Class 12 Engineering Maths - Vector Algebra

Find 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 mid-point

$R$ of the vector joining points

$P (2, 3, 4)$ and

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

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

$=\dfrac {6\hat {i}+4\hat {j}+2\hat {k}}{2}=3\hat {i}+2\hat {j}+\hat {k}$

Find the sum of the following vectors:
$\vec {a} = \hat{i}-2\hat{j}, \vec{b} = 2\hat{i}-3\hat{j}, \vec{c} = 2\hat{i} + 3\hat{k}$ .

Given that:

$\vec a=\hat i-2\hat j, \vec b=2\hat i-3\hat j$ and

$\vec c=2\hat i+3\hat k$

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

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

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

$=5\hat i-5\hat j+3\hat k$

If $\overline { a } ,\overline { b } ,\overline { c }$ are the position vectors of the points $A,B,C$ respectively and $2\overline { a } +3\overline { b } -5\overline { c } =\overline { 0 }$ , then find the ratio in which the point $C$ divides line segment $AB$ .

Let the ratio be

$\lambda :1$
Position vector of point

$(C)$

$\vec { c } =\cfrac { \lambda \vec { a } +\vec { b } }{ \lambda +1 }$
Put the value of

$\overrightarrow { c }$ in

$\quad 2\vec { a } +3\vec { b } -5\vec { c } =0$

$2\vec { a } +3\vec { b } -5\times \left( \cfrac { \lambda \vec { a } +\vec { b } }{ \lambda +1 } \right) =0$

$\Rightarrow \cfrac { \left( \lambda +1 \right) \left( 2\vec { a } +3\vec { b } \right) -5\lambda \vec { a } -5\vec { b } }{ \left( \lambda +1 \right) } =0\quad$

$\Rightarrow 2\vec { a } \lambda +3\vec { b } \lambda +2\vec { a } +3\vec { b } -5\lambda \vec { a } -5\vec { b } =0$

$\Rightarrow 3\vec { b } \lambda -3\vec { a } \lambda +2\vec { a } -2\vec { b } =0\quad$

$\Rightarrow 3\lambda (\vec { b } -\vec { a } )=2\vec { b } -2\vec { a }$

$\Rightarrow 3\lambda =\cfrac { 2(\vec { b } -\vec { a } ) }{ \left( \vec { b } -\vec { a } \right) } \quad \Rightarrow \cfrac { \lambda }{ 1 } =\cfrac { 3 }{ 2 }$
hence the point

$C$ divides line segment

$AB$ in the ratio

$2:3$ .

Solve ${ (A+B) }^{ 2 }=?$

${\left(A+B\right)}^{2}=\left(A+B\right)\left(A+B\right)$

$=A.A+A.B+B.A+B.B$

$={A}^{2}+A.B+B.A+{B}^{2}$

Case

$1:$ If

$A.B=B.A\Rightarrow A.B+A.B=2A.B$

$\Rightarrow {\left(A+B\right)}^{2}={A}^{2}+2A.B+{B}^{2}$

Case

$2:$ If

$A.B\neq B.A$ then

${\left(A+B\right)}^{2}={A}^{2}+A.B+B.A+{B}^{2}$

Case

$3:$ If

$A.B=-B.A$ then

${\left(A+B\right)}^{2}={A}^{2}+{B}^{2}$

Find the area of the parallelogram with adjacent sides formed by $\overrightarrow { P }$ and $\overrightarrow { P }$
where $\overrightarrow { P } =2\hat { i } +3\hat { j } +4\hat { k }$ and $\overrightarrow { Q } =3\hat { i } +2\hat { j } -2\hat { k }$ expressed in meter.

Area of parallelogram is magnitude of cross product of two vectors

so crossprodect is

$-14i-18j-5k$

$area=\sqrt{(-14)^2+(-18)^2+(-5)^2}=\sqrt{545}$

Let $\bar { a } ,\bar { b } ,\bar { c }$ be vectors of length $3,4$ and $5$ respectively. Find the length of the vector $|\bar { a } +\bar { b } +\bar { c }|$ . if they are mutually perpendicular.

$|\bar a|=3\\|\bar b|=4\\|\bar c|=5$

Since they are perendicular

$|\bar a+\bar b+\bar c|\\\sqrt{(a^2+b^2+c^2)}\\\sqrt{9+16+25}\\\sqrt{50}=5\sqrt2$

In Fig. $ABCD$ is a regular hexagon, which vectors are:
Coinitial

1404577_1662207_ans_a99e410805694ead9e622060610f8de2.jpg

Two forces 7 and 3 N simultaneously act on a body. What is the value of their (i) maximum resultant, (ii) minimum resultant, and (iii) what will be the resultant if the forces act at right angle to each other?

Given:

$F_1=7N\\F_2=3N$

a. Resultant is maximum when both vectors act in same direction.

$R_{max} = A + B = 7 + 3 = 10 \ N$

b. Resultant is minimum when both vectors act in opposite direction

$R_{min} = A - B = 7 - 3 = 4 \ N$

c. If both vectors act at right angle, then

$R = \sqrt{A^2 + B^2} = \sqrt{7^2 + 3^2} = \sqrt{58} N$

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

The position vector of point

$R$ dividing the line segment joining two points

$P$ and

$Q$ in the ratio

$m:n$ is given by:
i. Internally:

$\dfrac {m\vec {Q}+n\vec {P}}{m+n}$
ii. Externally:

$\dfrac {m\vec {Q}-n\vec {P}}{m-n}$
Position vectors of

$P$ and

$Q$ are given as:

$\vec {OP}=\hat {i}+2\hat {j}-\hat {k}$ and

$\vec {OQ}=-\hat {i}+\hat {j}+\hat {k}$
(i) The position vector of point

$R$ which divides the line joining two points

$P$ and

$Q$ internally in the ratio

$2:1$ is given by,

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

$=\dfrac {-\hat {i}+4\hat {j}+\hat {k}}{3}=-\dfrac {1}{3}\hat {i}+\dfrac {4}{3}\hat {j}+\dfrac {1}{3}\hat {k}$
(ii) The position vector of point

$R$ which divides the line joining two points

$P$ and

$Q$ externally in the ratio

$2:1$ is given by,

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

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

Show that $\left(\vec{a}-\vec{b}\right)\times \left(\vec{a}+\vec{b}\right)=2\left(\vec{a}\times \vec{b}\right)$ .

1168664_879170_ans_a6de079a1515414e999fdfe2117c7c1f.jpg

If $\left(\bar{x}+\bar{y}\right)\cdot \left(\bar{x}-\bar{y}\right)=63$ and $|x|=8|\bar{y}|$ then, find $|\bar{x}|$ .

$(\vec x+\vec y)\cdot (\vec x-\vec y)=63$

$\implies |\vec x|^2-|\vec y^2|=63$

$|\vec x|=8|\vec y|\implies |\vec y| =\cfrac{|\vec x|}{8}$

$\implies |\vec x|^2-\cfrac{|\vec x^2|}{64}=63$

$\implies |\vec x|=8$

If $\vec R = (\vec A + \vec B),$ Show that $R^2 = A^2 + B^2 + 2 \ AB \ cos \theta$

${ \left| \overline { R } \right| }^{ 2 }=\overline { R } .R\implies\quad { \left| \overline { R } \right| }^{ 2 }=(\overline { A } +\overline { B } ).(\overline { A } +\overline { B } )$

$\implies\quad { \left| \overline { R } \right| }^{ 2 }=\overline { A } .\overline { A } +\overline { A } .\overline { B } +\overline { B } .\overline { A } +\overline { B } .\overline { B }$

$(\overline { A } .\overline { B } =\overline { B } .\overline { A } (for\quad dot\quad product))$

$\implies\quad { \left| \overline { R } \right| }^{ 2 }=\left| \overline { A } \right| .\left| \overline { A } \right| \cos { (0) } +\left| \overline { A } \right| .\left| \overline { B } \right| \cos { \theta } +\left| \overline { B } \right| .\left| \overline { A } \right| \cos { \theta }$

$+\left| \overline { B } \right| .\left| \overline { B } \right| \cos { (0) }$

$\implies\quad { \left| \overline { R } \right| }^{ 2 }={ \left| \overline { A } \right| }^{ 2 }+{ \left| \overline { B } \right| }^{ 2 }+2\left| \overline { A } \right| .\left| \overline { B } \right| \cos { \theta }$

$(where\quad { \left| \overline { R } \right| }^{ 2 }\quad is\quad dot\quad product\quad of\quad R.R\quad or{ \left| \overline { R } \right| }^{ 2 })$

Express the vector $\vec {a}=5 \hat {i}+5\hat {k}$ as sum of two vector such that one is parallel to the $\vec {b}=3\hat {i}+\hat {k}$ and other is perpendicular to $\vec {b}$

Let the vector parallel to

$\vec{b} = 3 \hat{i} + \hat{k}$ is

$n \left( 3 \hat{i} + \hat{k} \right)$ and the vector perpendicular to

$\vec{b}$ be

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

Given that adding these two vectors will give

$\vec{a} = 5 \hat{i} + 5 \hat{k}$

$\therefore \; n \left( 3 \hat{i} + \hat{k} \right) + x\hat{i} + y \hat{j} + z \hat{k} = 5 \hat{i} + 5 \hat{k}$

$\Rightarrow \; 3n \hat{i} + n \hat{k} + x\hat{i} + y \hat{j} + z \hat{k} = 5 \hat{i} + 5 \hat{k}$

Therefore,

$x + 3n = 5 ..... \left( 1 \right)$

$y = 0$

$z + n = 5 ..... \left( 2 \right)$

Since vector

$x\hat{i} + y \hat{j} + z \hat{k}$ is perpendicular to

$\vec{b}$ , their dot product must be zero.

$\therefore \; \left( 3 \hat{i} + \hat{k} \right) . \left( x\hat{i} + y \hat{j} + z \hat{k} \right) = 0$

$\Rightarrow \; 3x + z = 0$

$\Rightarrow \; z = -3x ..... \left( 3 \right)$

Substituting the value of z in

${eq}^{n} \left( 2 \right)$ , we have

$-3x + n = 5 ..... \left( 4 \right)$

From

${eq}^{n} \left( 1 \right) \& \left( 4 \right)$ , we get

$n = 2$

$x = -1$

On substituting the value of x in

${eq}^{n} \left( 2 \right)$ , we get

$z = -3 \times \left( -1 \right) = 3$

Hence the vector parallel to

$\vec{b}$ will be

$2 \left( 3\hat{i} + \hat{k} \right) \Rightarrow 6 \hat{i} + 2 \hat{k}$ and the vector perpendicular to

$\vec{b}$ will be

$-1 \hat{i} + 0 \hat{j} + 3 \hat{k} \Rightarrow \; - \hat{i} + 3 \hat{k}$ .

Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b}=3\hat{i}+\hat{k}$ and the other is perpendicular to $\vec{b}$ .

$\begin{array}{l} \vec { a } =5\hat { i } -2\hat { j } +5\hat { k } \\ \\ \vec { b } =3\hat { i } +\hat { k } \end{array}$

Consider,

$\vec a=\vec {\beta _1}+\vec {\beta _2}$

Where

$\vec {\beta _1}$ is parallel to

$\vec b$ and

$\vec {\beta _2}$ is perpendicular to

$\vec b$

Since,

$\vec {\beta _1}$ is parallel to

$\vec b$

Then,

$\vec {\beta _1}=\lambda \vec b=\lambda (3\hat i+\hat k)=3\lambda \hat i+\lambda \hat k$ ------

$(1)$

Now,

$\vec {a}=\vec {\beta _1}+\vec {\beta _2}$

$\vec {\beta _2}=\vec a-\vec {\beta _1}$

$\vec {\beta _2}=(5\hat i-2\hat j+5\hat k)-(3\lambda \hat i+\lambda \hat k)$

$\vec {\beta _2}=(5-3\lambda )\hat i-2\hat j+(5-\lambda )\hat k$ -----

$(2)$

Since,

$\vec {\beta _2}$ is perpendicular to

$\vec b$

$\vec {\beta _2} .\vec b=0$

Then,

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

$\lambda =2$

Therefore,

$\vec {\beta _1}=6\hat i+2\hat k$ from

$(1)$

$\vec {\beta _2}=-\hat i-2\hat j+3\hat k$ -----

$(2)$

Hence,

$\vec a=(6\hat i+2\hat k)+(-\hat i-2\hat j+3\hat k)$

The position vectors of $A,B,C,D$ are $\vec a,\vec b,2\vec a + 3\vec b$ and $\vec a - 2\vec b$ respectively. Show that $\overrightarrow {DB} = 3\vec b - \vec a$ and $\overrightarrow {AC} = \vec a + 3\vec b$

$\vec {OA}=\vec a$

$\vec {OB}=\vec b$

$\vec {OC}=2\vec a+3\vec b$

$\vec {OD}=\vec a-2\vec b$

$\vec {DB}=\vec {OB}-\vec {OD}=\vec b-(\vec a-2\vec b)=3\vec b-\vec a$

$\vec {AC}=\vec {OC}-\vec {OA}=(2\vec a+3\vec b)-(\vec a)=\vec a+3\vec b$

Let position vector of the points A,B and C are $\overrightarrow a \,,\,\overrightarrow b$ and $\,\overrightarrow c$ respectively. Point D divides line segment BC internally in the ratio $2:1$ . Find vector AD

P.V of

$\vec{D}=\dfrac{2(\vec{(c)}+1(\vec{b})}{2+1}=\dfrac{2\vec{c}+\vec{b}}{3}$

we have to find

$\vec{AD}$

we know by triangular law of addition

$\vec{BA}+\vec{AD}=\vec{BD}$

$\vec{AD}=\vec{BD}-\vec{BA}$

P.V of

$\vec{BD}=$ P.V of

$D-$ P.V of

$B$

$=\dfrac{2\vec{c}+\vec{b}}{3}-\vec{b}=\dfrac{2\vec{c}-2\vec{b}}{3}$

P.V of

$\vec{BA}=(\vec{a}-\vec{b})$

$\therefore \vec{AD}=\dfrac{2\vec{c}-2\vec{b}}{3}-\vec{a}+\vec{b}$

$=\dfrac{2}{3}\vec{c}-\vec{a}+\dfrac{1}{3}\vec{b}$

1096227_1164880_ans_037ddee58ed44d4b8bf5fb97fb7a8900.png

let ABCD is trapezium such that $\overrightarrow {AB} = 3\overrightarrow {DE} \,\,$ , E divides line segement AB internally in the ratio 2:1 and F is mid point of DC. if position vector of A,B and C are $\overrightarrow a \,,\,\overrightarrow b \,\,\,and\,\overrightarrow c \,\,\,$ respectively then find the vector $\overrightarrow {FE} \,$ .

Solution :-

REF.Image

Given

$\vec{AB}=3\vec{DC}$

$(\vec{b}-\vec{a})=3(\vec{c}-\vec{d})-(1)$

$\Rightarrow 2\vec{AE}= EB$

$\therefore \vec{E}= (\dfrac{\vec{a}+2\vec{b}}{3})$ (By section formula)

Similarly

$\vec{F}=\dfrac{\vec{c}+\vec{d}}{2}$

$\vec{FE}=\vec{E}-\vec{F}=\dfrac{\vec{a}+2\vec{b}}{3}-\dfrac{\vec{c}+\vec{d}}{2}$

$=\dfrac{2\vec{a}+4\vec{b}-3\vec{c}-3\vec{d}}{6}$

From (1)

$3\vec{d}=-\vec{b}+\vec{a}+3\vec{c}$

$\therefore \vec{FE}=\dfrac{2\vec{a}+4\vec{b}-3\vec{c}-\vec{a}+\vec{b}-3\vec{c}}{6}$

$\vec{FE}=\dfrac{\vec{a}+5\vec{b}-6\vec{c}}{6}$ Required vector

1107597_1166081_ans_e20f05a5ecca46febc4e352ba8a3f689.png

Evaluate the following:

$[2\hat{i}\hat{j}\hat{k}]+[\hat{i}\hat{k}\hat{j}]+[\hat{k}\hat{j}2\hat{i}]$

$\left[2\hat{i}\hat{j}\hat{k}\right]+\left[\hat{i}\hat{k}\hat{j}\right]+\left[\hat{k}\hat{j}2\hat{i}\right]$

We know,

$\left[x\vec{a}y\vec{b}z\vec{c}\right]=xyz\left[\vec{a}\vec{b}\vec{c}\right]$

$\Rightarrow$

$2\left[\hat{i}\hat{j}\hat{k}\right]+\left[\hat{i}\hat{k}\hat{j}\right]+2\left[\hat{k}\hat{j}\hat{i}\right]$

$\Rightarrow$

$2\left(\vec{i}\times \vec{j}\right).\vec{k}+\left(\vec{i}\times \vec{k}\right).\vec{j}+2\left(\vec{k}\times \vec{j}\right).\vec{i}$

$\left[\because \left[\vec{a}\vec{b}\vec{c}\right]=\left(\vec{a}\times \vec{b}\right).\vec{c}\right]$

$\Rightarrow$

$2\left(\vec{k}.\vec{k}\right)+\left(-j.\vec{j}\right)+2\left(-\vec{i}.\vec{i}\right)$

$\Rightarrow$

$2(1)+(-1)+2(-1)$

$\Rightarrow$

$2-1-2$

$\Rightarrow$

$-1$

A pyramid with vertex at point P has a regular hexagonal base ABCDEF. Position vectors of points A & B are $\hat{i}$ and $\hat{i} + 2 \hat{j}$ , respectively. Center of the base has position vector $\hat{i} +\hat{j} + \sqrt{3}\hat{k}$ . Altitude drawn from P on the base meets the diagonal AD at point G. Find all the possible positions vectors of G. It is given that the volume of the pyramid is 6 $\sqrt{3}$ cubic units and AP = 5 units.

Let the center of the base (0). Therefore,

$\left | \underset{AB}{\rightarrow} \right |$ = 2

$\bigtriangleup OAB$ =

$\frac{1}{4} X 4 X \sqrt{3}$ =

$\sqrt{3}$

Base area =

$6\sqrt{3}$ sq. units

Let height of the pyramid be h. Therefore,

$\frac{1}{3} X 6 \sqrt{3}$ h =

$6 \sqrt{3}\Rightarrow$ h = 3 units

It is given that

$\left | \underset{AP}{\rightarrow} \right |$ = 5. Therefore,

AG =

$\sqrt{25-9}$ = 4 units

$\Rightarrow \left | \underset{AG}{\rightarrow} \right |$ = 4

Now

$\left | \underset{AG}{\rightarrow} \right |$ and

$\left | \underset{AO}{\rightarrow} \right |$ are collinear. Therefore,

$\underset{AG}{\rightarrow}$ =

$\lambda \underset{AO}{\rightarrow}\Rightarrow\left | \underset{AG}{\rightarrow} \right |$ =

$\left | \lambda \right |\left | \underset{AO}{\rightarrow} \right |\Rightarrow2\left | \lambda \right |$ =

$4\Rightarrow \left | \lambda \right |$ =

$2\Rightarrow\underset{AG}{\rightarrow}$ =

$\pm2(\hat{i}+\hat{j}+\sqrt{3}\hat{k})\Rightarrow\underset{OG}$ =

$\pm2 (\hat{i}+\hat{j}+\sqrt{3}\hat{k})+\hat{i}\underset{OG}{\rightarrow}$ =

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

1655615_1783217_ans_a2681f82a6314f72b8de3ef99edc1304.png

If $\left| \bar { a } \right| =a$ , then find the value of the following.
$\left| \overrightarrow { a } \times \hat { i } \right| ^{ 2 }+\left| \overrightarrow { a } \times \hat { j } \right| ^{ 2 }+\left| \overrightarrow { a } \times \hat { k } \right| ^{ 2 }$

let

$\overrightarrow {a}$ makes angle

$\alpha ,\beta ,\gamma$ with

$x , y$ and

$z$ axis.

$\therefore | \overrightarrow {a} \times \hat {i} | = | \overrightarrow {a} | . 1. sin \ \alpha = a. sin\ \alpha$

Similarly ,

$| \overrightarrow {a} \times \hat {j} | = a. sin\ \beta$

and

$| \overrightarrow {a} \times \hat {k} | =a. sin\ \gamma$

$\therefore |\overrightarrow {a} \times \hat {i} |^2 + | \overrightarrow {a} \times \hat {k}|^2 = a^2 sin^2 \alpha + a^2 sin \beta + a^2 sin^2 \gamma$

$= a^2 [ sin^2 \alpha + sin^2 \beta + sin^2 \gamma ]$

$= a^2 [ 1- cos^2 \alpha +1 -cos^2 \beta +1 -cos^2 \gamma ]$

$= a^2 [ 3- (cos^2 \alpha + cos^2 \beta +cos^2 g) ]$

$= a^2 (3- 1) [ \because l^2 +m^2 +n^2 = 1 \Rightarrow cos^2 \alpha + cos^2 \beta + cos^2 \gamma = 1]$

$= 2a^2$

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

$P.V.$ of mid point

$\frac { pv\ of\ P+pv\ of\ Q }{ 2 } =\frac { (2\overrightarrow { i } +3\overrightarrow { j } +4\overline { k } )+(4\overrightarrow { i } +\overrightarrow { j } -2\overline { k } ) }{ 2 }$

$3i+2j+\bar{k}$

Show that the points $A, B$ and $C$ with the position vectors.
$\overrightarrow {a} = 3 \hat {i} -4 \hat {j} - 4 \hat {j} -4 , \overrightarrow {b} = 2 \hat {i} - \hat {j} + \hat {k}$ and $\overrightarrow {c} = \hat {i} - 3 \hat {j} - 5 \hat {k}$

$\overline{AB}=-\overrightarrow{i}+3\overrightarrow{j}+5\bar{k},\quad \overrightarrow{BC}=-\overrightarrow{i}-2\overrightarrow{j}-6\bar{k},\overrightarrow{AC}=-2\overrightarrow{i}+\overrightarrow{j}-\bar{k}$

$\overline{AB}+\overline{BO}+\overline{CA}=(-\overrightarrow{i} -\overrightarrow{i}+2i)+(3\overrightarrow{j}-2\overrightarrow{j}-\bar{j})+(5\overrightarrow{k}-6\bar{k}+\bar{k})=0$

hence

$A,B,C$ are the vertices of triangle

$|AB| \sqrt{1+9+25}=\sqrt{35}$ ,

$|BC| \sqrt{4+1+36}=\sqrt{41},$

$|AC|\sqrt{4+1+1}=\sqrt{6} ,$

$|AB|^2+|AC|^2=|BC|^2$

hence

$A,B,C$ from right angled triangle.

$ABCD$ is a parrallelogram. If $L$ and $M$ are the middle points of $BC$ and $CD$ respectively, then find (i) $AL$ and $AM$ interns of $AB$ and $AD$ . (ii) $\lambda$ , if $AM=\lambda AD-LM$

$\textbf{Step 1:Use Triangle law of vectors to get AL and AM in terms of AB and AD.}$

$(i)$

$\text{Let}$

$A$

$\text{be the origin of vectors,}$

$\overline{AB}=\overline{a},\ \ \overline{AC}=\overline{b},\ \ \overline{AD}=\overline{c}$

$\overline{AB}=\overline{DC}, \overline{BC}=\overline{AD}$

$\text{opposite sides of parallelogram.}$

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

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

$\overline{AL}=\dfrac{\overline{a}+\overline{b}}{2}$

$\overline{AM}=\dfrac{\overline{b}+\overline{c}}{2}$

$\overline{AL}=\dfrac{\overline{AB}+\overline{AC}}{2}=\dfrac{\overline{AB}+\overline{AB}+\overline{BC}}{2}$

$\,=\dfrac{2\overline{AB}+\overline{AB}}{2}$

$\therefore \overline{AL}=\overline{AB}+\dfrac{1}{2}\overline{AB}$

$\overline{AM}=\dfrac{\overline{AC}+\overline{AD}}{2}=\dfrac{\overline{AB}+\overline{BC}+\overline{AD}}{2}$

$=\dfrac{\overline{AB}+\overline{AD}+\overline{AD}}{2}$ [

$\because\boldsymbol{ \overline{BC}=\overline{AD}}$ ]

$\overline{AB}=\overline{AB}+\overline{AD}$

$(ii)$

$\overline{AB}=\overline{AM}=\lambda \overline{AC}$

$\overline{AB}+\dfrac{1}{2}\overline{AD}+\dfrac{1}{2}\overline{AB}+\overline{AD}=\lambda\overline{AC}+\dfrac{3}{2}(\overline{AC}+\overline{BC})$

$=\lambda \overline{AC}$

$\dfrac{3}{2}[\overline{AB}+\overline{BC}]=\lambda \overline{AC}$

$\dfrac{3}{2}\overline{AC}=\lambda\overline{AC}$

$\therefore\lambda=\dfrac{3}{2}$

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

Add the vectors

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

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

$\vec { c } =\hat { i } -6\hat { j } -7\hat { k }$ by writing them in the component form that is

$\vec { a } =\left< 1,-2,1 \right> \\$ ,

$\vec { b } =\left< -2,4,5 \right> \\$ and

$\vec { c } =\left< 1,-6,-7 \right>$

Then the sum of the vectors can be computed as follows:

$\vec { a } +\vec { b } +\vec { c } =\left< 1-2+1,-2+4-6,1+5-7 \right> \\$

$=\left< 0,-4,-1 \right>$

Hence the sum of the vectors is

$\left< 0,-4,-1 \right> =-4\hat { j } -\hat { k }. \\.$

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 2\hat {j}, \vec {b} = 2\hat {i} - 3\hat {j}, \vec {c} = 2\hat {i} + 3\hat {k}$

Add the vectors

$\vec { a } =\hat { i } -2\hat { j } \\$ ,

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

$\vec { c } =2\hat { i } +3\hat { k }$ by writing them in the component form that is

$\vec { a } =\left< 1,-2,0 \right> \\$ ,

$\vec { b } =\left< 2,-3,0 \right> \\$ and

$\vec { c } =\left< 2,0,3 \right>$

Then the sum of the vectors can be computed as follows:

$\vec { a } +\vec { b } +\vec { c } =\left< 1+2+2,-2-3+0,0+0+3 \right> \\$

$=\left< 5,-5,3 \right>$

Hence the sum of the vectors is

$\left< 5,-5,3 \right>=5\hat{i}-5\hat{j}+3\hat{k}$ .

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 3\hat {k},. \vec {b} = 2\hat {j} - \hat {k}, \vec {c} = 2\hat {i} - 3\hat {j} + 2\hat {k}$

Add the vectors

$\vec { a } =\hat { i } -3\hat { k } \\$ ,

$\vec { b } =2\hat { j } -\hat { k } \\$ and

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

by writing them in the component form that is

$\vec { a } =\left< 1,0,-3 \right> \\$ ,

$\vec { b } =\left< 0,2,-1 \right> \\$ and

$\vec { c } =\left< 2,-3,2 \right>$

The sum of the vectors can be computed as follows:

$\vec { a } +\vec { b } +\vec { c } =\left( 1+0+2,0+2-3,-3-1+2 \right) \\$

$=\left< 3,-1,-2 \right>$

Hence the sum of the vectors is

$\left< 3,-1,-2 \right>$ .

Let $\vec{a}$ and $\vec{b}$ are two vectors in space such that $\vec{a}=\dfrac{3\hat{i}+5\hat{j}-\hat{k}}{\sqrt{35}}, \vec{b}=\dfrac{5i-j+10\hat{k}}{\sqrt{126}}$ , then the value of $[\vec{a}+\vec{b} \vec{a}+2\vec{b} \vec{a}\times \vec{b}]$ is equal to?

$\begin{matrix} \vec { a } =\frac { { 3\hat { i } +5\hat { j } -\hat { k } } }{ { \sqrt { 35 } } } ,\vec { b } =\frac { { 5\hat { i } -\hat { j } +10\hat { k } } }{ { \sqrt { 126 } } } \\ Then,\, \, \, \left[ { \vec { a } +\vec { b } \cdot \vec { a } +2\vec { b } \vec { a } \times \vec { b } } \right] \\ \Rightarrow \vec { b } \cdot \vec { a } =\left( { \frac { { 3\hat { i } +5\hat { j } -\hat { k } } }{ { \sqrt { 35 } } } } \right) \cdot \left( { \frac { { 5\hat { i } -\hat { j } +10\hat { k } } }{ { \sqrt { 126 } } } } \right) \\ \Rightarrow \vec { b } \cdot \vec { a } =0 \\ and\, \, 2\vec { b } \cdot \vec { a } \times \vec { b } =0 \\ \therefore \left[ { \vec { a } +\vec { b } \cdot \vec { a } +2\vec { b } \cdot \vec { a } \times \vec { b } } \right] =\left[ { \frac { { 3\hat { i } +5\hat { j } -\hat { k } } }{ { \sqrt { 35 } } } +0+0 } \right] \\ =\frac { { 3\hat { i } } }{ { \sqrt { 35 } } } +\frac { { 5\hat { j } } }{ { \sqrt { 35 } } } -\frac { { \hat { k } } }{ { \sqrt { 35 } } } \, \, \, \, \, \, Ans. \\ \end{matrix}$

Find $|\vec{b}|$ , if $\left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=8$ and $|\vec{a}|=8|\vec{b}|$ .

1296500_1579577_ans_959c6d4dc6d04f19860d5758c13e6795.jpg

In $\triangle ABC$ , a point P is taken on AB such that AP/BP = 1/3 and a point Q is taken BC such that CQ/BQ = 3/If R is the point of intersection of the lines AQ and CP, using vector method, find the area of $\triangle ABC$ if the area of $\triangle BRC$ is 1 unit.

Taking A as origin let

$\overrightarrow{b}$ and

$\overrightarrow {c}$ be the postion vectors of

$B$ and

$C$ , respectively,

The position vector of

$Q$ is

$\dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4}$ and that of

$p$ is

$\dfrac{\overrightarrow {b}}{4}$

if

$\dfrac{AR}{QR}=\dfrac{λ}{1}$ then position vector of

$R = \lambda\left(\dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4}\right)$

$..(i)$

if

$\dfrac{CR}{RP}=\dfrac{μ}{1}$ then position vector of

$R = \dfrac{\mu \frac{\overrightarrow {b}}{4}+\overrightarrow {c}}{\mu +1}$

$...(ii)$

comparing

$(i)$ and

$(ii)$ , we have,

$\dfrac {3 \lambda}{4}=\dfrac{\mu}{4(\mu+1)}$ and

$\dfrac{\lambda}{4}=\dfrac{1}{\mu +1}$

solving ,

$\lambda=\dfrac{4}{13}$ and

$\mu=12$

Therefore, position , vector

$R$ is

$\dfrac{3 \overrightarrow {b}+ \overrightarrow {c}}{13}$

$\triangle ABC$ and

$\triangle BRC$ have the same base so, their areas are proporotional to

$AQ$ and

$RQ$

$\dfrac{\triangle ABC} {\triangle BRC}=\dfrac{\left| \dfrac{3 \overrightarrow {b} + \overrightarrow {c}}{4}\right|}{\left| \dfrac{3 \overrightarrow {b}+\overrightarrow {c}}{4} - \left( \dfrac{3\overrightarrow {b}+\overrightarrow {c}}{13} \right) \right|}=\dfrac{13}{9}$

Area of

$\triangle ABC$ is

$\dfrac{13}{9}$ units
1654544_1783900_ans_21f16e77eb8146e59dca6e6517b2650d.PNG

Vector Algebra - Class 12 Engineering Maths - Extra Questions

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

Find the sum of the following vectors:
$\vec {a} = \hat{i}-2\hat{j}, \vec{b} = 2\hat{i}-3\hat{j}, \vec{c} = 2\hat{i} + 3\hat{k}$ .

If $\overline { a } ,\overline { b } ,\overline { c }$ are the position vectors of the points $A,B,C$ respectively and $2\overline { a } +3\overline { b } -5\overline { c } =\overline { 0 }$ , then find the ratio in which the point $C$ divides line segment $AB$ .

Solve ${ (A+B) }^{ 2 }=?$

Find the area of the parallelogram with adjacent sides formed by $\overrightarrow { P }$ and $\overrightarrow { P }$
where $\overrightarrow { P } =2\hat { i } +3\hat { j } +4\hat { k }$ and $\overrightarrow { Q } =3\hat { i } +2\hat { j } -2\hat { k }$ expressed in meter.

Let $\bar { a } ,\bar { b } ,\bar { c }$ be vectors of length $3,4$ and $5$ respectively. Find the length of the vector $|\bar { a } +\bar { b } +\bar { c }|$ . if they are mutually perpendicular.

In Fig. $ABCD$ is a regular hexagon, which vectors are:
Coinitial

Two forces 7 and 3 N simultaneously act on a body. What is the value of their (i) maximum resultant, (ii) minimum resultant, and (iii) what will be the resultant if the forces act at right angle to each other?

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

Show that $\left(\vec{a}-\vec{b}\right)\times \left(\vec{a}+\vec{b}\right)=2\left(\vec{a}\times \vec{b}\right)$ .

If $\left(\bar{x}+\bar{y}\right)\cdot \left(\bar{x}-\bar{y}\right)=63$ and $|x|=8|\bar{y}|$ then, find $|\bar{x}|$ .

If $\vec R = (\vec A + \vec B),$ Show that $R^2 = A^2 + B^2 + 2 \ AB \ cos \theta$

Express the vector $\vec {a}=5 \hat {i}+5\hat {k}$ as sum of two vector such that one is parallel to the $\vec {b}=3\hat {i}+\hat {k}$ and other is perpendicular to $\vec {b}$

Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b}=3\hat{i}+\hat{k}$ and the other is perpendicular to $\vec{b}$ .

The position vectors of $A,B,C,D$ are $\vec a,\vec b,2\vec a + 3\vec b$ and $\vec a - 2\vec b$ respectively. Show that $\overrightarrow {DB} = 3\vec b - \vec a$ and $\overrightarrow {AC} = \vec a + 3\vec b$

Let position vector of the points A,B and C are $\overrightarrow a \,,\,\overrightarrow b$ and $\,\overrightarrow c$ respectively. Point D divides line segment BC internally in the ratio $2:1$ . Find vector AD

Evaluate the following:

$[2\hat{i}\hat{j}\hat{k}]+[\hat{i}\hat{k}\hat{j}]+[\hat{k}\hat{j}2\hat{i}]$

If $\left| \bar { a } \right| =a$ , then find the value of the following.
$\left| \overrightarrow { a } \times \hat { i } \right| ^{ 2 }+\left| \overrightarrow { a } \times \hat { j } \right| ^{ 2 }+\left| \overrightarrow { a } \times \hat { k } \right| ^{ 2 }$

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

Show that the points $A, B$ and $C$ with the position vectors.
$\overrightarrow {a} = 3 \hat {i} -4 \hat {j} - 4 \hat {j} -4 , \overrightarrow {b} = 2 \hat {i} - \hat {j} + \hat {k}$ and $\overrightarrow {c} = \hat {i} - 3 \hat {j} - 5 \hat {k}$

$ABCD$ is a parrallelogram. If $L$ and $M$ are the middle points of $BC$ and $CD$ respectively, then find (i) $AL$ and $AM$ interns of $AB$ and $AD$ . (ii) $\lambda$ , if $AM=\lambda AD-LM$

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

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 2\hat {j}, \vec {b} = 2\hat {i} - 3\hat {j}, \vec {c} = 2\hat {i} + 3\hat {k}$

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 3\hat {k},. \vec {b} = 2\hat {j} - \hat {k}, \vec {c} = 2\hat {i} - 3\hat {j} + 2\hat {k}$

Let $\vec{a}$ and $\vec{b}$ are two vectors in space such that $\vec{a}=\dfrac{3\hat{i}+5\hat{j}-\hat{k}}{\sqrt{35}}, \vec{b}=\dfrac{5i-j+10\hat{k}}{\sqrt{126}}$ , then the value of $[\vec{a}+\vec{b} \vec{a}+2\vec{b} \vec{a}\times \vec{b}]$ is equal to?

Find $|\vec{b}|$ , if $\left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=8$ and $|\vec{a}|=8|\vec{b}|$ .

In $\triangle ABC$ , a point P is taken on AB such that AP/BP = 1/3 and a point Q is taken BC such that CQ/BQ = 3/If R is the point of intersection of the lines AQ and CP, using vector method, find the area of $\triangle ABC$ if the area of $\triangle BRC$ is 1 unit.

Class 12 Engineering Maths Extra Questions

Vector Algebra - Class 12 Engineering Maths - Extra Questions

Find 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).

Find the sum of the following vectors:→a=ˆi−2ˆj,→b=2ˆi−3ˆj,→c=2ˆi+3ˆk\vec {a} = \hat{i}-2\hat{j}, \vec{b} = 2\hat{i}-3\hat{j}, \vec{c} = 2\hat{i} + 3\hat{k}.

If ¯a,¯b,¯c\overline { a } ,\overline { b } ,\overline { c } are the position vectors of the points A,B,CA,B,C respectively and 2¯a+3¯b−5¯c=¯02\overline { a } +3\overline { b } -5\overline { c } =\overline { 0 } , then find the ratio in which the point CC divides line segment ABAB.

Solve (A+B)2=?{ (A+B) }^{ 2 }=?

Let ˉa,ˉb,ˉc\bar { a } ,\bar { b } ,\bar { c } be vectors of length 3,43,4 and 55 respectively. Find the length of the vector |ˉa+ˉb+ˉc||\bar { a } +\bar { b } +\bar { c }| . if they are mutually perpendicular.

In Fig. ABCDABCD is a regular hexagon, which vectors are:Coinitial

Two forces 7 and 3 N simultaneously act on a body. What is the value of their (i) maximum resultant, (ii) minimum resultant, and (iii) what will be the resultant if the forces act at right angle to each other?

Find the position vector of a point RR which divides the line joining two points PP and QQ whose position vectors are ˆi+2ˆj−ˆk\hat {i}+2\hat {j}-\hat {k} and −ˆi+ˆj+ˆk-\hat {i}+\hat {j}+\hat {k} respectively, in the ratio 2:12:1,(i) internally(ii) externally

Show that (→a−→b)×(→a+→b)=2(→a×→b)\left(\vec{a}-\vec{b}\right)\times \left(\vec{a}+\vec{b}\right)=2\left(\vec{a}\times \vec{b}\right).

If (ˉx+ˉy)⋅(ˉx−ˉy)=63\left(\bar{x}+\bar{y}\right)\cdot \left(\bar{x}-\bar{y}\right)=63 and |x|=8|ˉy||x|=8|\bar{y}| then, find |ˉx||\bar{x}|.

If →R=(→A+→B),\vec R = (\vec A + \vec B), Show that R2=A2+B2+2 AB cosθR^2 = A^2 + B^2 + 2 \ AB \ cos \theta

Express the vector →a=5ˆi+5ˆk\vec {a}=5 \hat {i}+5\hat {k} as sum of two vector such that one is parallel to the →b=3ˆi+ˆk\vec {b}=3\hat {i}+\hat {k} and other is perpendicular to →b\vec {b}

Express the vector →a=5ˆi−2ˆj+5ˆk\vec{a}=5\hat{i}-2\hat{j}+5\hat{k} as the sum of two vectors such that one is parallel to the vector →b=3ˆi+ˆk\vec{b}=3\hat{i}+\hat{k} and the other is perpendicular to →b\vec{b}.

The position vectors of A,B,C,DA,B,C,D are →a,→b,2→a+3→b\vec a,\vec b,2\vec a + 3\vec b and →a−2→b\vec a - 2\vec b respectively. Show that →DB=3→b−→a\overrightarrow {DB} = 3\vec b - \vec a and →AC=→a+3→b\overrightarrow {AC} = \vec a + 3\vec b

Let position vector of the points A,B and C are →a,→b\overrightarrow a \,,\,\overrightarrow b and →c\,\overrightarrow c respectively. Point D divides line segment BC internally in the ratio 2:12:1. Find vector AD

Evaluate the following:[2ˆiˆjˆk]+[ˆiˆkˆj]+[ˆkˆj2ˆi][2\hat{i}\hat{j}\hat{k}]+[\hat{i}\hat{k}\hat{j}]+[\hat{k}\hat{j}2\hat{i}]

Find 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) .

ABCDABCD is a parrallelogram. If LL and MM are the middle points of BCBC and CDCD respectively, then find (i) ALAL and AMAM interns of ABAB and ADAD. (ii) λ\lambda, if AM=λAD−LMAM=\lambda AD-LM

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

Find the sum of the following vectors:\vec {a} = \hat {i} - 2\hat {j}, \vec {b} = 2\hat {i} - 3\hat {j}, \vec {c} = 2\hat {i} + 3\hat {k}\vec {a} = \hat {i} - 2\hat {j}, \vec {b} = 2\hat {i} - 3\hat {j}, \vec {c} = 2\hat {i} + 3\hat {k}

Find the sum of the following vectors:\vec {a} = \hat {i} - 3\hat {k},. \vec {b} = 2\hat {j} - \hat {k}, \vec {c} = 2\hat {i} - 3\hat {j} + 2\hat {k}\vec {a} = \hat {i} - 3\hat {k},. \vec {b} = 2\hat {j} - \hat {k}, \vec {c} = 2\hat {i} - 3\hat {j} + 2\hat {k}

Find |\vec{b}||\vec{b}|, if \left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=8\left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=8 and |\vec{a}|=8|\vec{b}||\vec{a}|=8|\vec{b}|.

In \triangle ABC\triangle ABC, a point P is taken on AB such that AP/BP = 1/3 and a point Q is taken BC such that CQ/BQ = 3/If R is the point of intersection of the lines AQ and CP, using vector method, find the area of \triangle ABC\triangle ABC if the area of \triangle BRC\triangle BRC is 1 unit.

Class 12 Engineering Maths Extra Questions

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

Find the sum of the following vectors:
$\vec {a} = \hat{i}-2\hat{j}, \vec{b} = 2\hat{i}-3\hat{j}, \vec{c} = 2\hat{i} + 3\hat{k}$ .

If $\overline { a } ,\overline { b } ,\overline { c }$ are the position vectors of the points $A,B,C$ respectively and $2\overline { a } +3\overline { b } -5\overline { c } =\overline { 0 }$ , then find the ratio in which the point $C$ divides line segment $AB$ .

Solve ${ (A+B) }^{ 2 }=?$

Let $\bar { a } ,\bar { b } ,\bar { c }$ be vectors of length $3,4$ and $5$ respectively. Find the length of the vector $|\bar { a } +\bar { b } +\bar { c }|$ . if they are mutually perpendicular.

In Fig. $ABCD$ is a regular hexagon, which vectors are:
Coinitial

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

Show that $\left(\vec{a}-\vec{b}\right)\times \left(\vec{a}+\vec{b}\right)=2\left(\vec{a}\times \vec{b}\right)$ .

If $\left(\bar{x}+\bar{y}\right)\cdot \left(\bar{x}-\bar{y}\right)=63$ and $|x|=8|\bar{y}|$ then, find $|\bar{x}|$ .

If $\vec R = (\vec A + \vec B),$ Show that $R^2 = A^2 + B^2 + 2 \ AB \ cos \theta$

Express the vector $\vec {a}=5 \hat {i}+5\hat {k}$ as sum of two vector such that one is parallel to the $\vec {b}=3\hat {i}+\hat {k}$ and other is perpendicular to $\vec {b}$

Express the vector $\vec{a}=5\hat{i}-2\hat{j}+5\hat{k}$ as the sum of two vectors such that one is parallel to the vector $\vec{b}=3\hat{i}+\hat{k}$ and the other is perpendicular to $\vec{b}$ .

The position vectors of $A,B,C,D$ are $\vec a,\vec b,2\vec a + 3\vec b$ and $\vec a - 2\vec b$ respectively. Show that $\overrightarrow {DB} = 3\vec b - \vec a$ and $\overrightarrow {AC} = \vec a + 3\vec b$

Let position vector of the points A,B and C are $\overrightarrow a \,,\,\overrightarrow b$ and $\,\overrightarrow c$ respectively. Point D divides line segment BC internally in the ratio $2:1$ . Find vector AD

Evaluate the following:

$[2\hat{i}\hat{j}\hat{k}]+[\hat{i}\hat{k}\hat{j}]+[\hat{k}\hat{j}2\hat{i}]$

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

$ABCD$ is a parrallelogram. If $L$ and $M$ are the middle points of $BC$ and $CD$ respectively, then find (i) $AL$ and $AM$ interns of $AB$ and $AD$ . (ii) $\lambda$ , if $AM=\lambda AD-LM$

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

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 2\hat {j}, \vec {b} = 2\hat {i} - 3\hat {j}, \vec {c} = 2\hat {i} + 3\hat {k}$

Find the sum of the following vectors:
$\vec {a} = \hat {i} - 3\hat {k},. \vec {b} = 2\hat {j} - \hat {k}, \vec {c} = 2\hat {i} - 3\hat {j} + 2\hat {k}$

Find $|\vec{b}|$ , if $\left(\vec{a}+\vec{b}\right)\left(\vec{a}-\vec{b}\right)=8$ and $|\vec{a}|=8|\vec{b}|$ .

In $\triangle ABC$ , a point P is taken on AB such that AP/BP = 1/3 and a point Q is taken BC such that CQ/BQ = 3/If R is the point of intersection of the lines AQ and CP, using vector method, find the area of $\triangle ABC$ if the area of $\triangle BRC$ is 1 unit.