hat i)\hat i+(\vec a. \hat j)\hat j+(\vec a. \hat k)\hat k=a_1\hat i+a_2\hat j+a_3\hat k=\vec a$

The value of the expression $|\vec a\times \vec b|^2 +(\vec a, \vec b)^2$ is _______ .

$=|\vec a|^2 |\vec b|^2 (1-\cos^2 \theta)+(\vec a.\vec b)^2$

$=|\vec a|^2 |\vec b|^2 -|\vec a|^2 |\vec b|^2 \cos^2 (\vec a.\vec b)^2$

$=|\vec a|^2 |\vec b|^2 -(\vec a.\vec b)^2 +(\vec a. \vec b)^2$

$=|\vec a|^2|\vec b|^2$

$|\vec a\times \vec b|^2 +(\vec a.\vec b)^2 =|\vec a|^2 |\vec b|^2$

In the triangle $PQR,\bar {PQ}=\bar {2a},\bar {QR}=\bar{2b}$ . The midpoint of $PR$ is $M$ . Find the following vectors in terms of $\bar a$ and $\bar b$ :
(i) $\bar {PR}$ (ii) $\bar {PM}$ (iii) $\bar {QM}$ .

$\bar{PQ}=\bar{2a},\bar{QR}=\bar{2b}$
(i)

$\bar{PR}=\bar{PQ}+\bar{QR}$

$=2\bar a+2\bar b$
(ii)

$\because M$ is the midpoint of

$PR$

$\\bar{PM}=\dfrac{1}{2}\bar{PR}=\dfrac{1}{2}[2\bar a+2\bar b]$

$=\bar a+\bar b$
(iii)

$\bar{RM}=\dfrac{1}{2}(\bar{RP})=-\dfrac{1}{2}\bar{PR}=-\dfrac{1}{2}(2\bar a+2\bar b)$

$=-\bar a-\bar b$

$\therefore \bar{QM}=\bar{QR}+\bar{RM}$

$=2\bar b-\bar a-\bar b$

$=\bar b-\bar a$
[note: point (i) answer in the textbook is incorrect.]
1808525_1846213_ans_c19f17e5d133466e8fd1ee4b909e0db4.png

If $\bar{p}, \bar{q}$ and $\bar{r}$ are unit vectors and forms a triangle, find $\bar{p}. \bar{r}$ .

Let the triangle be denoted by

$ABC$ , where

$\bar{AB}=\bar{p}, \bar{AC}=\bar{q}$ and

$\bar{BC}=\bar{r}$

$\because \bar{p}, \bar{q}, \bar{r}$ are unit vectors.

$\therefore I(AB)=I(BC)=I(CA)=1$

$\therefore$ the triangle is equilateral

$\therefore \angle A=\angle B=\angle C=60^{o}$

Using the formula for angle between two vectors,

$\cos B=\dfrac{\bar{p}.\bar{r}}{|\bar{p}|.|\bar{r}|}$

we get

$\bar{p}.\bar{r}=\dfrac{1}{2}$

$\therefore \cos 60^{o} =\dfrac{\bar{p}.\bar{r}}{1\times 1}$

$\therefore \dfrac{1}{2}=\bar{p}.\bar{r}$

$\therefore \bar{p}.\bar{r}=\dfrac{1}{2}$

If $\bar{p}, \bar{q}$ and $\bar{r}$ are unit vectors and forms a triangle, find $\bar{p}.\bar{q}$ .

Let the triangle be denoted by

$ABC$ , where

$\bar{AB}=\bar{p}, \bar{AC}=\bar{q}$ and

$\bar{BC}=\bar{r}$

$\because \bar{p}, \bar{q}, \bar{r}$ are unit vectors.

$\therefore I(AB)=I(BC)=I(CA)=1$

$\therefore$ the triangle is equilateral

$\therefore \angle A=\angle B=\angle C=60^{o}$

Using the formula for angle between two vectors,

$\cos A=\dfrac{\bar{p}.\bar{q}}{|\bar{p}|.|\bar{q}|}$

$\therefore \cos 60^{o}=\dfrac{\bar{p}.\bar{q}}{1\times 1}$

$\therefore \dfrac{1}{2}=\bar{p}.\bar{q}$

$\therefore \bar{p}.\bar{q}=\dfrac{1}{2}$

Show that the sum of the length of projections of $p\hat{i}+q\hat{j}+r\hat{k}$ on the coordinates axes, where $p=2, q=3$ and $r=4$ is $9$ .

Let

$\bar{a}=p\hat{i}+q\hat{j}+r\hat{k}$
Projection of

$\bar{a}$ on

$X-$ axis

$=\dfrac{\bar{a}.\hat{i}}{|\hat{i}|}=\dfrac{(p\hat{i}+q\hat{i}+r\hat{k}).\hat{i}}{1}=p=2$
Similarly, projections of

$\bar{a}$ on

$Y-$ and

$Z-$ axes are

$3$ and

$4$ respectively.

$\therefore$ sum of these projections

$=2+3+4=9$

$D$ and $E$ divide sides $BC$ and $CA$ of a triangle $ABC$ in the ratio $2:3$ each. Find the positive vector of the point of intersection of $AD$ and $BE$ and the ratio in which this point divides $AD$ and $BE$ .

Let

$AD$ and

$BE$ intersect at

$P$ .
Let

$A, B, C, D, E, P$ have position vectos

$\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{p}$ respectively.

$D$ and

$E$ divide segment

$BC$ and

$CA$ internally in the ratio

$2:3$ .
By the section formula for internal division,

$\bar{d}=\dfrac{2\bar{c}+3\bar{b}}{2+3}$

$\therefore 5\bar{d}=2\bar{c}+3\bar{b}$ ....(1)
and

$\bar{e}=\dfrac{2\bar{a}+3\bar{c}}{2+3}$

$\therefore 5\bar{e}=2\bar{a}+3\bar{c}$ .....(2)
From

$(1), 5\bar{d}-3\bar{b}=2\bar{c}\therefore 15\bar{d}-9\bar{b}=6\bar{c}$
From

$(2), 5\bar{e}-2\bar{a}=3\bar{c}\therefore 10\bar{e}-4\bar{a}=6\bar{c}$
Equating both values of

$6\bar{c}$ , we get

$15\bar{d}-9\bar{b}=10\bar{e}-4\bar{a}$

$\therefore 15\bar{d}+4\bar{a}=10\bar{e}+9\bar{b}$

$\therefore \dfrac{15\bar{d}+4\bar{a}}{15+4}=\dfrac{10\bar{e}+9\bar{b}}{10+9}$

$LHS$ is the position vector of the point which divides segment

$AD$ internally in the ration

$15:4$ .

$RHS$ is the position vector of the point which divides segment

$BE$ internally in the ratio

$10:9$ .
But

$P$ is the point of intersection of

$AD$ and

$BE$ .

$\therefore P$ divides

$AD$ internally in the ratio

$15:4$ and

$P$ divides

$BE$ internally in the ratio

$10:9$ .
Hence, the position vector of the point of interaction of

$AD$ and

$BE$ is

$\bar{p}=\dfrac{15\bar{d}+4\bar{a}}{19}=\dfrac{10\bar{e}+9\bar{b}}{19}$
and it divides

$AD$ internally in the ratio

$15:4$ and

$BE$ internally in the ratio

$10:9$ .

1808546_1846264_ans_2bc47d66884a437ba16d550b7420f6b2.png

Find the position vector of point $R$ which divides the line joining the points $P$ and $Q$ whose position vectors are $2\hat{i}-\hat{j}-3\hat{k}$ and $-5\hat{i}+2\hat{j}-5\hat{k}$ in the ratio $3:2$ is internally.

It is given that the points

$P$ and

$Q$ have position vectors

$\bar{p}=2\hat{i}-\hat{j}+3\hat{k}$ and

$\bar{q}=-5\hat{i}+2\hat{j}-5\hat{k}$ respectively.

$R(\bar{r})$ divides the line segment

$PQ$ internally in the ratio

$3:2$ , by section formula for internal division,

$\bar{r}=\dfrac{3\bar{q}+2\bar{p}}{3+2}$

$=\dfrac{3(-5\hat{i}+2\hat{j}-5\hat{k})+2(2\hat{i}-\hat{j}+3\hat{k})}{5}$

$=\dfrac{-11\hat{i}+4\hat{j}-9\hat{k}}{5}$

$=\dfrac{1}{5}(-11\hat{i}+4\hat{j}-9\hat{k})$

$\therefore$ coordinates of

$R=\left(-\dfrac{11}{5}, \dfrac{4}{5}, -\dfrac{9}{5}\right)$

Hence, the position vector of

$R$ is

$\dfrac{1}{5}(-11\hat{i}+4\hat{j}-9\hat{k})$ and the coordinates of

$R$ are

$\left(-\dfrac{11}{5}, \dfrac{4}{5}, -\dfrac{9}{5}\right)$

In $\triangle OAB, E$ is the midpoint of $OB$ and $D$ is the point on $AB$ such that $AD: DB= 2:1$ . If $OD$ and $AE$ intersect at $P$ , then determine the ratio $OP:PD$ using vector methods.

Let

$A,\ B,\ D,\ E,\ P$ have position vectors

$\bar {a},\bar {b},\bar {d},\bar {e},\bar {p}$
w.r.t

$O.$

$\because AD:DB=2:1$

$\therefore D$ divides

$AB$ internally in the ratio

$2:1.$
Using section formula for internal division, we get

$\bar {d} = \dfrac {2\bar {b}+\bar {a}}{2+1}$

$\bar {d} = 2\bar {b}+\bar {a}$ .......

$(1)$
Since

$E$ is the midpoint of

$OB,\ \bar {e}= \bar {OE} = \dfrac {1}{2}\bar {OB}=\dfrac {\bar {b}}{2}$

$\therefore \bar {b} = 2 \bar {e}$ .....

$(2)$

$\therefore$ from

$(1),$

$3\bar {d} = 2 (2\bar {e}) + \bar {a}$ .....

$[By\ (2)]$

$4\bar {e} + \bar {a} \\ \therefore \dfrac {3\bar {d}+2.\bar {0}}{3+2} = \dfrac {4\bar {e}+\bar {a}}{4+1}$

$LHS$ is the poistion vector of the point which divides

$OD$ internally in the ration

$3:2.$

$RHS$ is the position vector of the point which divides

$AE$ internally in the ration

$4:1.$
But

$OD$ and

$AE$ intersect at

$P$

$\therefore P$ divides

$OD$ internally in the ration

$3:2.$
Hence,

$OP:PD=3:2.$
1808559_1846282_ans_64d74d59a07f416f9674fcddcce7f513.png

Find the position vector of point $R$ which divides the line joining the points $P$ and $Q$ whose position vectors are $2\hat{i}-\hat{j}-3\hat{k}$ and $-5\hat{i}+2\hat{j}-5\hat{k}$ in the ratio $3:2$ is externally.

It is given that that points

$P$ and

$Q$ have position vectors

$\bar{p}=2\hat{i}-\hat{j}+3\hat{k}$ and

$\bar{q}=-5\hat{i}+2\hat{j}-5\hat{k}$ respectively.

$R(\bar{r})$ divides the line segment joining

$P$ and

$Q$ externally in the ratio

$3:2$ , by section formula for external division,

$\bar{r}=\dfrac{3\bar{q}-2\bar{p}}{3-2}$

$=\dfrac{3(-5\hat{i}+2\hat{j}-5\hat{k})-2(2\hat{i}-\hat{j}+3\hat{k})}{3-2}$

$=-19\hat{i}+8\hat{j}-21\hat{k}$

$\therefore$ coordinates of

$R=(-19, 8, -21)$ .

Hence, the position vector of

$R$ is

$-19\hat{i}+8\hat{j}-21\hat{k}$ and coordinates of

$R$ are

$(-19, 8, -21)$ .

Find the position vector of midpoint $M$ joining the points $L(7,-6,12)$ and $N(5,4,-2)$ .

The position vectors

$\bar{l}$ and

$\bar{n}$ of the points

$L(7, -6, 12)$ and

$N(5,4,-2)$ are given by

$|bar{l}=7\hat{i}-6\hat{j}+12\hat{k}, \bar{n}=5\hat{i}+4\hat{j}-2\hat{k}$

$M(\bar{m})$ is the midpoint of

$LN$ , by midpoint formula,

$\bar{m}=\dfrac{\bar{l}+\bar{n}}{2}$

$=\dfrac{(7\hat{i}-6\hat{j}+12\hat{k})+(5\hat{i}+4\hat{j}-2\hat{k})}{2}$

$=\dfrac{1}{2}(12\hat{i}-2\hat{j}+10\hat{k})=6\hat{i}-\hat{j}+5\hat{k}$

$\therefore$ coordinates of

$M(6,-1,5)$

Hence, position vector of

$M$ is

$6\hat{i}-\hat{j}+5\hat{k}$ and the coordinates of

$M$ are

$(6,-1,5)$

The position vector of points $A$ and $B$ are $6\bar{a}+2\bar{b}$ and $\bar{a}-3\bar{b}$ . If the point $C$ divides $AB$ in the ratio $3:2$ , show that the position vector of $C$ is $3\bar{a}-\bar{b}$ .

Let

$\bar{c}$ be the position vector of

$C$ .

Since

$C$ divides

$AB$ in the ratio

$3:2$ ,

$\bar{c}=\dfrac{3(\bar{a}-3\bar{b})+2(6\bar{a}+2\bar{b})}{3+2}$

$=\dfrac{3\bar{a}-9\bar{b}+12\bar{a}+4\bar{b}}{5}$

$=\dfrac{1}{5}(15\bar{a}-5\bar{b})=3\bar{a}-\bar{b}$

Hence, the position vector of

$C$ is

$3\bar{a}-\bar{b}$ .

Find $\bar u.\bar v$ if $|\bar u|=2, |\bar v|=5,|\bar u\times \bar n|=8$

Let

$\theta$ be the angle between

$\vec u$ and

$\vec v$

Then

$| \vec u \times \vec v| =8$ gives

$| \vec u | | \vec v| \sin \theta =8$

$\therefore 2 \times 5 \times \sin \theta =8$

$\therefore \sin \theta =\dfrac 45$

$\cos \theta =\pm \sqrt{1-\sin^2 \theta }$ ....

$[ \because 0 \le \theta \le ]$

$=-\sqrt{1-\left( \dfrac 45 \right)^2}$

$=\pm \sqrt{1-\dfrac{16}{25}}$

$=-\sqrt{\dfrac{9}{25}}=\pm \dfrac 35$

Now,

$\vec u. \vec v=| \vec u| | \vec v| \cos \theta$

$\therefore \vec u. \vec v=2\times 5 \times \left( \pm \dfrac 35 \right) =\pm 6$

Prove that $\left[\bar{a}\ \ \ \ \bar{b}+\bar{c}\ \ \ \ \ \bar{a}+\bar{b}+\bar{c}\right]=0$

$\left[\bar{a}\ \ \ \bar{b}+\bar{c}\ \ \ \bar{a}+\bar{b}+\bar{c}\right]$ $

$=\bar{a}.\left[(\bar{b}+\bar{c})\times (\bar{a}+\bar{b}+\bar{c})\right]$

$=\bar{a}.\left(\bar{b}\times \bar{a}+\bar{b}\times \bar{b}+\bar{b}\times \bar{c}+\bar{c}\times \bar{a}+\bar{c}\times \bar{b}+\bar{c}\times \bar{c}\right)$

$=\bar{a}.(\bar{b}\times \bar{a})+\bar{a}(\bar{b}\times \bar{b})+\bar{a}.(\bar{b}\times \bar{c})+\bar{a}.(\bar{c}\times \bar{b})+\bar{a}.(\bar{c}\times \bar{c})$

$=0+0+\bar{a}.(\bar{b}\times \bar{c})+0-\bar{a}.(\bar{b}\times \bar{c})+0$

$=0$

Find $\bar{a}. (\bar{b}\times \bar{c})$ if $\bar{a}=3\hat{i}-\hat{j}+4\hat{k}, \bar{b}=2\hat{i}+3\hat{j}-\hat{k}$ and $\bar{c}=-5\hat{i}+2\hat{j}+3\hat{k}$

$\bar{a}.(\bar{b}\times \bar{c})=\begin{vmatrix} 3 & -1 & 4 \\ 2 & 3 & -1 \\ -5 & 2 & 3 \end{vmatrix}$

$=3(9+2)+1(6-5)+4(4+15)$

$=33+1+76$

$=110$

If $\bar a=\hat i+\hat j+\hat k$ and $\bar c=\hat j-\hat k$ , find $\bar a$ vector $\bar b$ satisfying $\bar a\times \bar b=\bar c$ and $\bar a.\bar b=3$

Given:

$\vec a =\hat i +\hat j +\hat k, \vec c=\hat j - \hat k$
Let

$\vec b=x\hat i +y\hat j +z\hat k$
Then

$\vec a. \vec b=3$ gives

$( \hat i +\hat j +\hat k). ( x \hat i+y\hat j +z\hat k)=3$

$\therefore (1) (x) +(1)(y) +(1)(z)=3$
Also,

$x+y+z=3$ ....(1)
Also,

$\vec c=\vec a \times \vec b$

$\therefore \hat j - \hat k =\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix}$

$=(z-y) \hat i -( z-x) \hat j +( y-x) \hat k$

$=(z-y) \hat i +( z-x) \hat j +( y-x) \hat k$
By equality of vectors,

$z-y=0$ ....(2)

$x-z=1$ ....(3)

$y-x=-1$ .....(4)
From (2),

$y=z$ .
From (3),

$x=1+z$
Substituting these values of

$x$ and

$y$ in (1), we get

$1+z+z+z=3$

$\therefore z=\dfrac 23$

$\therefore y=z=\dfrac 23$

$\therefore x=1+z=1+\dfrac 23 =\dfrac 53$

$\therefore \vec b =\dfrac 53 \hat i +\dfrac 23 \hat j+\dfrac 23 \hat k$
i.e.

$\vec b=\dfrac 13 ( 5\hat i +2\hat j +2\hat k)$

If $\bar{u}=\hat{i}-2\hat{j}+\hat{k}, \bar{r}=3\hat{i}+\hat{k}$ and $\bar{w}=\hat{j}-\hat{k}$ are given vectors, then find $(\bar{u}+\bar{w}).\left[(\bar{u}\times \bar{r})\times (\bar{r}\times \bar{w})\right]$

$\bar{u}+\bar{w}=(\hat{i}-2\hat{j}+\hat{k})+(\hat{j}-\hat{k})$

$=\hat{i}-\hat{j}$

$\bar{u}\times \bar{r}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 3 & 0 & 1 \end{vmatrix}$

$=(-2-0)\hat{i}-(1-3)\hat{j}+(0+6)\hat{k}$

$=-2\hat{i}+2\hat{j}+6\hat{k}$

$\bar{r}\times \bar{w}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 0 & 1 \\ 0 & 1 & -1 \end{vmatrix}$

$=(0-1)\hat{i}-(-3-0)\hat{j}+(3-0)\hat{k}$

$=-\hat{i}+3\hat{j}+3\hat{k}$
Now,

$(\bar{u}+\bar{w}).[(\bar{u}\times \bar{r}\times (\bar{r}\times \bar{w})]=\begin{vmatrix} 1 & -1 & 0 \\ -2 & 2 & 6 \\ -1 & 3 & 3 \end{vmatrix}$

$=1(6-18)+1(-6+6)+0$

$=-12+0+0=-12$

If $\bar{a}=\hat{i}+2\hat{j}+3\hat{k}, \bar{b}=3\hat{i}+2\hat{j}$ and $\bar{c}=2\hat{i}+\hat{j}+3\hat{k}$ , then verify that $\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}.\bar{c})\bar{b}-(\bar{a}.\bar{b})\bar{c}$

$\bar{b}\times \bar{c}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 0 \\ 3 & 1 & 3 \end{vmatrix}$

$=(6-0)\hat{i}-(9-0)\hat{j}+(3-4)\hat{k}$

$+6\hat{i}-9\hat{j}-\hat{k}$

$\therefore \bar{a}\times (\bar{b}\times \bar{c})=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 6 & -9 & -1 \end{vmatrix}$

$=(-2+27)\hat{i}-(-1-18)\hat{j}+(-9-12)\hat{k}$

$=25\hat{i}+19\hat{j}-21\hat{k}$ ....(1)

$\bar{a}.\bar{c}=(\hat{i}+2\hat{j}+3\hat{k}).(2\hat{i}+\hat{j}+3\hat{k})$

$=(1)(2)+(2)(1)+(3)(3)$

$=2+2+9=13$

$\therefore (\bar{a}.\bar{c}).\bar{b}=13(3\hat{i}+2\hat{j})=39\hat{i}+26\hat{j}$
Also,

$(\bar{a}.\bar{b})=(\hat{i}+2\hat{j}+3\hat{k}).(3\hat{i}+2\hat{j})$

$=(1)(3)+(2)(2)+(3)(0)$

$=3+4+0=7$

$\therefore (\bar{a}.\bar{b}).\bar{c}=7(2\hat{i}+\hat{j}+3\hat{k})=14\hat{i}+7\hat{j}+21\hat{k}$

$\therefore (\bar{a}.\bar{c}).\bar{b}-(\bar{a}.bar{b}).\bar{c}$

$=(39\hat{i}+26\hat{j})-(14\hat{i}+7\hat{j}+21\hat{k})$

$=25\hat{i}+19\hat{j}-21\hat{k}$ .....(2)
From

$(1)$ and

$(2)$ , we get

$\bar{a}\times (\bar{b}\times \bar{c})=(\bar{a}.\bar{c})\bar{b}-(\bar{a}.\bar{b})\bar{c}$

$ABCD$ is a trapezium with $AB$ parallel to $DC$ and $DC=3AB$ $.$ $M$ $is the midpoint of$ $DC.\ \overline {AB}= \overline {p},\ \overline {BC}=\overline {q}.$ $
Find in terms of $\overline {p}$ and $\overline {q}$ :
$\overline {DA}$

$DC$ is parallel to

$AB$ and

$DC=3AB$

$\because \overline {AB} = \overline {p}\ \therefore \overline {DC}=3AB.$

$M$ is the midpoint of

$DC$

$\therefore \overline {DM} = \overline {MC} = \dfrac {1}{2} \overline {DC}=\dfrac {3}{2} \overline {p}$

$(i)\ \overline {AM} = \overline {AM}+\overline {BC}+\overline {CM}$

$\overline {AB}+\overline {BC}-\overline {MC}$

$\overline {p}+\overline {q}-\overline {\dfrac {3p}{2}}$

$\overline {q}-\dfrac {1}{2}\overline {p}$

$(iv)\ \overline {DA}=\overline {DC}+\overline {CB}+\overline {BA}=\overline {DC}-\overline {BC}-\overline {AB}=3\overline {p}-\overline {q}-\overline {p}=2\overline {p}-\overline {q}$
1808592_1846453_ans_ebe2c016acc5409c8c93bba9a4acf313.png

1808592_1846453_ans_ebe2c016acc5409c8c93bba9a4acf313.png

$ABCD$ is a trapezium with $AB$ parallel to $DC$ and
$DC=3AB$ . $M$ is the midpoint of $DC.\ \overline {AB}= \overline {p},\ \overline {BC}=\overline {q}.$
Find in terms of $\overline {p}$ and $\overline {q}$ :
$\overline {MB}$

$DC$ is parallel to

$AB$ and

$DC=3AB$

$\because \overline {AB} = \overline {p}\ \therefore \overline {DC}=3AB.$

$M$ is the midpoint of

$DC$

$\therefore \overline {DM} = \overline {MC} = \dfrac {1}{2} \overline {DC}=\dfrac {3}{2} \overline {p}$

$(i)\ \overline {AM} = \overline {AM}+\overline {BC}+\overline {CM}$

$\overline {AB}+\overline {BC}-\overline {MC}$

$\overline {p}+\overline {q}-\overline {\dfrac {3p}{2}}$

$\overline {q}-\dfrac {1}{2}\overline {p}$

$(iii)\ \overline {MB}=\overline {MC}+\overline {CB}=\overline {MC}-\overline {BC}=\dfrac {3}{2}\overline {p}-\overline {q}$
1808589_1846451_ans_254608cd238c40bbbc68609af2833aaa.png

1808589_1846451_ans_254608cd238c40bbbc68609af2833aaa.png

If $\bar {a},\ \bar {b},\ \bar {c}$ are unit vectors such that $\bar {a}+\bar {b}+\bar {c}=\bar {0},$ then find the value of $\bar {a}. \bar {b}+\bar {b}.\bar {c}+\bar {c}.\bar {a}.$

$\bar {a},\ \bar {b},\ \bar {c}$ are unit vectors.

Also,

$\bar {a}.\bar {a} = \bar {b}.\bar {b} = \bar {c}.\bar {c} = 1$

Now,

$\bar {a}+\bar {b}+\bar {c}=\bar {0}$ ....

$(1)$

Taking scalar product of both sides with

$\bar {a}$ we get

$\bar {a}(\bar {a}+\bar {b}+\bar {c})=\bar {a}.\bar {0}$

$\therefore \bar {a}.\bar {a} + \bar {a}.\bar {b} + \bar {a}.\bar {c} = 0$

$\bar {a}.\bar {b} + \bar {a}.\bar {c} = - \bar {a}.\bar {a} = -1$ ....

$(2)$

Similarly taking scalar product of both side of

$(1)$ with

$\bar {b}$ and

$\bar {c},$ we get,

$\bar {b}.\bar {a} + \bar {b}.\bar {c} = -1$ ....

$(3)$

$\bar {c}.\bar {a} + \bar {c}.\bar {b} = -1$ ....

$(4)$

Adding

$(2),\ (3),\ (4)$ and using the fact that scalar product is commutative, we get

$2(\bar {a}.\bar {b} + \bar {b}.\bar {c} +\bar {c}.\bar {a}= -1)=-3$

$\therefore \bar {a}.\bar {b} + \bar {b}.\bar {c} +\bar {c}.\bar {a}= -\dfrac {3}{2}$

In a parallelogram $ABCD,$ diagonal vectors are $\overline {AC}=2\hat {i}+3\hat {j}+4\hat {k}$ and $\overline {BD}=-6\hat {i}+7\hat {j}-2\hat {k}$ then find the adjacent side vectors $\overline {AB}$ and $\overline {AD}$ .

$ABCD$ is a parallelogram.

$\therefore \overline {AB}=\overline {DC},\ \overline {AD}=\overline {BC}$

$\overline {AC}=\overline {AB}+\overline {BC}$

$\overline {AB}+\overline {AD} ...(1)$

$\overline {BD}=\overline {BA}+\overline {AD}=-\overline {AB}+\overline {AD} ...(2)$
Adding

$(1)$ and

$(2)$ , we get

$2\overline {AD}=\overline {AC}+\overline {BD}= (2\overline {i}+3\overline {j}+4\overline {k})+(-6\overline {i}+7\overline {j}-2\overline {k})$

$=-4\overline {i}+10\overline {j}+2\overline {k}$

$\therefore \overline {AD}=\dfrac {1}{2}(-4\overline {i}+10\overline {j}+2\overline {k})$

$=-2\overline {i}+5\overline {j}+\overline {k}$
From

$(1),\ \overline {AB}=\overline {AC}-\overline {AD}$

$=(2\overline {i}+3\overline {j}+4\overline {k})-(-2\overline {i}+5\overline {j}+\overline {k})$

$=4\overline {i}-2\overline {j}+3\overline {k}$
1808599_1846458_ans_29694e7c80b44850b57601ac0e04726d.png

A point $P$ with position vector $\dfrac {-14\hat {i}+39\hat {j}+28\hat {k}}{5}$ divides the line joining $A\ (1,\ 6,\ 5)$ and
$B$ in the ration $3:2,$ then find the point $B.$

Let

$A,\ B$ and

$P$ have position vectors

$a,\ b$ and

$p$ respectively.

Then

$\bar {a} = -\hat {i} + 6\hat {j}+ 5\hat {k},$

$\bar {p} = \dfrac {-14\hat {i} + 39\hat {j}+ 28\hat {k}}{5}$

Now,

$P$ divides

$AB$ internally in the ratio

$3:2$

$\therefore \bar {p} = \dfrac {3\bar {b}+2\bar {a}}{5} \\ \therefore 5\bar {p} = 3\bar {b}+2\bar {a} \\ \therefore 3\bar {b} = 5\bar {p}-2\bar {a}$

$\therefore 3\bar {b} = 5 \left(\dfrac {-14\hat {i} + 39\hat {j}+ 28\hat {k}}{5} \right) -2(-\hat {i} + 6\hat {j}+ 5\hat {k})$

$=-14\hat {i} + 39\hat {j}+ 28\hat {k}+2\hat {i} - 12\hat {j}- 10\hat {k}$

$-12\hat {i} + 27\hat {j}+ 18\hat {k}$

$\therefore = -4\hat {i} + 9\hat {j}+ 6\hat {k}$

$\therefore$ coordinates of

$B$ are

$(-4,\ 9,\ 6).$

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $\overline {a}. (\overline {b} . \overline {c})$

This is meaningless because

$\bar {a}$ is a vector,

$\bar {b}.\bar {c}$ is a scalar and the scalar product of vector and scalar is not defined.

If $\left| \bar { a } \right| = \left| \bar { b } \right| =1,\ \bar{a}.\bar {b}=0,\ \bar{a}+\bar{b}+\bar{c}=\bar{0},$ find $\left| \bar { c } \right|.$

$\overline {a}+\overline {b}+\overline {c}=\overline {0}$

$\therefore -\overline {c}=\overline {a}+\overline {b}$
Taking dot product of both sides with itself, we get

$(\overline {-c}).(\overline {-c}) = (\overline {a}+\overline {b}).(\overline {a}+\overline {b})$

$\therefore { \left| \overline { c } \right| }^{ 2 }=\overline {a}.(\overline {a}+\overline {b})+\overline {b}.(\overline {a}+\overline {b})$

$\overline {a}.\overline {a} + \overline {a}.\overline {b}+\overline {b}\overline {a}+\overline {b}\overline {b}$

$={ \left| \overline { a } \right| }^{ 2 }+0+0+{ \left| \overline { b } \right| }^{ 2 }\ ... [\overline { a }.\overline { b }=\overline { b }.\overline { a }=0]$

$1+1=2\ ...[{ \left| \overline { a } \right| }={ \left| \overline { b } \right| }=1]$

$\therefore \left| \overline { c } \right| = \sqrt {2}$

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $\overline {a}.(\overline {b} \times \overline {c})$

This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

$ABCD$ is a trapezium with $AB$ parallel to $DC$ and $DC=3AB$ . $M$ is the midpoint of $DC.\ \overline {AB}=\overline {p},\ \overline {BC}=\overline {q}.$
Find in terms of $\overline {p}$ and $\overline {q}$ :
$\overline {BD}$

$DC$ is parallel to

$AB$ and

$DC=3AB$

$\because \overline {AB} = \overline {p}\ \therefore \overline {DC}=3AB.$

$M$ is the midpoint of

$DC$

$\therefore \overline {DM} = \overline {MC} = \dfrac {1}{2} \overline {DC}=\dfrac {3}{2} \overline {p}$

$(i)\ \overline {AM} = \overline {AM}+\overline {BC}+\overline {CM}$

$\overline {AB}+\overline {BC}-\overline {MC}$

$\overline {p}+\overline {q}-\overline {\dfrac {3p}{2}}$

$\overline {q}-\dfrac {1}{2}\overline {p}$

$(ii)\ \overline {BD}=\overline {BC}+\overline {CD}=\overline {BC}-\overline {CD}=\overline {q}-3\overline {p}$
1808587_1846450_ans_c76fe9e2805f4ba4a468ca3e47782860.png

1808587_1846450_ans_c76fe9e2805f4ba4a468ca3e47782860.png

If $D,\ E,\ F$ are the midpoints of the sides $BC,\ CA,\ AB$ of a
triangle $ABC,$ prove that $\overline {AD}+\overline {BE}+ \overline {CF}=\overline {0}.$

Let

$\bar {a},\ \bar {b},\ \bar {c},\ \bar {d},\ \bar {e},\ \bar {f}$ be the position vectors of the points

$A,\ B,\ C,\ D,\ E,\ F$ respectively.
Since

$D,\ E,\ F$ are the midpoints of

$BC,\ CA,\ AB$ respectively, by the midpoint formula

$\bar {d} = \dfrac {\bar {b}+\bar {c}}{2},\ \dfrac {\bar {c}+\bar {a}}{2},\ \dfrac {\bar {a}+\bar {b}}{2},$

$\therefore \overline {AD} + \overline {BE} + \overline {CF} = (\bar {d} - \bar {a})+(\bar {e} - \bar {b})+(\bar {f} - \bar {c})$

$\left(\dfrac {\bar {b}+\bar {c}}{2} - \bar {a}\right) + \left(\dfrac {\bar {c}+\bar {a}}{2} - \bar {b}\right) + \left(\dfrac {\bar {a}+\bar {b}}{2} - \bar {c}\right)$

$=\dfrac {1}{2} \bar {b} + \dfrac {1}{2} \bar {c} - \bar {a} + \dfrac {1}{2} \bar {c} + \dfrac {1}{2} \bar {a} - \bar {b} + \dfrac {1}{2} \bar {a} + \dfrac {1}{2} \bar {b} - \bar {c}$

$=(\bar {a}+\bar {b}+\bar {c})-(\bar {a}+\bar {b}+\bar {c})=\bar {0}.$

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $\left| \bar { a } \right| (\overline {b} . \overline {c})$

This is the product of two scalars. Therefore, this expression is meaningful and it is a scalar.

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $\bar { a }.\bar { b }+\bar { c }$

This is the sum of scalar and vector which is not defined. Therefore, this expression is meaningless.

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $(\overline {a} . \overline {b}). \overline {c}$

This is meaningless because

$\bar {c}$ is a vector,

$\bar {a}.\bar {b}$ scalar and scalar product of vector and scalar is not defined.

Prove that $(\overline {a} \times \overline {b}).(\overline {c} \times \overline {d}) = \begin{vmatrix} \bar { a } .\bar { c } & \bar { b } .\bar { c } \\ \bar { a } .\bar { d } & \bar { b } .\bar { d } \end{vmatrix}.$

Let

$\bar {a} \times \bar {b} = \bar {m}$

$LHS= (\bar {a} \times \bar {b}) . (\bar {c} \times \bar {d})$

$= \bar {m}. (\bar {c} \times \bar {d})$

$. (\bar {m} \times \bar {c}). \bar {d}$

$\left[(\bar {a} \times \bar {b})\times \bar {c}\right].\bar {d}$

$\left[(\bar {c} . \bar {a})\bar {b} - (\bar {c} . \bar {b})\bar {a}\right].\bar {d}$

$(\bar {c}.\bar {a})(\bar {b}.\bar {d})-(\bar {c}.\bar {b})(\bar {a}.\bar {d})$

$= \begin{vmatrix}\bar {c}.\bar {a} & \bar {c}.\bar {b} \\ \bar {a}.\bar {d} & \bar {b}.\bar {d}\end{vmatrix} \cdot$

$= \begin{vmatrix}\bar {a}.\bar {c} & \bar {b}.\bar {c}\\ \bar {a}.\bar {d} & \bar {b}.\bar {d}\end{vmatrix} \cdot$ ...[dot product is commutative]

$=RHS.$

A person going for a walk follows the path shown in above figure. The total trip consists of four straight line paths. At the end of the walk, what is the persons resultant displacement measured from the starting point.

On our version of the diagram we have drawn in the resultant from the tail of the first arrow to the head of the last arrow.
The resultant displacement

$\vec{R}$ is equal to the sum of the four individual displacements,

$\vec{R}=\vec{d}_{1}+\vec{d}_{2}+\vec{d}_{3}+\vec{d}_{4}$ . We translate from the pictorial representation to a mathematical representation by writing the individual displacements in
unit-vector notation:

$\vec{d}_{1} = 100\hat{i} m$

$\vec{d}_{2}= −300\hat{j} m$

$\vec{d}_{3} = (−150 \cos 30^{0})\hat{i} m+ ( − 150 \sin 30^{0})\hat{j} m = –130\hat{i} m − 75\hat{j} m$

$\vec{d}_{4}= (− 200 \cos60^{0})\hat{i} m + (200 \sin60^{0})\hat{j} m = − 100\hat{i} m + 173\hat{j} m$
Summing the components together, we find

$R_{x} = d_{1x} + d_{2x} + d_{3x} + d_{4x} = (100 + 0 − 130 − 100) m = − 130m$

$R_{y} = d_{1y} + d_{2y} + d_{3y} + d_{4y} = (0 − 300 − 75 + 173) m = –202 m$
so altogether

$\vec{R}=\vec{d}_{1}+\vec{d}_{2}+\vec{d}_{3}+\vec{d}_{4} = (−130\hat{i} − 202\hat{j})m$
Its magnitude is

$|\vec{R}|=\sqrt{(-130)^{2}+(-202)^{2}}=240m$
We calculate the angle

$\phi=\tan^{-1}\left ( \dfrac{R_{y}}{R_{x}} \right )=\tan^{-1}\left ( \dfrac{-202}{-130} \right )=57.2^{0}$
The resultant points into the third quadrant instead of the first quadrant. The angle counter clockwise from the

$+x$ axis is

$\theta = 180 +\phi = 237^{0}$
1799892_1859696_ans_36381d20d5db42008a66a8c52990b9f3.png

Suppose $\overline {a}\neq \overline {0}:$ If $\overline {a}.\overline {b}=\overline {a}.\overline {c},$ then is
$\overline {b}=\overline {c} \ ?$ .

Take

$\bar a=\hat i, \bar {b}=\hat {j},\ \bar {c} = \hat {k}$

Then

$\bar {a} . \bar {b} = \bar {a} . \bar {c} = 0,$ but

$\bar {b} \neq \bar {c}$

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $(\overline {a} \times \overline {b}) . (\overline {c} \times \overline {d})$

This is scalar products of two vectors. Therefore, this expression is meaningful and it is a scalar.

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $(\overline {a} . \overline {b}) \overline {c}$

This is a scalar multiplication of a vector. Therefore, this expression is meaningful and it is a vector.

State whether the expression is meaningful. If not, explain why? If so,
state whether it is a vector or a scalar : $\bar { a }.(\overline {b} + \overline {c})$

This is the scalar product of two vectors. Therefore, this expression is meaningful and it is a scalar.

Suppose $\overline {a}=\overline {0}:$ If $\overline {a}.\overline {b}=\overline {a}.\overline {c},$ and $\overline {a}\times \overline {b}=\overline {a}\times \overline {c}$
then is $\overline {b}=\overline {c} \ ?$ .

Take

$\bar {a}=\bar {0},\ \bar {b} = \hat {i},\ \bar {c} = \hat {j}$

Then

$\bar {a} . \bar {b} = \bar {a} . \bar {c} = 0,$ and

$\bar {a} \times \bar {b} = \bar {a} \times \bar {c} = 0,$ but

$\bar {b} \neq \bar {c}$

Find the sum of the vector
$\overrightarrow {a} = \hat {i} - 2 \hat {j} + \hat {k},\ \ \overrightarrow {b} = -2\hat {i} + 4 \hat {j} + 5 \hat {k}$ and $\overrightarrow {c} = \hat {i} - 6 \hat {j} - 7 \hat {k}$

$\bar{a} +\bar{b}+\bar{c}=i(1-2+1)+j(-2+4-6)+k(1+5-7) =-4\hat{j}-\overrightarrow {k}$

Find the position vector of a vector of a point R which divides the line joining two points P and Q whose position vectors are $( 2a + b)$ and $9 \overrightarrow {a} - 3 \overrightarrow {b})$ externally in the ratio $1 : 2$ also show that P is the mid point of the line segment $RQ$

$( 2a + \bar {b}) \quad \quad ( \bar {a} - 3 \bar {b})$

$\frac { 1( 7bar {a} - 3 \bar {b}) - 2( 2 \bar {a} + \bar {b}) }{ 1 -2 } , \frac { -3 \bar {a} - 5 \bar {b} }{-1} = 3 \bar {a} + 5 \bar {b}$

$\therefore$ p.v. of R

$( 3 \bar {a} + 5 \bar {b} )$
1954929_1860696_ans_a1e1882d248c4c8c88050551664c4be7.png

Find $| \overrightarrow {a} |$ and $| \overrightarrow {b} |$ , if $( \overrightarrow {a} + \overrightarrow {b} ) . ( \overrightarrow {a} - \overrightarrow {b} ) = 8$ and $| \overrightarrow {a} | = 8 | \overrightarrow {b} |$

${\textbf{Step-1: Solve given equation.}}$

${\text{Given,}}$

$\Rightarrow (\bar a +\bar b).(\bar a −\bar b)=8$

$\Rightarrow \bar a(\bar a -\bar b)+ \bar b.(\bar a −\bar b)=8$

$\Rightarrow \bar a. \bar a- \bar a. \bar b+ \bar b. \bar a= \bar b. \bar b=8$

$\Rightarrow \bar a. \bar a- \bar a. \bar b+ \bar a. \bar b- \bar b. \bar b=8$

${\text{(Using property:}}$

$\bar a. \bar b = \bar b . \bar a)$

$\Rightarrow \bar a. \bar a- \bar b. \bar b=8$

$\Rightarrow |\bar a|^2 - |\bar b|^2 =8$

${\text{(Using property:}}$

$\bar a. \bar a = | \bar a|^2)......(1)$

${\textbf{Step-2: Apply given condition.}}$

$|\bar a| = 8 |\bar b|$

${\text{Put in equation (1)}}$

$\Rightarrow |\bar a|^2 - |\bar b|^2 =8$

$\Rightarrow (8 |\bar b|)^2 - |\bar b|^2 =8$

$\Rightarrow 64 |\bar b|^2 - |\bar b|^2 =8$

$\Rightarrow 63 |\bar b|^2=8$

$\Rightarrow |\bar b|^2=\dfrac{8}{63}$

$\Rightarrow |\bar b|=\dfrac{\sqrt 8}{\sqrt 63}$

$\Rightarrow |\bar b|=\dfrac{2\sqrt 2}{3\sqrt 7}$

${\text{Also,}}$

$|\bar a| = 8 |\bar b|$

$\Rightarrow |\bar a| = 8\times \dfrac{2\sqrt 2}{3\sqrt 7}$

$\Rightarrow |\bar a| = \dfrac{16\sqrt 2}{3\sqrt 7}$

${\textbf{ Hence,}}$

${\mathbf{ |\bar a| = \dfrac{16\sqrt 2}{3\sqrt 7}}}$

${\textbf{and}}$

${\mathbf{|\bar b|=\dfrac{2\sqrt 2}{3\sqrt 7}}}.$

$\overrightarrow {a} = \overrightarrow {b} + \overrightarrow {c}$ then is it true that $| \overrightarrow {a} | = | \overrightarrow {b} |+ | \overrightarrow {c} |$ justify your answer.

${\textbf{Step-1: Find relation with help of diagram.}}$

${\text{Given,}}$

${\text{In}}$

$ΔABC,$

${\text{let}}$

$\bar CB=\bar a, \bar CA=\bar b,$

${\text{and}}$

$\bar AB=\bar c$

${\text{(as shown in the following figure).}}$

${\text{Now, by the triangle law of vector addition, we have}}$

$\bar a= \bar b+\bar c.$

${\textbf{Step-2: Show true or false for that given condition.}}$

${\text{It is clearly known that}}$

$∣\bar a∣,∣\bar b∣,$

${\text{and}}$

$∣\bar c∣$

${\text{represent the sides of ΔABC.}}$

${\text{Also, it is known that the sum of the lengths of any two sides}}$

${\text{of a triangle is greater than the third side.}}$

$∴∣\bar a∣<∣\bar b∣+∣\bar c|$

${\text{Hence, it is not true that}}$

$∣\bar a∣=∣\bar b∣+∣\bar c∣.$

${\textbf{ Hence, it is not true that}}$

${\mathbf{∣\bar a∣=∣\bar b∣+∣\bar c∣.}}$

if either vector $\overrightarrow {a} = \overrightarrow {0}$ or $\overrightarrow {b} = \overrightarrow {0}$ then $\overrightarrow {a}. \overrightarrow {b} = \overrightarrow {0}$ but the converse need not be true. justify your answer with an example.

$\overrightarrow {i}.\overrightarrow {j} =|i|.|j|. cos 90$

$\overrightarrow {i}.\overrightarrow {j}=0$ but

$|i|=1, |j|=1$

i.e.: neither

$\overrightarrow {i} =0$ nor

$\overrightarrow {j}=0$

Given that $\overrightarrow {a} . \overrightarrow {b} = 0$ and $\overrightarrow {a} \times \overrightarrow {b} = \overrightarrow {0 }$ when can you conclude about the vectors $\overrightarrow {a}$ and $\overrightarrow {b}$

${\textbf{Step-1: Conclude about the given vectors(First condition).}}$

${\text{Given,}}$

$\bar a⋅\bar b=0$

${\text{Then,}}$

${\text{(i) Either}}$

$∣\bar a∣=0$

${\text{or}}$

$∣\bar b∣=0,$

${\text{or}}$

$a ⊥ b$

${\text{(in case a and b are non-zero)}}$

$\bar a×\bar b=0$

${\textbf{Step-2: Conclude about the given vectors(Second condition).}}$

${\text{(ii) Either ∣ a∣=0 or ∣ b∣=0, or a∥b (in case a and b are non-zero)}}$

${\text{But, a and b cannot be perpendicular and parallel simultaneously.}}$

${\text{Hence,}}$

$∣\bar a∣=0\text{ or }∣\bar b∣=0.$

${\textbf{Hence , }}$

${\mathbf{\bar a⋅\bar b=0}}$

${\textbf{and}}$

${\mathbf{\bar a×\bar b=0}}$

${\textbf{are conclude that}}$

${\mathbf{\bar a}}$

${\textbf{and}}$

${\mathbf{\bar b}}$

${\textbf{are}}$

${\mathbf{∣\bar a∣=0\textbf{ or }∣\bar b∣=0.}}$

Evaluate the product
$(3\overrightarrow { a } -5\overrightarrow { b } ).(2\overrightarrow { a } +7\overrightarrow { b } )$

$3\times 2 \bar{a}.\bar{a}+3\times 7 \bar{a}.\bar{b} -5 \times 2 \bar{b}.\bar{a}-5\times 7 \bar{b}.\bar{b}$

$6|a|^2+11 \bar{a}.\bar{b}-35|b|^2$

Find the $| \overrightarrow {x} |$ if a is a unit vector $\overrightarrow {a} , ( \overrightarrow {x} - \overrightarrow {a} )$ . $( \overrightarrow {x} + \overrightarrow {a}) = 12$

${\textbf{Step-1: Find x.}}$

${\text{Given,}}$

$( \bar x− \bar a)⋅(\bar x+\bar a)=12$

$⇒ \bar x.\bar x+\bar x⋅\bar a−\bar a⋅\bar x−\bar a⋅\bar a=12$

$⇒∣\bar x∣^2−∣\bar a∣^2=12$

$⇒∣ x∣^2−1=12,$

${\text{[∵∣a∣=1 as a is a unit vector]}}$

$⇒∣ x∣^2=13$

$⇒∣ \bar x∣=\sqrt13$

${\textbf{ Hence, }}$

${\mathbf{∣ \bar x∣=\sqrt{13}.}}$

If magnitude of two vectors be $4$ and $5$ units, then find scalar product of them, where angle between them be:
$60^{o}$

Let two vectors are

$\vec{a}$ and

$\vec{b}$ and

$\theta$ be the angle between them, then

$\cos \theta=\dfrac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}$

when

$\theta=60^{o}$ and

$|\vec{a}|=4, |\vec{b}|=5$

then

$\cos 60^{o}=\dfrac{\vec{a}.\vec{b}}{4\times 5}$

$\Rightarrow \dfrac{1}{2}=\dfrac{\vec{a}.\vec{b}}{20}$

$\Rightarrow \vec{a}.\vec{b}=\dfrac{20}{2}$

$\Rightarrow \vec{a}.\vec{b}=10$

$\therefore$ Scalar product of vectors

$=10$ .

Find $\overrightarrow {a}. \vec{b}$ , if $\overrightarrow{a}$ and $\vec{b}$ are as follows:
$5\hat{i}+\hat{j}-2\hat{k}; 2\hat{i}-3\hat{j}$

Given that

$\vec{a}=5\hat{i}+\hat{j}-2\hat{k}$
and

$\vec{b}=2\hat{i}-\hat{j}$
Given vectors can be written as

$\vec{a}=5\hat{i}+\hat{j}-2\hat{k}$
and

$\vec{b}=2\hat{i}-3\hat{j}+0\hat{k}$
Now

$\vec{a}.\vec{b}=(5\hat{i}+\hat{j}-2\hat{k}).(2\hat{i}-3\hat{j}+0\hat{k})$

$=(5\times 2)+\left\{1\times (-3)\right\}+\left\{(-2)\times 0\right\}$

$=10-3+0$

$\therefore \vec{a}.\vec{b}=7$ .

Find $\bar {a}. \bar {b}$ of $| \bar {a}| = 3 , | \bar {b} | = 4 , | \bar {a} \times b| = 6$

$| \bar {a} \times \bar {b}| = 6 \Rightarrow | \bar {a} \times \bar {b}|^2 = | a|^2 |b|^2 sin^2 \theta$

$= |a|^2 \quad |b|^2 - |a|^2 \quad |b|^2 cos^2 \theta$

$= |a|^2 \quad |b|^2- ( \bar {a} . \bar {b})^2 \Rightarrow 36 = 9 \times 16 - ( \bar {a} . \bar {b})^2$

$\Rightarrow ( \bar {a}.\bar {b})^2 = 144-36= 108$

$\Rightarrow ( \bar {a} - \bar {b})^2 = \sqrt {108} = 6 \sqrt {3}$

Find $\overrightarrow {a}. \vec{b}$ , if $\overrightarrow{a}$ and $\vec{b}$ are as follows:
$4\hat{i}+3\hat{k}; \hat{i}-\hat{j}+\hat{k}$

Given that

$\vec{a}=4\hat{i}+3\hat{k}$
and

$\hat{b}=\hat{i}-\hat{j}+\hat{k}$
Given vectors can be written as

$\vec{a}=4\hat{i}+0\hat{j}+3\hat{k}$
and

$\vec{b}=\hat{i}-\hat{j}+\hat{k}$
Now

$\vec{a}.\vec{b}=(4\hat{i}+0\hat{j}+3\hat{k}).(\hat{i}-\hat{j}+\hat{k})$

$=(4\times 1)+\left\{0\times (-1)\right\}+(3\times 1)$

$=4+0+3$

$\therefore \vec{a}.\vec{b}=7$ .

If magnitude of two vectors be $4$ and $5$ units, then find scalar product of them, where angle between them be:
$90^{o}$

Let two vectors are

$\vec{a}$ and

$\vec{b}$ and

$\theta$ be the angle between them, then

$\cos \theta=\dfrac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}$

when

$\theta=90^{o}$ and

$|\vec{a}|=4, |\vec{b}|=5$

then

$\cos 90^{o}=\dfrac{\vec{a}.\vec{b}}{4\times 5}$

$\Rightarrow 0=\dfrac{\vec{a}.\vec{b}}{10}$

$\Rightarrow \vec{a}.\vec{b}=0$

$\therefore$ Scalar product of vectors

$=0$

Write the projection of $i - j$ on $i + j$

projection of

$\overrightarrow {a}$ on

$\overrightarrow {b}$ is

$\frac {\overrightarrow {a} . \overrightarrow {b}}{|\overrightarrow {b}|} = \frac {1 \times - 1 \times 1}{\sqrt { 1+1 }} = 0$

Write the value of $( i \times j) . \bar {k} + i . j$

$\overrightarrow {i} \times \overrightarrow {j} = \overrightarrow {k}$ and

$\overrightarrow {i} . \overrightarrow {j} = 0$ and

$\overrightarrow {k} . \overrightarrow {k} = | k|^2$

$\therefore ( \overrightarrow {i} \times \overrightarrow {j}). \overrightarrow {k} + \overrightarrow {i} . \overrightarrow{j} =( \overrightarrow {k} . \overrightarrow {k}) +0 = | k|^2 = 1$

If magnitude of two vectors be $4$ and $5$ units, then find scalar product of them, where angle between them be:
$30^{o}$

Let two vectors are

$\vec{a}$ and

$\vec{b}$ and

$\theta$ be the angle between them, then

$\cos \theta=\dfrac{\vec{a}.\vec{b}}{|\vec{a}|.|\vec{b}|}$
When

$\theta=30^{o}$ and

$|\vec{a}|=4, |\vec{b}|=5$
then

$\cos 30^{o}=\dfrac{\vec{a}.\vec{b}}{4\times 4}$

$\Rightarrow \dfrac{\sqrt{3}}{2}=\dfrac{\vec{a}.\vec{b}}{20}$

$\Rightarrow \vec{a}.\vec{b}=\dfrac{20\sqrt{3}}{2}$

$\Rightarrow \vec{a}.\vec{b}=10\sqrt{3}$

$\therefore$ Scalar product of vectors

$=10\sqrt{3}$

Find $\overrightarrow {a}. \vec{b}$ , if $\overrightarrow{a}$ and $\vec{b}$ are as follows:
$2\hat{i}+5\hat{j}; 3\hat{i}-2\hat{j}$

$\vec{a}=2\hat{i}+5\hat{j}$ and

$\vec{b}=3\hat{i}-2\hat{j}$

The given vectors can be written as

$\vec{a}=2\hat{i}+5\hat{j}+0\hat{k}$

and

$\vec{b}=3\hat{i}-2\hat{j}+0\hat{k}$

Now

$\vec{a}.\vec{b}=(2\hat{i}+5\hat{j}+0\hat{k}).(3\hat{i}-2\hat{j}+0\hat{k})$

$=(2\times 3)+\left\{5\times (-2)\right\}+(0\times 0)$

$=6-10+0$

$\therefore \vec{a}.\vec{b}=-4$

Prove that:
$|\vec{a}.\vec{b}|^{2}\le |\vec{a}|^{2} |\vec{b}|^{2}$

$\vec{a}=0$ or

$\vec{b}=0$ , then
identify

$|\vec{a}.\vec{b}|=0=|\vec{a}||\vec{b}|$
But here let

$|\vec{a}|\neq 0\neq |\vec{b}|$
Then we know that

$\vec{a}.\vec{b}=|\vec{a}||\vec{b}|\cos \theta$
[where

$\theta$ be the angle between

$\vec{a}$ and

$\vec{b}$ ]
or

$\dfrac{|\vec{a}.\vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta|\le 1[\because -1\le \cos \theta \le 1]$

$\therefore |\vec{a}.\vec{b}|\le |\vec{a}||\vec{b}|$
Squaring on both sides

$|\vec{a}.\vec{b}|^{2}\le |\vec{a}|^{2}|\vec{b}|^{2}$
1860266_1872857_ans_fcc7fa3f70e64a6185dcb0929c938ca3.png

Prove that
$\begin{bmatrix} \hat i & \hat j & \hat k\end{bmatrix}+\begin{bmatrix} \hat i & \hat k & \hat j\end{bmatrix}=0$

$\begin{bmatrix} \hat i & \hat j & \hat k\end{bmatrix}+\begin{bmatrix} \hat i & \hat k & \hat j\end{bmatrix}=0$

$L.H.S =\begin{bmatrix} \hat i & \hat j & \hat k\end{bmatrix}+\begin{bmatrix} \hat i & \hat k & \hat j\end{bmatrix}$

$=\hat i. ( \hat j \times \hat k )+\hat i . ( \hat k \times \hat j )$

$=\hat i. \hat i +\hat i. (-\hat i)$

$=1-1$

$=0$

$=R.H.S$

If for a vetoer $\vec a, (\vec x-\vec a). (\vec x +\vec a)=12$ , then $|\vec x|$ .

According to question,

$|\vec a|=1$
and

$(\vec x-\vec a).(\vec x+\vec a)=12$

$\Rightarrow \vec x.(\vec x+\vec a)-\vec a(\vec x+\vec a)=12$

$\Rightarrow \vec x.\vec x+\vec x.\vec a-\vec a.\vec x-\vec a.\vec a=12$

$\Rightarrow \ \vec |\vec x|^2 -|\vec a|^2 =12\ (\because \vec x. \vec a =\vec a. \vec x)$

$\Rightarrow |\vec x|^2 -1=12$

$\Rightarrow |\vec x|^2 =12+1=13$

$\therefore |\vec x|=\sqrt {13}$

If vertices $\vec a, \vec b, \vec c$ are such that
$\vec a+\vec b+\vec c=\vec 0$
then, find the value of $\vec a. \vec b+\vec b. \vec c+\vec c. \vec a$ .

According to question

$\vec a+\vec b+\vec c=\vec 0$

$\therefore \vec a. (\vec a+\vec b+\vec c)=\vec a. \vec 0$

$\Rightarrow \vec a.\vec a+\vec a.\vec b+\vec a.\vec c=0$

$\Rightarrow |\vec a|^2 +\vec a.\vec b+\vec a.\vec c=0$

$\Rightarrow 1+\vec a.\vec b+\vec a.\vec c=0 \ (\because |\vec a|=1)....(1)$

Again

$\vec b, (\vec a+\vec b+\vec c)=\vec b, \vec 0$

$\Rightarrow \vec b. \vec a+\vec b. \vec b+\vec b.\vec c=0$

$\Rightarrow \vec a.\vec b+|\vec b|^2 +\vec b. \vec c=0$

$\Rightarrow \vec a. \vec b+1+\vec b. \vec c=0\quad \begin{bmatrix} \because |\vec { b } |=1 \\ { \vec { a } .\vec { b } = }\vec { b } .\vec { a } \end{bmatrix}$

and

$\Rightarrow \vec c. (\vec a+\vec b+\vec c)=\vec c. \vec 0$

$\Rightarrow \vec c. \vec a+\vec c. \vec b+ \vec c. \vec c=0$

$\Rightarrow \vec a.\vec c+\vec b.\vec c+|\vec c|^2 =0$

$\Rightarrow \vec a.\vec c+\vec b.\vec c+1=0\quad \begin{bmatrix} \because \vec { a } .\vec { c } =\vec { c } .\vec { a } \\ { \vec { c } .\vec { b } = }\vec { b } .\vec { c } ,\vec { c } =1 \end{bmatrix}.....(3)$

On adding (1), (2) and (3)

$1+\vec a.\vec b+\vec a.\vec c+\vec a.\vec b+1+\vec b.\vec c+\vec a.\vec c+\vec b.\vec c+1=0$

$\Rightarrow 2[\vec a.\vec b+\vec b.\vec c+\vec a.\vec c]+3=0$

$\Rightarrow \vec a.\vec b+\vec b.\vec c+\vec c.\vec a=-\dfrac 32\ (\because \vec a.\vec a\vec c=\vec c.\vec a)$

Hence

$\vec a.\vec b+\vec b.\vec c+\vec c.\vec a=-\dfrac 32$

Find the projection of the vector
$4\hat{i}-2\hat{j}+\hat{k}$
on the vector
$3\hat{i}+6\hat{j}-2\hat{k}$

Let

$\vec{a}=4\hat{i}-2\hat{j}+\hat{k}$

and

$\vec{b}=3\hat{i}+6\hat{j}-2\hat{k}$

$\therefore$ Projection of

$\vec{a}$ on

$\vec{b}$

$=\dfrac{\vec{a}.\vec{b}}{|\vec{b}|}$

$=\dfrac{(4\hat{i}-2\hat{j}+\hat{k}).(3\hat{i}+6\hat{j}-2\hat{k})}{|3\hat{i}+6\hat{j}-2\hat{k}|}$

$=\dfrac{(4\times 3)+\left\{ (-2)\times 6\right\}+\left\{1+(-2)\right\}}{\sqrt{(3)^{2}+(6)^{2}+(-2)^{2}}}$

$=\dfrac{12-12-2}{\sqrt{9+36+4}}$

$=\dfrac{-2}{\sqrt{49}}$

$=\dfrac{2}{7}$ (

$\because$ Projection is never negative)

Hence, projection of

$4\hat{i}-2\hat{j}+\hat{k}$ on

$3\hat{i}+6\hat{j}-2\hat{k}$ is

$\dfrac{2}{7}$ .

Find projection vector on vector
$\hat i +3\hat j +7\hat k$
on vector
$7\hat i -\hat j+8\hat k$

Let

$\vec a=\hat i+3\hat j+7\hat k$
and

$\vec b=7\hat i-\hat j+8\hat k$
Then projection of

$\vec a$ on

$\vec b$

$=\vec a. \dfrac {\vec b}{|\vec b|}$

$=\dfrac {(\hat i+3\hat j+7\hat k). (7\hat i-\hat j+8\hat k)}{|7\hat i+\hat j+8\hat k|}$

$=\dfrac {1\times 7+3\times (-1)+7\times 8}{\sqrt {7^2 +(-1)^2 +8^2}}$

$=\dfrac {7-3+56}{\sqrt {49+1+64}}=\dfrac {60}{\sqrt {114}}$

$\therefore$ Projection of

$\hat i+3\hat j+7\hat k$ on

$7\hat i-\hat j+8\hat k=\dfrac {60}{\sqrt {114}}$

Find the value of
$(3\vec a -5\vec b). (2\vec a +7\vec b)$

$(3\vec a-5\vec b).(2\vec a+7\vec b)$

$=3\vec a.(2\vec a+7\vec b)-5\vec b.(2\vec a+7\vec b)$

$=6(\vec a.\vec a)+21\vec a.\vec b-10\vec b.\vec a-35\vec b.\vec b$

$=6|\vec a|^2 +11\vec a.\vec b-35|\vec b|^2\quad \begin{bmatrix} \because \vec { a } \vec { a } ={ |\vec { a } | }^{ 2 } \\ \vec { b } .\vec { b } ={ |\vec { a } | }^{ 2 } \end{bmatrix}$

$\therefore (3\vec a-5\vec b).(2\vec a+7\vec b)=6|\vec a|^2 +11 \vec a.\vec b-35 |\vec b|^2$

Find the projection of vector
$\hat i-\hat j$ on $\hat i +\hat j$

We know that projection of

$\vec a$ on

$\vec b$

$=\vec a. \dfrac {\vec b}{|\vec b|}$
Let

$\vec a=\hat i +\hat j$ and

$\vec b=\hat i -\hat j$
Then projection of

$\vec b=\hat i -\hat j$ on

$\vec a=\hat i +\hat j$

$=\vec b. \dfrac {\vec a}{|\vec a|}=\dfrac {(\hat i -\hat j)(\hat i +\hat j)}{|\hat i +\hat j|}$

$=\dfrac {\hat i.\hat i+\hat i.\hat j-\hat j.\hat i-\hat j.\hat j}{\sqrt {|2+2|^2}}$

$=\dfrac {|\hat i|^2 +0-0-|\hat j|^2}{\sqrt 2}$

$=\dfrac {1-1}{\sqrt 2}\quad \left[\because |\hat i|=1\\ |\hat j|=1\right]$

$=0$ .

$\therefore$ Projection of

$\hat i-\hat j$ on

$\hat i+\hat j$ is zero.

Use vectors to prove the sum of square of diagonal of a parallelogram is equal to be the sum of square of their side.

Let

$OACB$ is a parallelogram. Taking

$O$ as origin, the position vectors of

$A$ and

$B$ are

$\vec{a}$ and

$\vec{b}$ respectively.
Let

$\vec{OA}=\vec{BC}=\vec{a}$
and

$\vec{OB}=\vec{AC}=\vec{b}$
then

$\vec{AB}=\vec{b}-\vec{a}$

$\vec{OC}=\vec{a}+\vec{b}$

$\therefore \vec{OC}^{2}+\vec{AB}^{2}=(\vec{a}+\vec{b})^{2}+(\vec{b}-\vec{a})^{2}$

$=2(\vec{a^{2}}+\vec{b^{2}})$

$=(\vec{OA^{2}}+\vec{BC^{2}})+(\vec{OB^{2}}+\vec{AC^{2}})$

$\therefore$ Sum of the squares of diagonals= sum of the squares of sides.
Hence the sum of square of diagonals of a parallelogram is equal to the sum of the square of their sides.
1860277_1872891_ans_d156f25cf04d4dea8320a45e6df2f9c5.png

1860277_1872891_ans_d156f25cf04d4dea8320a45e6df2f9c5.png

For any vector $\vec{a}$ , prove that
$\vec{a}=(\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k}$ .

Let

$\vec{a}=a_{1}\hat{i}+a_2 \hat{j}+a_3 \hat{k}$ ....(1)

then

$\vec{a}.\hat{i}=(a_{1}\hat{i}+a_2\hat{j}+a_3\hat{k}).\hat{i}=a_1$ ....(2)

$\begin{bmatrix} \because \hat{i}.\hat{i}=1=\hat{j}.\hat{j} \\ \hat{i}.\hat{j}=\hat{i}.\hat{k=0} \end{bmatrix}$

Similarly

$\vec{a}.\hat{j}=(a_{1}\hat{i}+a_{2}\hat{j}+a_3\hat{k}).\hat{j}=a_2$ .....(3)

and $$\vec{a}.\hat{k}=(a_1\hat{i}+a_2 \hat{j}+a_3 \hat{k}).\hat{k}=a_{3} ....(4)

Now putting the values of

$a_{1}, a_{2}$ and

$a_{3}$ in equation

$(1)$ .

$\vec{a}=(\vec{a}.\hat{i})\hat{i}+(\hat{a}.\hat{j})+(\vec{a}.\hat{k})\hat{k}$

Prove that
$\begin{bmatrix} 2\hat i & \hat j & \hat k\end{bmatrix}+\begin{bmatrix} \hat i & \hat k & \hat j\end{bmatrix}+\begin{bmatrix} \hat k & \hat j & 2\hat i\end{bmatrix}=0$

$L.H.S=\begin{bmatrix} 2\hat i & \hat j & \hat k\end{bmatrix}+\begin{bmatrix} \hat i & \hat k & \hat j\end{bmatrix}+\begin{bmatrix} \hat k & \hat j & 2\hat i\end{bmatrix}$

$=2\hat i . ( \hat j \times \hat k )+\hat i.( \hat k\times \hat j)+\hat k.( \hat j \times 2\hat i)$

$=2\hat i . \hat i+\hat i. (-\hat i)+\hat k.(-2\hat k)$

$[ \because \hat k \times \hat j=-\hat i$ and

$\hat j \times \hat i=-\hat k]$

$=2-\hat i. \hat i -2\hat k. \hat k$

$=2-1-2$

$=-1$

$=R.H.S$

If $\vec{a} = 7 \hat{i} + \hat{j} - 4 \hat{k}$ and $\vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k}$ , then find the projection of $\vec{a}$ on $\vec{b}$ .

Given: \vec{a} = 7\hat{i} + \hat{j} - 4\hat{k}$$

$\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$

The projection of

$\vec{a}$ on

$\vec{b}$ is given by

$P = \dfrac{1}{|\vec{b}|} (\vec{a} \cdot \vec{b} )$

$= \dfrac{14+6-12}{\sqrt{(2)^2 + (6)^2 + (3)^2}}$

$= \dfrac{8}{\sqrt{4+36+9}}$

$\Rightarrow \boxed{P = \dfrac{8}{7}}$ Ans

If two adjacent sides of a triangle are represented by vectors
$\hat i+2\hat j+2\hat k$
and
$3\hat i-2\hat j+\hat k$ ,
then find the area of triangle.

$\dfrac{1}{2}|(\hat{i}+2\hat{j}+2\hat{k})\times (3\hat{i}-2\hat{j}+\hat{k})|$

$=\dfrac{1}{2}\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & 2 \\ 3 & -2 & 1 \end{vmatrix}$

$=\dfrac{1}{2}[\hat{i}(2+4)-\hat{j}(1-6)+\hat{k}(-2-6)]$

$=\dfrac{1}{2}|(6\hat{i}+5\hat{j}-8\hat{k})|$

$\therefore$ Magnitude of vector area of triangle

$=\dfrac{1}{2}\sqrt{(6)^{2}+(5)^{2}+(-8)^{2}}$

$=\dfrac{1}{2}\sqrt{36+25+64}$

$=\dfrac{1}{2}\sqrt{125}$

$\therefore$ Area of triangle

$=\dfrac{5}{2}\sqrt{5}sq.units$ .

For any vector $\vec a$ , prove that
$|\vec a\times \hat i|^2+|\vec a\times \hat j|^2+|\vec a\times \hat k|^2=2|\vec a|^2$

$\because (\vec{a}\times \hat{i})^{2}=(\vec{a}\times \hat{i}).(\vec{a}\times \hat{i})$

$=\begin{vmatrix} \vec{a}.\vec{a} & \vec{a}.\hat{i} \\ \hat{i}.\vec{a} & 1 \end{vmatrix}$

$=|\vec{a}|^{2}-a_{1}^{2}$
where

$\vec{a}=a_{1}\hat{i}+a_{2}\hat{j}+a_{3}\hat{k}$
and

$|\vec{a}|^{2}=a_{1}^{2}+a_{2}^{2}+a_{3}^{2}$
Similarly

$(\vec{a}\times \vec{j})^{2}=|\vec{a}|^{2}-a_{2}^{2}$
and

$(\vec{a}\times \vec{k})^{2}=|\vec{a}|^{2}-a_{3}^{2}$

$L.H.S.=(\vec{a}\times \hat{i})^{2}+(\vec{a}\times \hat{j})^{2}+(\vec{a}\times \hat{k})^{2}$

$=|\vec{a}|^{2}-a_{1}^{2}+|\vec{a}|^{2}-a_{2}^{2}+|\vec{a}|^{2}-a_{3}^{2}$

$=3|\vec{a}|^{2}-(a_{1}^{2}+a_{2}^{2}+a_{3}^{2})$

$=3|\vec{a}|^{2}-|\vec{a}|^{2}=2|\vec{a}|^{2}$

$=2|\vec{a}|^{2}=R.H.S.$
So,

$(\vec{a}\times \hat{i})^{2}+(\vec{a}\times \hat{j})^{2}+(\vec{a}\times \hat{k})^{2}=2|\vec{a}|^{2}$

If $\vec a=2\hat i -3\hat j +4\hat k, \vec b=\hat i+2\hat j -\hat k$ and $\vec c=3\hat i -\hat j+2\hat k$ , then find $\begin{bmatrix} \vec a & \vec b & \vec c \end{bmatrix}$ .

Given

$\vec a=2\hat i -3\hat j +4\hat k$ ,

$\vec b=\hat i+2\hat j -\hat k$

$\vec c=3\hat i -\hat j+2\hat k$

$\therefore \begin{bmatrix} \vec { a } & \vec { b } & \vec { c } \end{bmatrix}=\begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix}$

$=2(4-1)+3(2+3)+4(-1-6)$

$=6+15-28$

$=21-28$

$=-7$
Hence

$\begin{bmatrix} \vec { a } & \vec { b } & \vec { c } \end{bmatrix}=-7$ .

Match the statement of Column I with values of Column II.

$a \cdot b = b \cdot c = c \cdot a = 0$
Also,

$|a| = |b| = |c| = \lambda$
Consider

$a \times [(x - b) \times a]$

$= (a \cdot a) (x - b) - \{ a \cdot (x - b) \} a$

$= (a \cdot a) x - (a \cdot a) b - \{ (a \cdot x) - a \cdot b \} a$

$= (a \cdot a) x - (a \cdot a) b - (a \cdot x) a + \underset{zero}{\underbrace{(a\cdot b})}a$

$a \times [(x - b) \times a] = (a \cdot a) x - (a \cdot a) b - (a \cdot x) a$
Similarly,

$b \times [(x - c) \times b] = (b \cdot b) x - (b \cdot b) c - (b \cdot x) b$
and

$c \times [(x - a) \times c] = (c \cdot c) x - (c \cdot c) c - (c \cdot x) c$

$\therefore LHS = 3 \lambda^2 x - \lambda^2 (a + b + c) - [(a \cdot x) a + (b \cdot x) b + (c \cdot x)c] = 0$ (i)
Now, a, b and c are non-planar.

$\therefore x = x_1 a + x_2 b + x_3 c$

$\displaystyle \frac{a \cdot x}{\lambda^2} =x_1 , \displaystyle \frac{b \cdot x}{\lambda^2} =x_2$ and

$\displaystyle \frac{c \cdot x}{\lambda^2} = x_3$

$\therefore \lambda^2 x = (a \cdot x) a + (b \cdot x) b + (c \cdot x) c$
On substituting the value in equation (i), we get

$x \displaystyle = \frac{a + b + c}{2}$

$\vec{a}$ and $\vec{b}$ form the consecutive sides of a regular hexagon ABCDEF.
$\vec{AB} = \vec{a}, \vec{BC} = \vec{b}$

$\vec{AB} = \vec{a}, \vec{BC} = \vec{b}$

$\vec{AC} =\vec{AB} + \vec{BC} = \vec{a} + \vec{a}$ (i)

$\vec{AD} = 2 \vec{BC} = 2 \vec{b}$ (ii)
(because AD is parallel to BC and twice its length).

$\vec{CD} = \vec{AD} - \vec{AC} = 2 \vec{b} - (\vec{a} + \vec{b}) = \vec{b} -\vec{a}$

$\vec{FA} = - \vec{CD} = \vec{a} - \vec{b}$ (iii)

$\vec{DE} = - \vec{AB} = - \vec{a}$ (iv)

$\vec{EF} = - \vec{BC} = -\vec{b}$ (v)

$\vec{AE} = \vec{AD} + \vec{DE} = 2 \vec{b} - \vec{a}$ (vi)

$\vec{CE} = \vec{CD} + \vec{DE} = \vec{b} - \vec{A} - \vec{A} = \vec{b} - 2 \vec{a}$ (vii)

Write unit vector in the direction of the sum of vectors:
$\overrightarrow{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{b} =-\hat{i}+\hat{j}+3\hat{k}$ .

Add the vectors

$\vec { a } =2\hat { i } -\hat { j } +2\hat { k } \\$ and

$\vec { b } =--\hat { i } +\hat { j } +3\hat { k } \\$ by writing them in the component form that is

$\vec { a } =\left< 2,-1,2 \right> \\$ and

$\vec { b } =\left< -1,1,3 \right> \\$

Then the sum of the vectors can be computed as follows:

$\vec { a } +\vec { b } =\left< 2-1,-1+1,2+3 \right> \\$

$=\left< 1,0,5 \right>$

Therefore, the sum of the vectors is

$\left< 1,0,5 \right>$ .

Now the unit vector can be obtained as:

$\left\| \vec { a } +\vec { b } \right\| =\sqrt { { \left( 1 \right) }^{ 2 }+{ \left( 0 \right) }^{ 2 }+{ \left( 5 \right) }^{ 2 } } =\sqrt { 1+25 } =\sqrt { 26 }$

Unit Vector:

$\left<\dfrac1{\sqrt{26}}, 0, \dfrac5{\sqrt{26}}\right>$

Find the shortest distance between the lines $\overrightarrow{r}=\hat{i}+2\hat{j}+3\hat{k}+\lambda (2\hat{i}+3\hat{j}+4\hat{k})$ and $\overrightarrow{r}=2\hat{i}+4\hat{j}+5\hat{k}+\mu (4\hat{i}+6\hat{j}+8\hat{k})$ .

Let

$l_1$ and

$l_2$ be the given lines whose equations are

$\overrightarrow{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b}$
and

$\overrightarrow{r}=\overrightarrow{a_2}+2\mu \overrightarrow{b}$

Through the points

$\overrightarrow{a_1}=\hat{i}+2\hat{j}+3\hat{k}$ and

$\overrightarrow{a_2}=2\hat{i}+4\hat{j}+5\hat{k}$ are parallel to

$\overrightarrow{b}=2\hat{i}+3\hat{j}+4\hat{k}$ .

Hence the distance between the lines using the formula :

$\dfrac{|\overrightarrow{b}\times (a_2-a_1)|}{|\overrightarrow{b}|}=\dfrac{|(2\hat{i}+3\hat{j}+4\hat{k})\times (\hat{i}+2\hat{j}+2\hat{k})|}{|\overrightarrow{b}|}$

$=\dfrac{|(2\hat{i}+3\hat{j}+4\hat{k})\times (\hat{i}+2\hat{j}+2\hat{k})|}{|2\hat{i}+3\hat{j}+4\hat{k}|}$

$=\dfrac{\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\1&2&2\end{vmatrix}}{\sqrt{4+9+16}}=\dfrac{|-2\hat{i}+\hat{k}|}{\sqrt{29}}$

Ans.

$=\dfrac{\sqrt{5}}{\sqrt29}$

If $a+b+c=\alpha d, b+c+d=\beta a$ and a, b, c are non-coplanar vectors, then show the $a+b+c+d=0$ .

$\textbf{Hint: Parallel vectors can be depicted as }\mathbf{\vec a=\alpha \vec b}$

$\textbf{Step 1: Calculating}$

$\text{Given that,}$

$a+b+c=\alpha d$

$\implies b+c=\alpha d-a ---(1)$

$\text{Given that,}$

$b+c+d=\beta a$

$\implies b+c=\beta a-d ---(2)$

$\textbf{Step 2: Calculating}$

$\implies \alpha d-a=\beta a-d$

$\implies (\alpha+1)d=(\beta +1)a ---(3)$

$\implies \vec d || \vec a$

$\text{Now let }\alpha =-1 , \beta =-1$

$\implies \text{Equation(3) will be 0}$

$\therefore a+b+c+d=0$

$\textbf{Final Answer: Hence proved}$

Show that the line joining the points P and Q with position vectors $\vec { p } ={ p }_{ 1 }\hat { l } +{ p }_{ 2 }\hat { j } +{ p }_{ 3 }\hat { k }\ and\ \vec { q } ={ q }_{ 1 }\hat { i } +{ q }_{ 2 }\hat { j } +{ q }_{ 3 }\hat { k }$ passes through the origin if $\vec { p } \cdot \vec { q } =\left| \vec { p } \right| \left| \vec { q } \right|.$

$\vec{p} = p_1\hat{i} + p_2\hat{j} + p_3\hat{k}$

$\vec{q} = q_1\hat{i} + q_2\hat{j} + q_3\hat{k}$

Now given,

$\vec{p}.\vec{q} = \left| \vec{p} \right| \left| \vec{q} \right|$

$\Rightarrow \dfrac{\vec{p}.\vec{q}}{\left| \vec{p} \right| \left| \vec{q} \right|} = 1$

$\dfrac{\vec{p}.\vec{q}}{\left| \vec{p} \right| \left| \vec{q} \right|}$ represnts angle between

$\vec{p}$ and

$\vec{q}$ ,

so the angle between them is

$\theta$ then

$cos\theta = 1$

$\theta = 0$

Hence

$\vec{p}$ is parallel to

$\vec{q}$

As position vectors are defined with respect to origin hence point

$\vec{p}$ and point

$\vec{q}$ and origin will be collinear.

Hence, the line joining

$\vec{p}$ and

$\vec{q}$ will pass through origin.

Find $\, | a \, - \, b |$ if the vectors a and b makes an angle of $\displaystyle 120^{\circ} \,and \ | a | \, = \, 3, \, and \, | b | \, = \, 5 \,.$

Here

$|a| = 3$

$|b| = 5$

and angle between them

$\theta = 120^{\circ}$

$\vec {a} . \vec {b} = ab\cos \theta$

$= |a| |b| \cos 120^{\circ}$

$= 3\times 5\times \cos (90 + 30)$

$= 15\times (-\sin 30)$

$= 15\left (-\dfrac {1}{2}\right ) = \dfrac {-15}{2}$

$|a - b| = \sqrt {a^{2} + b^{2} - 2a.b}$

$= \sqrt {9 + 25 - 2\times \left (\dfrac {-15}{2}\right )}$

$= \sqrt {34 + 15}$

$= 7$ .

Calculate the scalar product of the following vectors.
$\displaystyle a \, = \, \left \{- 2, 3, 11 \right \} \, and \, b \, = \, \left \{ 5, 7, -4 \right \}$

Scalar product of

$\vec {a} = -2\hat {i} + 3\hat {j} + 11\hat {k}$

and

$\vec {b} = 5\hat {i} + 7\hat {j} + -4\hat {k}$ is given by:

$\vec {a} . \vec {b} = (-2\hat {i} + 3\hat {j} + 11\hat {k}) . (5\hat {i} + 7\hat {j} - 4\hat {k})$

$= -2.5 + 3.(7) + 11(-4)$

$= -10 + 21 - 44$ [

$\hat {i} .\hat {i} = 1, \hat {j} . \hat {j} = 1. \hat {k} . \hat {k} = 1]$

$= -54 + 21$

$= -33$ .

Find the value of $\displaystyle (2\alpha \, + \, 3\beta) \, . \, (4\alpha \, - \, 6\beta), \, where \, \alpha, \, \beta$ are mutually perpendicular unit vectors.

Given

$|\alpha| = 1$

and

$|\beta | = 1$

and

$\alpha, \beta$ are mutually perpendicular

$\vec {\alpha} .\vec {\beta} = 0$

So here

$= (2\alpha + 3\beta) . (4\alpha - 6\beta)$

$= 8(\alpha . \alpha) - 12(\alpha . \beta) + 12 (\alpha . \beta) - 18\beta^{2}$

$= 8\alpha^{2} - 12\alpha . \beta + 12 (\alpha . \beta) - 18\beta^{2}$

$= 8|\alpha|^{2} - 0 - 18|\beta|^{2}$

$= 8 - 18$

$= -10$ .

Find the sin angle between the vectors $A = 3i - 4j + 5k$ and $B = i - j + k$ ?

Consider the problem

The resultant of two vectors is the vector sum of two vectors

$\vec R = \vec a + \vec b$

$\begin{array}{l} =3\vec { i } +4\vec { j } +5\vec { k } +5\vec { i } +3\vec { j } +4\vec { k } \\ =8\vec { i } +7\vec { j } +9\vec { k } .......1> \end{array}$

And magnitude of

$R$

$\begin{array}{l} \left| { \vec { R } } \right| =\sqrt { { 8^{ 2 } }+{ 7^{ 2 } }+{ 9^{ 2 } } } \\ =\sqrt { 64+49+81 } \\ =\sqrt { 194 } \\ =13.9 \end{array}$

Now, using the formula to find the angle between two vectors

$\cos \theta = \frac{{\vec a.\vec b}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$

$\begin{array}{l} \cos \theta =\frac { { \left( { 8\vec { i } +7\vec { j } +9\vec { k } } \right) .\vec { i } } }{ { \left| { \left( { 8\vec { i } +7\vec { j } +9\vec { k } } \right) } \right| \left| { \vec { i } } \right| } } \\ =\frac { 8 }{ { 13.9\times 1 } } \\ =0.57 \end{array}$

$\cos \theta = 0.57$

$\begin{array}{l} \Rightarrow \sin \theta =\sqrt { { 1^{ 2 } }-{ { \left( { 0.57 } \right) }^{ 2 } } } \\ =\sqrt { 0.6751 } \\ =0.82 \end{array}$

Prove that for any three vectors a, b, c [b + c c + a a + b] = 2 [abc].

For any vector

$\therefore \vec{a}, \vec{b}$ and

$\vec{c}$ , prove that

$[\vec{a} + \vec{b} \,\, \vec{b} + \vec{c} \,\, \vec{c} + \vec{a}] = 2 [\vec{a} \, \vec{b} \, \vec{c}]$

sowing LHS :-

$[\vec{a} + \vec{b} \, \vec{b} + \vec{c} \, \vec{c} + \vec{a}] = (\vec{a} + \vec{b}) . [(\vec{b} + \vec{c}) \times (\vec{c} + \vec{a})]$

$[\vec{a} \, \vec{b} \, \vec{c} = \vec{a}. (\vec{b} \times \vec{c}]$

$= (\vec{a} + \vec{b}). [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{c}) + (\vec{c} \times \vec{a})]$

$\because [\vec{c} \times \vec{c} = |\vec{c}||\vec{c}| \sin \, O \hat{n} = o$

$= (\vec{a} + \vec{b}). [(\vec{b} \times \vec{c}) + (\vec{b} \times \vec{a}) + o + ( \vec{c} \times \vec{a})]$

$= \vec{a} . (\vec{b} \times \vec{c}) + \vec{b} . (\vec{b} \times \vec{c}) + \vec{a} . (\vec{b} \times \vec{a}) + \vec{b} . (\vec{b} \times \vec{a}) + \vec{a} .(\vec{c} \times \vec{a}) + \vec{b}. (\vec{c} \times \vec{a})$

$= [\vec{a}, \vec{b}, \vec{c}] + [\vec{b} \vec{b} \vec{c}] + [\vec{a} \, \vec{b} \, \vec{a}] + [\vec{a} \, \vec{c} \, \vec{a}] + [\vec{b} \, \vec{c} \, \vec{a}]$

$\because [\vec{b}\, \vec{b} \, \vec{c} ] = [\vec{c} \, \vec{b} \, \vec{b}]$

$= \vec{c} . (\vec{b} \times \vec{b})$

$(\vec{b} \times \vec{b}) = \vec{o}$

$= \vec{c} . o$

$=\vec{o}$

$\because$ Using property

$[\vec{b} \, \vec{b} \, \vec{c}] = 0$

$[\vec{a} \, \vec{b} \, \vec{a}] = 0$

$[\vec{b} \, \vec{b} \, \vec{a}] = 0$

$[\vec{a} \, \vec{c} \, \vec{a}]= 0$

$= [\vec{a} \, \vec{b} \, \vec{c}] + o + o + o + o +[\vec{b} \, \vec{c} \, \vec{a}]$

$= [\vec{a} \, \vec{b} \, \vec{c}] + [\vec{b} \, \vec{c} \, \vec{a}]$

$= [\vec{a} \, \vec{b} \, \vec{c}] + [\vec{a} \, \vec{b} \, \vec{c}]$

$= 2 [\vec{a} \, \vec{b} \, \vec{c}] = R.H.S$

If $\hat {i}+\hat {j}+\hat {k},\ 2\hat {i}+5\hat {j},\ 3\hat {i}+2\hat {j}-3\hat {k}$ and $\hat {i}-6\hat {j}-\hat {k}$ are the position vectors of points $A,B,C$ and $D$ respectively, then find the angel between $\overline {AB}$ and $\overline {CD}$ . Deduce that $\overline {AB}$ and $\overline {CD}$ are collinear.

$\begin{array}{l} Position\, \, Vector\, \, of\, \, \overrightarrow { AB } =\hat { i } +4\hat { j } -\hat { k } \\ and\, \, Position\, \, Vector\, \, of\, \, \overrightarrow { CD } =-2\hat { i } -8\hat { j } +2\hat { k } \\ Then,\, \, angle\, \, between\, \, \overrightarrow { AB } \, \, and\, \, \overrightarrow { CD } =\frac { { \left( { \hat { i } +4\hat { j } -\hat { k } } \right) .\left( { 2\hat { i } -8\hat { j } +2\hat { k } } \right) } }{ { \left| { \hat { i } +4\hat { j } -\hat { k } } \right| .\left| { -2\hat { i } -8\hat { j } +2\hat { k } } \right| } } \\ =\frac { { \left( { -2-32-2 } \right) } }{ { \sqrt { 1+16+1 } .\sqrt { 4+64+4 } } } =\frac { { -36 } }{ { \sqrt { 20 } .\sqrt { 72 } } } \\ =\frac { { -36 } }{ { 3\sqrt { 2 } .\sqrt { 9\times 2 } } } =\frac { { -36 } }{ { 3\sqrt { 2 } \times 6\sqrt { 2 } } } =\frac { { -36 } }{ { 36 } } { }=-1 \\ Angle\, \, between\, \, \overrightarrow { AB } \, \, and\, \, \overrightarrow { CD } \, \, is=\pi \left( { { { 180 }^{ ^{ \circ } } } } \right) \\ and\, \, \overrightarrow { AB } =\lambda \, \, c\, \vec { D } \\ \overrightarrow { AB } =-2\left( { \hat { i } +4\hat { j } -\hat { k } } \right) \\ Hence,\, \, It\, \, is\, collinear. \end{array}$

Find $|\vec{a}\times \vec{b}|$ if $|\vec{a}|=10$ , $|\vec{b}|=2$ and $\vec{a}\cdot \vec{b}=12$ .

1190490_1236453_ans_c7a6e48dc92f4293a25906e0c9e666b7.jpg

If $\vec { a }$ and $\vec { b }$ are non-collinear and $\overline { A } = ( x + 4 y ) \overline { a } + ( 2 x + y + 1 ) \vec { b }$ and $\vec { \mathrm { B } } = ( y - 2 x + 2 ) \vec { a } + ( 2 x - 3 y - 1 )b$ Find $x$ and $y$ such that $3 \vec { A } = 2 \vec { B }$

1197173_1223633_ans_eb37910cc30440d095e1f08cd314dc49.jpg

The values of $\lambda$ such than $(x,y,z)\neq (0,0,0)\\$ and $(\hat { i+ } \hat { j } +3\hat { k } )x+(3\hat { i } -3\hat { j } +\hat { k } )y+(-4\hat { i } +5\hat { j } )z=\lambda (x\hat { i } +y\hat { j } +z\hat { k } )$ are

$\begin{array}{l} Given\, that \\ \left( { \widehat { i } +\widehat { j } +\widehat { k } } \right) x+\left( { 3\widehat { i } -3\widehat { j } +\widehat { k } } \right) y+\left( { -4\widehat { i } +5\widehat { j } } \right) z=\lambda \left( { x\widehat { i } +y\widehat { j } +z\widehat { k } } \right) \\ \Rightarrow \left( { x+3y-4z } \right) \widehat { i } +\left( { x-3y+5z } \right) \widehat { j } +\left( { 3x+y } \right) \widehat { k } =\left( { \lambda x\widehat { i } +\lambda y\widehat { j } +\lambda z\widehat { k } } \right) \\ \Rightarrow x+3y-4z=\lambda x\, \Rightarrow \left( { 1-\lambda } \right) x+3y-4z=0\, \, \, \, \, ----\left( 1 \right) \\ x-3y+5z=\lambda y\, \, \, \Rightarrow x-\left( { \lambda +3 } \right) y+5z=0\, \, \, \, \, \, -----\left( 2 \right) \\ and, \\ 3x+y=\lambda z\, \, \Rightarrow 3x+y-\lambda z=0\, \, \, \, \, -----\left( 3 \right) \\ now, \\ \left[ { \begin{array} { *{ 20 }{ c } }{ 1-\lambda } & 3 & { -4 } \\ 1 & { -\left( { \lambda +3 } \right) } & 5 \\ 3 & 1 & { -\lambda } \end{array} } \right] \left[ { \begin{array} { *{ 20 }{ c } }x \\ y \\ z \end{array} } \right] =\left[ { \begin{array} { *{ 20 }{ c } }0 \\ 0 \\ 0 \end{array} } \right] \\ \Rightarrow \left[ { \begin{array} { *{ 20 }{ c } }{ 1-\lambda } & 3 & { -4 } \\ 1 & { -\left( { \lambda +3 } \right) } & 5 \\ 3 & 1 & { -\lambda } \end{array} } \right] =0\, \, \, \, \, \left[ { \left[ { \begin{array} { *{ 20 }{ c } }x \\ y \\ z \end{array} } \right] \ne \left[ { \begin{array} { *{ 20 }{ c } }0 \\ 0 \\ 0 \end{array} } \right] } \right] \\ 3x+4y=9\, \, \, \, ---\left( 1 \right) \\ y=mx+1\, \, \, \, ----\left( 2 \right) \\ To\, \, \, get\, \, x-cordinate\, of\, po{ { int } }\, of\, { { int } }er\sec tion\, of\, \left( 1 \right) \& \left( 2 \right) \\ 3x+4\left( { mx+1 } \right) =9 \\ \Rightarrow x=\frac { { 9-4 } }{ { 3+4m } } =\frac { 5 }{ { 4m+3 } } \\ x-co-ordiante\, is\, { { int } }eger \\ 4m+3=-1,\, \, +1,\, -5,\, +5\, possible \\ take \\ 4m+3=-1\Rightarrow m=\left( { -1 } \right) \, \, it\, \, is\, true \\ 4m+3=1\, \, \, \, \Rightarrow m=\frac { { -1 } }{ 2 } \, \, \, it\, is\, wrong \\ 4m+3=-5\, \, \, \Rightarrow m=-2\, \, it\, is\, true \\ 4m+3=+5\Rightarrow m=\frac { 1 }{ 2 } \, \, \, it\, is\, wrong \\ so,\, { { int } }erger\, value\, of\, m\, \, are\, \left( { -1,\, -2 } \right) \\ 2{ { int } }eger \end{array}$

If $\overline{a}, \vec{b}, \vec{c}$ are unit vectors such that $\overline{a}+ \vec{b}+ \vec{c}$ , find the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\overline{c} \cdot \vec{c}$ .

2040908_1453455_ans_eb97b913f0bf4645a8915edec9d628d6.jpg

Prove that: $\left[ \begin{array} { l l l } { \overline { x } + \overline { y } } & { \overline { y } + \overline { z } } & { \overline { z } + \overline { x } } \end{array} \right] = 2 \left[ \begin{array} { l l } { \overline { x } \overline { y } \overline { z } } \end{array} \right]$

2184145_1481845_ans_d2c96bfe80b24f6d8ed17437e2460228.jpeg

Show, by vector method, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

1968988_1783270_ans_1e81767fe3924cd6b59995d2581bb273.jpg

Let OACB be a parallelogram with O at the origin and OC a diagonal. Let D be the midpoint of OA Using vector methods prove that BD and CO intersect in the same ratio. Determine this ratio. (IIT - JEE, 1993)

1968986_1783263_ans_b5756fce58af4e7587d4d59e32f6b372.jpg

Show that the four points P, Q, R, S with position vectors $\vec{p},\vec{q},\vec{r},\vec{s}$ respectively such that $5\vec{p}-2\vec{q}+6\vec{r}-9\vec{s} = \vec{0},$ are coplanar. Also, find the position vector of the point of intersection of the line segments PR and QS.

Given

$\vec p,\vec q,\vec r,\vec s$ are the position vectors

$P,Q,R,S$

Also given that

$5\vec p-2\vec q+6\vec r-9\vec s=\vec 0$

$\implies 5\vec p+6\vec r=2\vec q+9\vec s$

Divide by

$11$ on both sides

$\implies \dfrac{5\vec p+6\vec r}{11}=\dfrac{2\vec q+9\vec s}{11}$

$\implies \dfrac{5\vec p+6\vec r}{6+5}=\dfrac{2\vec q+9\vec s}{9+2}\cdots(1)$

As we know that

If a point

$C$ divides the line segments

$AB$ in the ratio

$m:n$ then

$\vec {OC}=\dfrac{n\vec {OA}+m\vec {OB}}{m+n}$

Let

$A$ be the point which divides

$PR$ in the ratio

$6:5$ and

$B$ be the point which divides

$QS$ in the ratio

$9:2$

Hence the position vectors of

$A,B$ are

$\dfrac{5\vec p+6\vec r}{11},\dfrac{2\vec q+9\vec s}{11}$ respectively

From

$(1)$

The position vector of

$A,B$ is same

$\implies$ Points

$A\ \&\ B$ are the same point

$\implies PR$ and

$QS$ intersect at

$A$

If two line segments intersects then the line must be in same plane

$\implies P,Q,R,S$ are coplanar

The position vector of point of intersection of

$PR$ and

$QS$ is

$\dfrac{5 \vec p+6\vec r}{11}\text{ or }\dfrac{2 \vec q+9\vec s}{11}$

If $\vec{a}, \vec{b}$ are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2 BA.

Given

$\vec a,\vec b$ are the position vectors of

$A,B$

Let

$\vec c,\vec d$ be the position vectors of

$C,D$

Given

$AC=3 AB\cdots(1)$

and

$C$ is a point on

$AB$

$\implies AC\parallel AB\cdots(2)$

From

$(1)\ \& (2)$

$\vec {AC}=3\vec {AB}$

$\implies \vec c-\vec a=3(\vec b-\vec a)$

$\implies \vec c=3 \vec b-3 \vec a+\vec a=3\vec b-2 \vec a$

Given

$BD=2 BA\cdots(3)$

$D$ is a point on

$BA$

$\implies BD\parallel BA\cdots(4)$

From

$(3)\ \& (4)$

$\vec {BD}=2\vec {BA}$

$\implies \vec d-\vec b=2(\vec a-\vec b)$

$\implies \vec d=2 \vec a-2 \vec b+\vec b=2\vec a- \vec b$

So the position vectors of

$C$ and

$D$ are

$3 \vec b-2 \vec a,2\vec a-\vec b$ respectively

Prove, by vector method or otherwice, that the point of intersection of the diagonals of a trapezium lies on the line passing through the midpoint of the parallel sides (you may assume that the trapezium is not a parallelogram)

1968987_1783269_ans_f12a208383d44962ae9767add3d22284.jpg

If $\vec{e}_1, \vec{e}_2, \vec{e}_3$ and $\vec{E}_1, \vec{E}_2, \vec{E}_3$ are two sets of vectors such that $\vec{e}_i . \vec{E}_j$ , if i = j and $\vec{e}_i . \vec{E}_j = 0$ and if $i \neq j$ , then prove that $[\vec{e}_1 \vec{e}_2 \vec{e}_3][\vec{E}_1 \vec{E}_2 \vec{E}_3] = 1$

1969292_1783980_ans_396719f8fc524f0b97b0a9c367d0fa91.jpg

Find the vector equation of the plane passing through the points $\hat{i}+\hat{j}-3\hat{k},\hat{i}+2\hat{j}+\hat{k},2\hat{i}-\hat{j}+\hat{k}$

2161335_1984841_ans_f6c0720800e147e39fbc060574f1439d.jpeg

Vector Algebra - Class 12 Commerce Maths - Extra Questions

Match the column

A point is on the xx-axis. What are its yy-coordinate and zz-coordinates?

Find the projection of the vector ˆi+3ˆj+7ˆk\hat {i}+3\hat {j}+7\hat {k} on the vector 7ˆi−ˆj+8ˆk7\hat {i}-\hat {j}+8\hat {k} .

If either vector →a=→0\vec {a}=\vec {0} or →b=→0\vec {b}=\vec {0}, then →a⋅→b=0\vec {a}\cdot \vec {b}=0. But the converse need not be true. Justify your answer with an example.

Find the sum of the vectors →a=ˆi−2ˆj+ˆk,→b=−2ˆi+4ˆj+5ˆk\vec {a}=\hat {i}-2\hat {j}+\hat {k}, \vec {b}=-2\hat {i}+4\hat {j}+5\hat {k} and →c=ˆi−6ˆj−7ˆk\vec {c}=\hat {i}-6\hat {j}-7\hat {k}.

Find |→x||\vec {x}|, if for a unit vector →a,(→x−→a)⋅(→x+→a)=12\vec {a}, (\vec {x}-\vec {a})\cdot (\vec {x}+\vec {a})=12Answer required

If →a,→b,→c\vec {a}, \vec {b}, \vec {c} are unit vectors such that →a+→b+→c=→0\vec {a}+\vec {b}+\vec {c}=\vec {0}, find the value of →a⋅→b+→b⋅→c+→c⋅→a\vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a}

If →a=−ˆi−ˆj+ˆk\vec{a}=-\hat{i}-\hat{j}+\hat{k} and →b=−ˆi+ˆj−ˆk\vec{b}=-\hat{i}+\hat{j}-\hat{k}, then find the value of |→a⋅→b||\vec{a}\cdot \vec{b}|.

Projection of →b\vec{b} in the direction of →a\vec{a} ___________.

If vectors →a=ˆi−ˆj+ˆk\vec{a}=\hat{i}-\hat{j}+\hat{k} and →b=ˆi+ˆj+ˆk\vec{b}=\hat{i}+\hat{j}+\hat{k}, then find the value of →a⋅→b\vec{a}\cdot\vec{b}.

For any two vectors →a,→b\vec { a } ,\vec { b } , prove that |→a×→b|2=|→a.→a→a.→b→b.→a→b.→b|{ \left| \vec { a } \times \vec { b } \right| }^{ 2 }=\begin{vmatrix} \vec { a } .\vec { a } & \vec { a } .\vec { b } \\ \vec { b } .\vec { a } & \vec { b }.\vec { b } \end{vmatrix}

If ¯OA=3ˆi+ˆj−2ˆk\bar{OA} = 3 \hat{i} + \hat{j} - 2\hat{k} and ¯OB=5ˆi+4ˆj−ˆk\bar{OB} = 5 \hat{i} + 4 \hat{j} - \hat{k} then find the value of ¯AB\bar{AB}

Let ε1\varepsilon_1 and ε2\varepsilon_2 be the angles made by →A\vec{A} and −→A-\overrightarrow{A} with the positive X-axis. Show that tanε1=tanε2\tan \varepsilon_1=\tan\varepsilon_2. Thus, given tanε\tan\varepsilon does not uniquely determine the direction of →A\overrightarrow{A}.

Calculate the scalar product of the following vectorsa=3i+2j+2kandb=18i−22j−5k\displaystyle a \, = \, 3i \, + \, 2j \, + \, 2k \, and \, b \, = \, 18i \, - \, 22j \, - \, 5k

In a Δ\Delta ABC, →AB=6→i+3→j+3→k\vec{AB}=6\vec{i}+3\vec{j}+3\vec{k} ; →AC=3→i−3→j+6→k\vec{AC}=3\vec{i}-3\vec{j}+6\vec{k}. D and D' are points of tri-sections of sides BC. Find →AD\vec{AD} and →AD′\vec{AD'}.

Given →AB=3i+2j−k,→OB=i+2j+k\vec{AB} = 3i + 2j - k \, , \vec{OB} = i + 2j + k. find →OA\vec{OA}

If vector →a=a1ˆi+a2ˆj+a3ˆk\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k then, find unit vector.

The sides of a parallelogram are 2ˆi−4ˆj+5ˆk2\hat i - 4\hat j +5\hat k and ˆi−2ˆj−3ˆk\hat i - 2\hat j - 3\hat k . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.

If r=t2i−tj+(2t+1)kr=t^2i-tj+(2t+1)k, find |dr/dt||dr/dt|.

If a vector →AB=2ˆi−ˆj+ˆk\vec{AB}=2\hat{i}-\hat{j}+\hat{k} and →OB=3ˆi−4ˆj+4ˆk\vec{OB}=3\hat{i}-4\hat{j}+4\hat{k} find the position vector →OA\vec{OA}.

Show that:[ˉa×ˉbˉb×ˉcˉc×ˉa]=[abc]2[\bar{a} \times \bar{b} \, \bar{b} \times \bar{c} \,\bar{c} \times \bar{a}] = [abc]^2

Points X & Y are taken on the sides QR & RS respectively of a parallelogram PQRS, so that →QX=4→XR\vec{QX}=4\vec{XR} & →RY=4→YS\vec{RY}=4\vec{YS}. The line XY cuts the line PR at Z. Prove that →PZ=(2125)→PR\vec{PZ}=\left(\dfrac{21}{25}\right)\vec{PR}.

Component of (1,1)(1,1) along (3,4)(3,4) is ?

Find the component of ¯i+¯kon¯i−¯j\overline i + \overline k \,on\,\overline i - \overline j and its magnitude.

If ¯a=(1,−1)\overline a = \left( {1, - 1} \right) and ¯b=(1,0)\overline b = \left( {1,0} \right) then comp ¯a¯b=..........\overline a \overline b = ..........

ABCD is parallelogram. If →AB=→a,→BC=→b\vec { AB } =\vec { a } ,\vec { BC } =\vec { b } then show that →AB=→a,+→b\vec { AB } =\vec { a } ,+\vec { b } and →BD=→b,−→a\vec { BD } =\vec { b } ,-\vec { a } ?

Find the value or values of mm for which m(ˆi+ˆj+ˆk)m\left( \hat { i } +\hat { j } +\hat { k } \right) is a unit vector.

If →a\vec{a} and →b\vec{b} are position vectors of point s A and B respectively, then the position vector of points of trisection of AB

Find a unit vector which is along the direction of vector ˆi+ˆj+2ˆk\hat{i}+\hat{j}+2\hat{k}.

If →a||→b×→c\vec{a}||\vec{b}\times \vec{c} then (→a×→b)⋅(→a×→c)\left(\vec{a}\times \vec{b}\right)\cdot (\vec{a}\times \vec{c}) is equal to?

Find the projection of (a,b,c)(a,b,c) on X-axis and its magnitude.

Let a=i+2j+3ka=i+2j+3k and b=3i+jb=3i+j. Find the unit vector in the direction of a+ba+b.

If \vec {a}=\vec {b}+\vec {c}\vec {a}=\vec {b}+\vec {c}, then is true that |\vec {a}|=|\vec {b}|+|\vec {c}||\vec {a}|=|\vec {b}|+|\vec {c}|? Justify your answer.

If the projections of the line segment \underset{PQ}{\rightarrow}\underset{PQ}{\rightarrow} on the axes are 3,4,12 then the length of \underset{PQ}{\rightarrow}\underset{PQ}{\rightarrow} is

If \hat { \mathbf { a } }\hat { \mathbf { a } } is a unit vector and ( x - \hat { a } ) \cdot ( \vec { x } + \hat { a } ) = 1.5 ( x - \hat { a } ) \cdot ( \vec { x } + \hat { a } ) = 1.5 then find the value of | \vec { x } | | \vec { x } |.

Find a\times (b\times c)a\times (b\times c) and (a\times b)\times c(a\times b)\times c where a=(1, -1, -6), b=(1, -3, 4), c=(2, -5, 3)a=(1, -1, -6), b=(1, -3, 4), c=(2, -5, 3).

Find a unit vector in the direction of \hat { i } + \hat { j }\hat { i } + \hat { j }.

The vector product of two vector is \sqrt{3}\sqrt{3} times of their of two scalar produced. The angle between the two vector is-

Write the unit vector in the direction of \overset { \rightarrow }{ A } =5\overset { \rightarrow }{ i } + \overset { \rightarrow }{ j } - 2\overset { \rightarrow }{ k } \overset { \rightarrow }{ A } =5\overset { \rightarrow }{ i } + \overset { \rightarrow }{ j } - 2\overset { \rightarrow }{ k } .

Simplify:(i)\vec { D E }\vec { D E }+( - \vec { B E } )( - \vec { B E } )(ii)\vec { AC }\vec { AC }+( - \vec { B C })( - \vec { B C })(iii)\vec { C D }\vec { C D }+\vec { B A }\vec { B A }( - \vec { B D } )( - \vec { B D } )

The dot product of a vector with vectors \hat{i}+\hat{j}-3\hat{k}\hat{i}+\hat{j}-3\hat{k}, \hat{ i}+ 3\hat{ j}-2\hat{k}\hat{ i}+ 3\hat{ j}-2\hat{k} ,2\hat{i} + \hat{j} + 4k2\hat{i} + \hat{j} + 4k are 0,50,5 and 88 respectively. Find the vector .

Forces measuring 5,35,3 and 11 unit act in the directions (6,2,3),(3,-2,6)(6,2,3),(3,-2,6) and (5,-1,1)(5,-1,1). Find the resultant force.

Find \vec{a}.\vec{b}\vec{a}.\vec{b}, when \vec{a}=\hat{j}+2\hat{k}\vec{a}=\hat{j}+2\hat{k} and \vec{b}=2\hat{i}+\hat{k}\vec{b}=2\hat{i}+\hat{k}.

Find the projection of the vector \widehat{i}-\widehat{j}\widehat{i}-\widehat{j} on the vector \widehat{i}+\widehat{j}\widehat{i}+\widehat{j}.

Find the dot product of the following vectors:\vec{a}=2\hat{i}+\hat{j}+3\hat{k}\vec{a}=2\hat{i}+\hat{j}+3\hat{k} and \vec{b}=3\hat{i}+5\hat{j}-2\hat{k}\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}.

Compute the magnitude of the following vector :\overrightarrow a = 2\widehat{i}-7\widehat{j}-3\widehat{k}\overrightarrow a = 2\widehat{i}-7\widehat{j}-3\widehat{k}

Find the value of x for which x(\widehat{i}+\widehat{j}+\widehat{k})x(\widehat{i}+\widehat{j}+\widehat{k}) is a unit vector.

Find the distance between the points A and B whose position vectors \widehat{i}-\widehat{j}\widehat{i}-\widehat{j} and 2\widehat{i}+\widehat{j}+2\widehat{k}2\widehat{i}+\widehat{j}+2\widehat{k}.

Answer the following as true or false: Zero vector is unique.

Find the projection of the vector 7\widehat{i}+\widehat{j}-4\widehat{k}7\widehat{i}+\widehat{j}-4\widehat{k} and 2\widehat{i}+6\widehat{j}+3\widehat{k}2\widehat{i}+6\widehat{j}+3\widehat{k}.

Enter 11 if true else 00. 3\hat{i}+2\hat{j}+\hat{k} 3\hat{i}+2\hat{j}+\hat{k} is the position vector of the mid-point of the vector joining the points P(2,3,4) P(2,3,4) and Q(4,1,-2). Q(4,1,-2).

If a\hat{i}+b\hat{j}+c\hat{k} a\hat{i}+b\hat{j}+c\hat{k} is the position vector of the mid - point of the vector joining the points P(2\hat{i}-3\hat{j}+4\hat{k}) P(2\hat{i}-3\hat{j}+4\hat{k}) and Q(4\hat{i}+\hat{j}-2\hat{k}) Q(4\hat{i}+\hat{j}-2\hat{k}) then a+b+ca+b+c is equal to ___.

\hat{a} \space and \space \hat{b}\hat{a} \space and \space \hat{b} form the consecutive sides of a regular hexagon ABCDEF.

Find the projection of the vector \displaystyle i-2j+k\displaystyle i-2j+k on the vector \displaystyle 4i-4j+7k\displaystyle 4i-4j+7k

If \displaystyle r=xi+yj+zk\displaystyle r=xi+yj+zk then show that r can also be expressed as \displaystyle r= \left ( r.i \right )i+\left ( r.j \right )j+\left ( r.k \right )k\displaystyle r= \left ( r.i \right )i+\left ( r.j \right )j+\left ( r.k \right )k.

If D, \ E, \ FD, \ E, \ F are the midpoints of the sides BC, \ CA, \ AB,BC, \ CA, \ AB, respectively, of a triangle ABCABC and OO is any point, then

Write the three times of the projection of the vector \overset{\rightarrow}{a} = 2\hat{i} - \hat{j} + \hat{k}\overset{\rightarrow}{a} = 2\hat{i} - \hat{j} + \hat{k} on the vector \overset{\rightarrow}{b} = \hat{i} + 2\hat{j} + 2\hat{k}\overset{\rightarrow}{b} = \hat{i} + 2\hat{j} + 2\hat{k}

Evaluate the product (3\vec {a}-5\vec {b})\cdot (2\vec {a}+7\vec {b})(3\vec {a}-5\vec {b})\cdot (2\vec {a}+7\vec {b}) .

If \vec {a}=\vec {b}+\vec {c}\vec {a}=\vec {b}+\vec {c}, then is it true that |\vec {a}|=|\vec {b}|+|\vec {c}||\vec {a}|=|\vec {b}|+|\vec {c}|? Justify your answer.

Find the projection of the vector \hat {i}-\hat {j}\hat {i}-\hat {j} on the vector \hat {i}+\hat {j}\hat {i}+\hat {j} .

The two vectors \hat j + \hat k\hat j + \hat k and 3\hat i - \hat j + 4\hat k3\hat i - \hat j + 4\hat k represent the two sides ABAB and ACAC, respectively of a \triangle ABC\triangle ABC. Find the length of the median through AA

Find '\lambda''\lambda' when the projection of \vec {a} = \lambda \hat{i} + \hat{j} + 4\hat{k}\vec {a} = \lambda \hat{i} + \hat{j} + 4\hat{k} on \vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k} is 4 units.

Find the scalar projection of the vector \hat {i} + 3\hat {j} + 7\hat {k}\hat {i} + 3\hat {j} + 7\hat {k} on the vector 2\hat {i} - 3\hat {j} + 6\hat {k}.2\hat {i} - 3\hat {j} + 6\hat {k}.

Find the projection of vector \overrightarrow { a } =2\hat { i } +3\hat { j } +2\hat { k } \overrightarrow { a } =2\hat { i } +3\hat { j } +2\hat { k } on a vector \overrightarrow { b } =\hat { i } +2\hat { j } +\hat { k } \overrightarrow { b } =\hat { i } +2\hat { j } +\hat { k }

If \vec {a}\vec {a} is a non-zero vector of modulus aa and mm is a non-zero scalar then m\vec {a}m\vec {a} is a unit vector if

ACAC and BDBD are the diagonals of a quadrilateral ABCDABCD, prove that: \vec{AB}+\vec{DC}=\vec{AC}+\vec{DB}\vec{AB}+\vec{DC}=\vec{AC}+\vec{DB}.

In a triangle ABC, D, E, FABC, D, E, F are the mid-points of the sides BC, CABC, CA and ABAB respectively then prove that, \vec {AD} = - (\vec {BE} + \vec {CF})\vec {AD} = - (\vec {BE} + \vec {CF}).

If the line segments joining the points A(a,b)A(a,b) and B(c,d)B(c,d) subtends and angle \theta\theta at the origin, then find \cos{\theta}\cos{\theta}

Write the associative law for addition of vectors.

Calculate the scalar product of the following vectors:\displaystyle a \, = \, \left \{ 2, 4, 1 \right \} \, and\ \, b \, = \, \left \{ 3, 5, 7 \right \}\displaystyle a \, = \, \left \{ 2, 4, 1 \right \} \, and\ \, b \, = \, \left \{ 3, 5, 7 \right \}

Three forces 2\hat { i } -\hat { j } +3\hat { k } ,\hat { j } +\hat { k } 2\hat { i } -\hat { j } +3\hat { k } ,\hat { j } +\hat { k } and -2\hat { i } -2\hat { j } +\hat { k } -2\hat { i } -2\hat { j } +\hat { k } are applied on a body. Is the body in equilibrium?

Find the value \left[ \hat { i } \quad \hat { j } \quad \hat { k } \right] \left[ \hat { i } \quad \hat { j } \quad \hat { k } \right] .

Find the magnitude of each of the two vectors \overset{\rightarrow}{a}\overset{\rightarrow}{a} and \overset{\rightarrow}{b}\overset{\rightarrow}{b}, having the same magnitude such that the angle between them is 60^o60^o and their scalar product is \displaystyle\frac{9}{2}\displaystyle\frac{9}{2}.

Evaluate \overset { \wedge }{ j } \left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right) \overset { \wedge }{ j } \left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right)

Prove that:\vec{P}\times \vec{Q}=(\hat{k})P_xQ_y+(-\hat{j})P_xQ_z+(-\hat{k})P_yQ_x+O+\hat{i}P_yQ_z+(\hat{j})P_zQ_x+(-\hat{i})P_zQ_y\vec{P}\times \vec{Q}=(\hat{k})P_xQ_y+(-\hat{j})P_xQ_z+(-\hat{k})P_yQ_x+O+\hat{i}P_yQ_z+(\hat{j})P_zQ_x+(-\hat{i})P_zQ_y.

Find the projection of the vector \hat {i} + 3\hat {j} + \hat {k}\hat {i} + 3\hat {j} + \hat {k} on the vector 7\hat {i} - \hat {j} + 8\hat {k}7\hat {i} - \hat {j} + 8\hat {k}.

If \vec{r}.\hat{i} = \vec{r}.\hat{j} = \vec{r}.\hat{k}\vec{r}.\hat{i} = \vec{r}.\hat{j} = \vec{r}.\hat{k} and |\vec{r}| = 3|\vec{r}| = 3, then find vector \vec{r}\vec{r}.

Show that magnitude of vector product of two vectors is numerically equal to the area of a parallelogram formed by two vectors.

For any two vectors \vec{a}\vec{a} and \vec{b}\vec{b}, If (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0.(\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0. Then | \vec{a}|-|\vec{b}|=? | \vec{a}|-|\vec{b}|=?.

A point is on the $x$ -axis. What are its $y$ -coordinate and $z$ -coordinates?

Find the projection of the vector $\hat {i}+3\hat {j}+7\hat {k}$ on the vector $7\hat {i}-\hat {j}+8\hat {k}$ .

If either vector $\vec {a}=\vec {0}$ or $\vec {b}=\vec {0}$ , then $\vec {a}\cdot \vec {b}=0$ . But the converse need not be true. Justify your answer with an example.

Find the sum of the vectors $\vec {a}=\hat {i}-2\hat {j}+\hat {k}, \vec {b}=-2\hat {i}+4\hat {j}+5\hat {k}$ and $\vec {c}=\hat {i}-6\hat {j}-7\hat {k}$ .

Find $|\vec {x}|$ , if for a unit vector $\vec {a}, (\vec {x}-\vec {a})\cdot (\vec {x}+\vec {a})=12$

Answer required

If $\vec {a}, \vec {b}, \vec {c}$ are unit vectors such that $\vec {a}+\vec {b}+\vec {c}=\vec {0}$ , find the value of $\vec {a}\cdot \vec {b}+\vec {b}\cdot \vec {c}+\vec {c}\cdot \vec {a}$

If $\vec{a}=-\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}-\hat{k}$ , then find the value of $|\vec{a}\cdot \vec{b}|$ .

Projection of $\vec{b}$ in the direction of $\vec{a}$ ___________.

If vectors $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ , then find the value of $\vec{a}\cdot\vec{b}$ .

For any two vectors $\vec { a } ,\vec { b }$ , prove that ${ \left| \vec { a } \times \vec { b } \right| }^{ 2 }=\begin{vmatrix} \vec { a } .\vec { a } & \vec { a } .\vec { b } \\ \vec { b } .\vec { a } & \vec { b }.\vec { b } \end{vmatrix}$

If $\bar{OA} = 3 \hat{i} + \hat{j} - 2\hat{k}$ and $\bar{OB} = 5 \hat{i} + 4 \hat{j} - \hat{k}$ then find the value of $\bar{AB}$

Let $\varepsilon_1$ and $\varepsilon_2$ be the angles made by $\vec{A}$ and $-\overrightarrow{A}$ with the positive X-axis. Show that $\tan \varepsilon_1=\tan\varepsilon_2$ . Thus, given $\tan\varepsilon$ does not uniquely determine the direction of $\overrightarrow{A}$ .

Calculate the scalar product of the following vectors
$\displaystyle a \, = \, 3i \, + \, 2j \, + \, 2k \, and \, b \, = \, 18i \, - \, 22j \, - \, 5k$

In a $\Delta$ ABC, $\vec{AB}=6\vec{i}+3\vec{j}+3\vec{k}$ ; $\vec{AC}=3\vec{i}-3\vec{j}+6\vec{k}$ . D and D' are points of tri-sections of sides BC. Find $\vec{AD}$ and $\vec{AD'}$ .

Given $\vec{AB} = 3i + 2j - k \, , \vec{OB} = i + 2j + k$ . find $\vec{OA}$

If vector $\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ then, find unit vector.

The sides of a parallelogram are $2\hat i - 4\hat j +5\hat k$ and $\hat i - 2\hat j - 3\hat k$ . Find the unit vectors parallel to the diagonals and also find the Area of parallelogram.

If $r=t^2i-tj+(2t+1)k$ , find $|dr/dt|$ .

If a vector $\vec{AB}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{OB}=3\hat{i}-4\hat{j}+4\hat{k}$ find the position vector $\vec{OA}$ .

Show that:
$[\bar{a} \times \bar{b} \, \bar{b} \times \bar{c} \,\bar{c} \times \bar{a}] = [abc]^2$

Points X & Y are taken on the sides QR & RS respectively of a parallelogram PQRS, so that $\vec{QX}=4\vec{XR}$ & $\vec{RY}=4\vec{YS}$ . The line XY cuts the line PR at Z. Prove that $\vec{PZ}=\left(\dfrac{21}{25}\right)\vec{PR}$ .

Component of $(1,1)$ along $(3,4)$ is ?

Find the component of $\overline i + \overline k \,on\,\overline i - \overline j$ and its magnitude.

If $\overline a = \left( {1, - 1} \right)$ and $\overline b = \left( {1,0} \right)$ then comp $\overline a \overline b = ..........$

ABCD is parallelogram. If $\vec { AB } =\vec { a } ,\vec { BC } =\vec { b }$ then show that $\vec { AB } =\vec { a } ,+\vec { b }$ and $\vec { BD } =\vec { b } ,-\vec { a }$ ?

Find the value or values of $m$ for which $m\left( \hat { i } +\hat { j } +\hat { k } \right)$ is a unit vector.

If $\vec{a}$ and $\vec{b}$ are position vectors of point s A and B respectively, then the position vector of points of trisection of AB

Find a unit vector which is along the direction of vector $\hat{i}+\hat{j}+2\hat{k}$ .

If $\vec{a}||\vec{b}\times \vec{c}$ then $\left(\vec{a}\times \vec{b}\right)\cdot (\vec{a}\times \vec{c})$ is equal to?

Find the projection of $(a,b,c)$ on X-axis and its magnitude.

Let $a=i+2j+3k$ and $b=3i+j$ . Find the unit vector in the direction of $a+b$ .

If $\vec {a}=\vec {b}+\vec {c}$ , then is true that $|\vec {a}|=|\vec {b}|+|\vec {c}|$ ? Justify your answer.

If the projections of the line segment $\underset{PQ}{\rightarrow}$ on the axes are 3,4,12 then the length of $\underset{PQ}{\rightarrow}$ is

If $\hat { \mathbf { a } }$ is a unit vector and $( x - \hat { a } ) \cdot ( \vec { x } + \hat { a } ) = 1.5$ then find the value of $| \vec { x } |$ .

Find $a\times (b\times c)$ and $(a\times b)\times c$ where $a=(1, -1, -6), b=(1, -3, 4), c=(2, -5, 3)$ .

Find a unit vector in the direction of $\hat { i } + \hat { j }$ .

The vector product of two vector is $\sqrt{3}$ times of their of two scalar produced. The angle between the two vector is-

Write the unit vector in the direction of $\overset { \rightarrow }{ A } =5\overset { \rightarrow }{ i } + \overset { \rightarrow }{ j } - 2\overset { \rightarrow }{ k }$ .

Simplify:
(i) $\vec { D E }$ + $( - \vec { B E } )$
(ii) $\vec { AC }$ + $( - \vec { B C })$
(iii) $\vec { C D }$ + $\vec { B A }$ $( - \vec { B D } )$

The dot product of a vector with vectors $\hat{i}+\hat{j}-3\hat{k}$ , $\hat{ i}+ 3\hat{ j}-2\hat{k}$ , $2\hat{i} + \hat{j} + 4k$ are $0,5$ and $8$ respectively. Find the vector .

Forces measuring $5,3$ and $1$ unit act in the directions $(6,2,3),(3,-2,6)$ and $(5,-1,1)$ . Find the resultant force.

Find $\vec{a}.\vec{b}$ , when $\vec{a}=\hat{j}+2\hat{k}$ and $\vec{b}=2\hat{i}+\hat{k}$ .

Find the projection of the vector $\widehat{i}-\widehat{j}$ on the vector $\widehat{i}+\widehat{j}$ .

Find the dot product of the following vectors:
$\vec{a}=2\hat{i}+\hat{j}+3\hat{k}$ and $\vec{b}=3\hat{i}+5\hat{j}-2\hat{k}$ .

Compute the magnitude of the following vector :
$\overrightarrow a = 2\widehat{i}-7\widehat{j}-3\widehat{k}$

Find the value of x for which $x(\widehat{i}+\widehat{j}+\widehat{k})$ is a unit vector.

Find the distance between the points A and B whose position vectors $\widehat{i}-\widehat{j}$ and $2\widehat{i}+\widehat{j}+2\widehat{k}$ .

Answer the following as true or false:
Zero vector is unique.

Find the projection of the vector $7\widehat{i}+\widehat{j}-4\widehat{k}$ and $2\widehat{i}+6\widehat{j}+3\widehat{k}$ .

Enter $1$ if true else $0$ .
$3\hat{i}+2\hat{j}+\hat{k}$ is the position vector of the mid-point of the vector joining the points $P(2,3,4)$ and $Q(4,1,-2).$

If $a\hat{i}+b\hat{j}+c\hat{k}$ is the position vector of the mid - point of the vector joining the points $P(2\hat{i}-3\hat{j}+4\hat{k})$ and $Q(4\hat{i}+\hat{j}-2\hat{k})$ then $a+b+c$ is equal to ___.

$\hat{a} \space and \space \hat{b}$ form the consecutive sides of a regular hexagon ABCDEF.

Find the projection of the vector $\displaystyle i-2j+k$ on the vector $\displaystyle 4i-4j+7k$

If $\displaystyle r=xi+yj+zk$ then show that r can also be expressed as $\displaystyle r= \left ( r.i \right )i+\left ( r.j \right )j+\left ( r.k \right )k$ .

If $D, \ E, \ F$ are the midpoints of the sides $BC, \ CA, \ AB,$ respectively, of a triangle $ABC$ and $O$ is any point, then

Write the three times of the projection of the vector $\overset{\rightarrow}{a} = 2\hat{i} - \hat{j} + \hat{k}$ on the vector $\overset{\rightarrow}{b} = \hat{i} + 2\hat{j} + 2\hat{k}$

Evaluate the product $(3\vec {a}-5\vec {b})\cdot (2\vec {a}+7\vec {b})$ .

If $\vec {a}=\vec {b}+\vec {c}$ , then is it true that $|\vec {a}|=|\vec {b}|+|\vec {c}|$ ? Justify your answer.

Find the projection of the vector $\hat {i}-\hat {j}$ on the vector $\hat {i}+\hat {j}$ .

The two vectors $\hat j + \hat k$ and $3\hat i - \hat j + 4\hat k$ represent the two sides $AB$ and $AC$ , respectively of a $\triangle ABC$ . Find the length of the median through $A$

Find $'\lambda'$ when the projection of $\vec {a} = \lambda \hat{i} + \hat{j} + 4\hat{k}$ on $\vec{b} = 2\hat{i} + 6\hat{j} + 3\hat{k}$ is 4 units.

Find the scalar projection of the vector $\hat {i} + 3\hat {j} + 7\hat {k}$ on the vector $2\hat {i} - 3\hat {j} + 6\hat {k}.$

Find the projection of vector $\overrightarrow { a } =2\hat { i } +3\hat { j } +2\hat { k }$ on a vector $\overrightarrow { b } =\hat { i } +2\hat { j } +\hat { k }$

If $\vec {a}$ is a non-zero vector of modulus $a$ and $m$ is a non-zero scalar then $m\vec {a}$ is a unit vector if

$AC$ and $BD$ are the diagonals of a quadrilateral $ABCD$ , prove that: $\vec{AB}+\vec{DC}=\vec{AC}+\vec{DB}$ .

In a triangle $ABC, D, E, F$ are the mid-points of the sides $BC, CA$ and $AB$ respectively then prove that, $\vec {AD} = - (\vec {BE} + \vec {CF})$ .

If the line segments joining the points $A(a,b)$ and $B(c,d)$ subtends and angle $\theta$ at the origin, then find $\cos{\theta}$

Calculate the scalar product of the following vectors:
$\displaystyle a \, = \, \left \{ 2, 4, 1 \right \} \, and\ \, b \, = \, \left \{ 3, 5, 7 \right \}$

Three forces $2\hat { i } -\hat { j } +3\hat { k } ,\hat { j } +\hat { k }$ and $-2\hat { i } -2\hat { j } +\hat { k }$ are applied on a body. Is the body in equilibrium?

Find the value $\left[ \hat { i } \quad \hat { j } \quad \hat { k } \right]$ .

Find the magnitude of each of the two vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ , having the same magnitude such that the angle between them is $60^o$ and their scalar product is $\displaystyle\frac{9}{2}$ .

Evaluate $\overset { \wedge }{ j } \left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right)$

Prove that:
$\vec{P}\times \vec{Q}=(\hat{k})P_xQ_y+(-\hat{j})P_xQ_z+(-\hat{k})P_yQ_x+O+\hat{i}P_yQ_z+(\hat{j})P_zQ_x+(-\hat{i})P_zQ_y$ .

Find the projection of the vector $\hat {i} + 3\hat {j} + \hat {k}$ on the vector $7\hat {i} - \hat {j} + 8\hat {k}$ .

If $\vec{r}.\hat{i} = \vec{r}.\hat{j} = \vec{r}.\hat{k}$ and $|\vec{r}| = 3$ , then find vector $\vec{r}$ .

For any two vectors $\vec{a}$ and $\vec{b}$ , If $(\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0.$
Then $| \vec{a}|-|\vec{b}|=?$ .

Find the value of $\lambda$ such that points $P(1, 2), Q(-2, 3)$ and $R(\lambda +1, \lambda)$ are not forming a triangle?

Two vertices of a triangle are at $-\hat{i}+3\hat{j}$ and $2\hat{i}+5\hat{j}$ and its orthocentre is at $\hat{i}+2\hat{j}$ . Find the position vector of thrid vertex.

Angle between $\overrightarrow{a}$ & $\overrightarrow{b}$ . Find a vector along the angle bisector of $\overrightarrow{a}$ & $\overrightarrow{b}$ .

Find the projection of $3\hat i + 4\hat j + \hat k$ on $\hat i + \hat j - \hat k$ and its magnitude.