MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 8

Find

$k$ if magnitude of vectors joining

$(0,k,0)$ and

$(1,1,1)$ is

$\sqrt {11}$

0%
$-3$
0%
$2$
0%
$-2$
0%
$0$

Explanation

Using definition of magnitude, we have

$11 = 1 +$

$(k-1)^{2}$

$+1$

$(k-1)^{2}$

$= 9$

$\Rightarrow k^2-2k+1-9=0$

$\Rightarrow k = +4$ or

$-2$

So, the answer is

$-2$ .

Let

$ABC$ be an acute scalene triangle, and

$O$ and

$H$ be its circumcentre and orthocentre respectively. Further let

$N$ be the midpoint of

$OH$ . The value of the vector sum

$\vec {NA} + \vec {NB} + \vec {NC}$ is

0%
$\vec {0}$ (zero vector)
0%
$\vec {HO}$
0%
$\dfrac {1}{2}\vec {HO}$
0%
$\dfrac {1}{2}\vec {OH}$

Explanation

$\overrightarrow{HE}=\overrightarrow{NE}-\overrightarrow{NH}$

$\Rightarrow \dfrac{1}2\overrightarrow{AH}=\dfrac{1}{2}[\overrightarrow{NB}+\overrightarrow{NC}]-\dfrac{1}{2}\overrightarrow{OH}$ ....

$(\because HE\perp BC,AH\perp BC$

$\Rightarrow \overrightarrow{HE}\parallel \overrightarrow{AH}$

$HE=R\cos(A),AH=2R\cos(A)\\\Rightarrow \overrightarrow{HE}=\dfrac{1}{2}\overrightarrow{AH}$

$E$ midpoint of

$BE$

$\Rightarrow \overrightarrow{NE}=\dfrac{1}{2}[\overrightarrow{NB}+\overrightarrow{NC}]$

$N$ midpoint of

$OH$

$\Rightarrow \overrightarrow{NH}=\dfrac{1}{2}\overrightarrow{OH} )\\\Rightarrow \overrightarrow{NH}-\overrightarrow{NA}=\overrightarrow{NB}+\overrightarrow{NC}+\overrightarrow{H)}$

$\Rightarrow \overrightarrow{NA}+\overrightarrow{NB}+\overrightarrow{NC}=\overrightarrow{HO}+\dfrac{1}{2}\overrightarrow{OH}=\dfrac{1}{2}\overrightarrow{HO}$

The projection of the vector

$\hat {i} - 2\hat {j} + \hat {k}$ on the vector

$4\hat {i} - 4\hat {j} + 7\hat {k}$ is

0%
$\dfrac {5}{19}\sqrt {5}$
0%
$\dfrac {19}{9}$
0%
$\dfrac {9}{19}$
0%
$\dfrac {1}{19}\sqrt {6}$

Explanation

Projection of

$\vec a$ on

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

Projection of

$\hat i-2\hat j+\hat k$ on vector

$4\hat i-4\hat j+7\hat k=\dfrac {4+8+7}{\sqrt {4^2+4^2+7^2}}=\dfrac {19}9$

So, Option B is correct.

The position vectors of A, B are a, 6 respectively. The position vector of C is

$\dfrac {5\bar{a}}{3} -\bar{b}$ . Then 3

0%
C is inside the $\Delta OAB$
0%
C is outside the $\Delta OAB$ but inside the angle OAB
0%
C is outside the $\Delta OAB$ but inside the angle OBA
0%
None of these

Explanation

We have given that

The position vector of

$A$ and

$B$ are

$a$ and

$6$ respectively.

$\therefore$

Now ,

The position vector of

$C$ is

$\dfrac{{5\overline a }}{3} - \overline b$ ,

Then, the position vector at

$C$ os inside the

$\Delta OAB$

Hence, the option

$(A)$ is correct.

$\overline { a } ,\overline { b } ,\overline { c }$ are three vectors such that

$\left| \overline { a } \right| =1,\left| \overline { b } \right| =2,\left| \overline { c } \right| =3$ and

$\overline { b } ,\overline { c }$ are perpendicular. IF projection of

$\overline { b }$ on

$\overline { a }$ is the same as the projection of

$\overline { c }$ on

$\overline { a }$ , then

$\left| \overline { a } -\overline { b } +\overline { c } \right|$

0%
$\sqrt { 2 }$
0%
$\sqrt { 7 }$
0%
$\sqrt { 14 }$
0%
$\sqrt { 21 }$

Explanation

$\textbf{Step1-Apply condition of Dot product of two vectors}$

$\text{It is given that projection of }$

$\vec{b}$

$\text{on}$

$\vec{a}$

$\text{is same as the projection of}$

$\vec{c}$ on

$\vec{a}$ .

$\implies \vec{b}.\vec{a} = \vec{c}.\vec{a}$ ...[1]

$\text{Also,}$

$\vec{b}$

$\text{and}$

$\vec{c}$

$\text{are perpendicular}$

$\implies \vec{b}.\vec{c} = 0$ ...[2]

$\text{Now,}$

$|\vec{a} - \vec{b} + \vec{c}|^2 = |\vec{a}|^2 +|\vec{b}|^2 +|\vec{c}|^2 -2\vec{b}.\vec{a} +2\vec{c}.\vec{a} -2\vec{b}.\vec{c}$

$|\vec{a} - \vec{b} + \vec{c}|^2 = |\vec{a}|^2 +|\vec{b}|^2 +|\vec{c}|^2 +0$ ...(Using [1] and [2])

$|\vec{a} - \vec{b} + \vec{c}|^2 = 1+4+9 = 14$

$\implies |\vec{a} - \vec{b} + \vec{c}| = \sqrt{14}$

$\textbf{Hence , option C is correct}$

$\bar{a},\bar{b}, \bar{c}$ are unit vectors such that

$\bar{a}+ \bar{b}+ \bar{c}=\bar{0}$ , then

$\bar{a}.\bar{b}+ \bar{b}.\bar{c}+ \bar{c}.\bar{a}=$

0%
$\dfrac{3}{2}$
0%
$-\dfrac{3}{2}$
0%
$\dfrac{1}{2}$
0%
$-\dfrac{1}{2}$

Explanation

Given,

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

Squaring both sides,

$\left(\bar{a}\right)^2+\left(\bar{b}\right)^2+\left(\bar{c}\right)^2+2.\bar{a}.\bar{b}+2.\bar{b}.\bar{c}+2.\bar{c}.\bar{a}=0$

$1+1+1+2\left(\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}\right)=0$

$\Rightarrow\left(\bar{a}.\bar{b}+\bar{b}.\bar{c}+\bar{c}.\bar{a}\right)=-\dfrac{3}{2}$ is the correct answer.

Let

$\vec {a} = x\hat {i} + 12\hat {j} - \hat {k}, \vec {b} = 2\hat {i} + 2x\hat {j} + \hat {k}$ and

$\vec {c} = \hat {i} + \hat {k}$ . If ordered set

$[\vec {b} \vec {c} \vec {a}]$ is left handed, then.

0%
$x\epsilon (2, \infty)$
0%
$x\epsilon (-\infty, -3)$
0%
$x\epsilon (-3, 2)$
0%
$x\epsilon \left \{3, 2\right \}$

Explanation

Given,

$\vec { a } =x\hat { i } +12\hat { j } -\hat { k } \\ \vec { b } =2\hat { i } +2x\hat { j } +\hat { k } \\ \vec { c } =\hat { i } +\hat { k }$

the ordered set

$[\vec { b } \vec { c } \vec { a } ]$ is left handed if and only if

$[\vec { b } \vec { c } \vec { a } ]<0$

Now,

$[\vec { b } \vec { c } \vec { a } ]=\vec { b } .(\vec { c } \times \vec { d } )=2{ x }^{ 2 }+2x-12=(2x-4)(x+3)$

the expression

$2{ x }^{ 2 }+2x-12$ is negative when

$x\quad \epsilon (-3,2)$

Hence

$[\vec { b } \vec { c } \vec { a } ]$ is left handed for

$x\quad \epsilon (-3,2)$

1151551_873478_ans_2845d129d78d49ebbd8b35ae1effa0cf.png

$\vec {a}$ and

$\vec {b}$ are unit vectors, then angle between

$\vec {a}$ and

$\vec {b}$ for

$\sqrt {3} \vec {a} - \vec {b}$ to be unit vector is

0%
$60^{\circ}$
0%
$90^{\circ}$
0%
$45^{\circ}$
0%
$30^{\circ}$

Explanation

$\textbf{Step 1: Dot product of vectors}$

Let angle between

$\vec{a}$ and

$\vec{b}$ is

$\theta$

Dot product between them:

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

$\vec{a}$ and

$\vec{b}$ are unit vetors therefore

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

$\Rightarrow \vec{a}\cdot \vec{b} = \cos \theta$

$....(1)$

$\textbf{Step 2: Equation formation}$

For

$\sqrt{3}\vec{a}-\vec{b}$ to be a unit vector,

$|\sqrt{3}\vec{a} - \vec{b}|^2 = 1$

$\Rightarrow (\sqrt{3})^2 |\vec{a}|^2 + |\vec{b}|^2 - 2\sqrt{3} \vec{a}\cdot \vec{b} = 1$

$\Rightarrow (\sqrt{3})^2 (1)^2 + (1)^2 - 2\sqrt{3} \cos \theta=1$ [ Using

$(1)$ ]

$\Rightarrow 4 - 2\sqrt{3} \cos\theta=1$

$\Rightarrow$

$\dfrac{-3}{-2\sqrt{3}} = \cos \theta$

$\Rightarrow$

$\cos \theta = \dfrac{\sqrt{3}}{2}$

$\Rightarrow$

$\theta = 30^o$

Hence option

$D$ is correct.

2112476_879428_ans_e4c4c6e42cc84251ab7dbc9ddc424dbe.png

Which is a unit vector?

0%
$(Cos \alpha, 2 Sin \alpha)$
0%
$(Sin \alpha, Cos \alpha)$
0%
$(1, -1)$
0%
$(2Cos \alpha, Sin \alpha)$

Explanation

$(\sin \alpha/\cos\alpha)$

Let

$\vec{a}=\sin\alpha\hat{i}$

$+\cos\alpha\hat{j}\implies |\vec{x}|=\sqrt{\sin^2\alpha+\cos^2\alpha}=1$

$\therefore (\sin\alpha,\cos\alpha)$ is a unit vector.

$\vec a+2\vec b+3 \vec c =0$ , then

$\vec{b}\times \vec{c}+\vec{c}\times \vec{a}+\vec{a} \times \vec{b}$ equals to:

0%
$6\left( b\times c \right)$
0%
$\left( a\times b \right)$
0%
$6\left( c\times a \right)$
0%
0

Explanation

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

$(\vec a+2\vec b+3\vec c)\times \vec c=0\implies \vec c\times \vec a=2(\vec b\times \vec c)$

$(\vec a+2\vec b+3\vec c)\times \vec b=0\implies \vec a\times \vec b=3(\vec b\times \vec c)$

$\vec b\times \vec c+\vec c\times \vec a+\vec a\times \vec b=(\vec b\times \vec c)+2(\vec b\times \vec c)+3(\vec b\times \vec c)=6(\vec b\times \vec c)=3(\vec c\times \vec a)=2(\vec a\times \vec b)$

Let

$\overline a ,\overline b ,\overline c$ be vectors of length

$3, 4, 5$ respectively. Let

$\overline a$ be perpendicular to

$\overline b + \overline c ,\overline b \,to\,\overline c + \overline a \,{\text{and}}\,\overline c \,{\text{to}}\,\overline a + \overline b .\,{\text{Then}}|\overline a + \overline b + \overline c |$ is equals to:

0%
$2\sqrt 5$
0%
$2\sqrt 2$
0%
$10\sqrt 5$
0%
$5\sqrt 2$

Explanation

According to question,

$\vec { a } .\left( \vec { b } +\vec { c } \right) =0$

$\vec { b } .\left( \vec { c } +\vec { a } \right) =0$

$\vec { c } .\left( \vec { a } +\vec { b } \right) =0$

$\Rightarrow \quad \vec { a } .\vec { b } +\vec { a } .\vec { c } =0\quad \quad \quad \rightarrow \left( 1 \right)$

$\Rightarrow \quad \vec { b } .\vec { c } +\vec { b } .\vec { a } =0\quad \quad \quad \rightarrow \left( 2 \right)$

$\Rightarrow \quad \vec { c } .\vec { a } +\vec { c } .\vec { b } =0\quad \quad \quad \rightarrow \left( 3 \right)$

${ \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }={ \left| a \right| }^{ 2 }+{ \left| b \right| }^{ 2 }+{ \left| c \right| }^{ 2 }+2\left( \vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { a } .\vec { c } \right) \quad \quad \quad \rightarrow (4)$

Adding (1), (2) and (3),

$2\left( \vec { a } .\vec { b } +\vec { b } .\vec { c } +\vec { a } .\vec { c } \right) =0\quad \quad \quad (\because \vec { a } .\vec { b } =\vec { b } .\vec { a } )$

$\Rightarrow { \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }={ \left| a \right| }^{ 2 }+{ \left| b \right| }^{ 2 }+{ \left| c \right| }^{ 2 }+0$

$\Rightarrow \quad { \left| \vec { a } +\vec { b } +\vec { c } \right| }^{ 2 }=9+16+25=50$

$\Rightarrow \quad { \left| \vec { a } +\vec { b } +\vec { c } \right| }=5\sqrt { 2 }$

If the projection of

$\vec{a}$ on

$\vec{b}$ and the projection of

$\vec{b}$ on

$\vec{a}$ are equal then the angle between

$\vec{a}+\vec{b}$ and

$\vec{a}-\vec{b}$ is

0%
$\large{\frac{\pi}{3}}$
0%
$\large{\frac{\pi}{2}}$
0%
$\large{\frac{\pi}{4}}$
0%
$\large{\frac{2\pi}{3}}$

Explanation

Projection of

$\vec{a}$ on

$\vec{b}$

$=\large{\frac{\vec{a}.\vec{b}}{|\vec{b}|}}$ , and projection of

$\vec{b}$ on

$\vec{a}$

$=\large{\frac{\vec{a}.\vec{b}}{|\vec{a}|}}$

$\therefore \large{\frac{\vec{a}.\vec{b}}{|\vec{b}|}}=\large{\frac{\vec{a}.\vec{b}}{|\vec{a}|}}$

$\therefore |\vec{b}|=|\vec{a}|$ .
Now, angle between

$\vec{a}+\vec{b}$ and

$\vec{a}-\vec{b}$ be

$\theta$ .
Then

$\cos \theta=\large{\frac{(\vec{a}+\vec{b}).(\vec{a}-\vec{b})}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}}$
or,

$\cos \theta =\large{\frac{|\vec{a}|^2-|\vec{b}|^2}{|\vec{a}+\vec{b}||\vec{a}-\vec{b}|}}$
or,

$\cos \theta =0$ [Since,

$|\vec{b}|=|\vec{a}|$ ]
or,

$\theta=\large{\frac{\pi}{2}}$

The value of

$|\overrightarrow A + \overrightarrow B - \overrightarrow C + \overrightarrow D |$ can be zero if :-

0%
$|\overrightarrow A | = 5,\,|\overrightarrow B | = 3,|\overrightarrow C | = 4;|\overrightarrow D | = 12$
0%
$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 5$
0%
$|\overrightarrow A | = 2\sqrt 2 ,\,|\overrightarrow B | = 2,|\overrightarrow C | = 2;|\overrightarrow D | = 10$
0%
$|\overrightarrow A | = 5,\,|\overrightarrow B | = 4,|\overrightarrow C | = 3;|\overrightarrow D | = 8$

Let

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

$\vec{c} = \vec{i} + \vec{j} - 2\vec{k}$ be three vectors. A vector of the type

$\vec{b} + \lambda \vec{c}$ for some scalar

$\lambda$ , whose projection on

$\vec{a}$ is of magnitude

$\sqrt {\frac{2}{3}}$ . Thenthe value of

$\lambda$ is

0%
$1$
0%
$0$
0%
$-1$
0%
$2$

Explanation

Consider the given vectors,

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

Given that projection of $\overrightarrow{b}+\lambda \overrightarrow{c}$ on $\overrightarrow{a}$ = $\sqrt{\dfrac{2}{3}}$

Hence,

$\dfrac{\left[ \overrightarrow{b}+\lambda \overrightarrow{c} \right].\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$ = $\sqrt{\dfrac{2}{3}}$

$\dfrac{\left[ \widehat{i}+2\widehat{j}-\widehat{k}+\lambda \left( \widehat{i}+\widehat{j}-2\widehat{k} \right) \right].\left( 2\widehat{i}-\widehat{j}+\widehat{k} \right)}{\left| 2\widehat{i}-\widehat{j}+\widehat{k} \right|}=\sqrt{\dfrac{2}{3}}$

$\dfrac{\left[ \left( 1+\lambda \right)\widehat{i}+\left( 2+\lambda \right)\widehat{j}+\left( 1-2\lambda \right)\widehat{k} \right].\left( 2\widehat{i}-\widehat{j}+\widehat{k} \right)}{\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}}=\sqrt{\dfrac{2}{3}}$

$\dfrac{\left( 1+\lambda \right).2+\left( 2+\lambda \right)\left( -1 \right)+\left( 1-2\lambda \right)\left( 1 \right)}{\sqrt{6}}=\sqrt{\dfrac{2}{3}}$

$2+2\lambda -2-\lambda +1-2\lambda =\sqrt{\dfrac{2}{3}}.\sqrt{6}$

$-\lambda +1=2$

$-\lambda =1$

$\lambda =-1$

Hence, this is the answer.

Given that P = 12, Q = 5 and R = 13 also

$\vec P + \vec Q = \vec R$ , then the angle between

$\vec P$ and

$\vec Q$ will be :

0%
${\pi}$
0%
${\pi}\over{2}$
0%
zero
0%
${\pi}\over{4}$

Explanation

Consider the problem

Let

$x$ is the angle between

$P$ and

$Q$ .

$\vec R=\vec P+\vec Q$

$\therefore 13=\sqrt { ({ 5^{ 2 } }+{ { 12 }^{ 2 } }+(2{ { \times } }5{ { \times } }12\times Cosx)) }$ [

$\because|\vec a+\vec b|=\sqrt{a^2+b^2+2ab\cos\theta}$ ]

$\Rightarrow 169=169{ { + } }120\cos x$

$\Rightarrow 0=120\cos x$

$\Rightarrow \cos x=0$

$\therefore x=90^{\circ}=\dfrac{\pi}{2}$

Hence option

$B$ is the correct answer.

The vertices of a triangle are

$A(1,1,2),B(4,3,1)$ and

$C(2,3,5).$ A vector representing the internal bisector of the angle

$A$ id

0%
$\hat{i}+\hat{j}+2\hat{k}$
0%
$2\hat{i}-2\hat{j}+\hat{k}$
0%
$2\hat{i}+2\hat{j}-\hat{k}$
0%
$2\hat{i}+2\hat{j}+\hat{k}$

Explanation

$\overrightarrow{AB}=(4-1)\hat i +(3-1)\hat j+(1-2)\hat k=3\hat i+2\hat j -1\hat k$

$\overrightarrow{AC}=(2-1)\hat i +(3-1)\hat j+(5-2)\hat k=1\hat i+2\hat j+3\hat k$

Bisector of two vectors is given by

$\overrightarrow{c}=||b||\overrightarrow{a}+||a||\overrightarrow{b}$

$||AB||=\sqrt{3^2+2^2+1^2}=\sqrt{9+4+1}=\sqrt{15}$

$||AC||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{15}$

Bisector of

$\overrightarrow{AB}$ and

$\overrightarrow{AC}=||AC||\overrightarrow{AB}+||AB||\overrightarrow{AC}$

$\Rightarrow\sqrt{15}(3\hat i+2\hat j -1\hat k)+\sqrt{15}(1\hat i+2\hat j+3\hat k)$

$\Rightarrow \sqrt{15}(4\hat i+4\hat j +2\hat k)$

Simplifying the vector

$\dfrac{\sqrt{15}(4\hat i+4\hat j +2\hat k)}{\sqrt{15}\times 2}$

$\Rightarrow 2\hat i+2\hat j +1\hat k$

A unit vector along the direction

$\hat i + \hat j + \hat k$ has a magnitude:

0%
$\sqrt 3$
0%
$\sqrt 2$
0%
$1$
0%
$0$

Explanation

$A\quad unit\quad vector\quad along\quad any\quad direction\quad always\quad has\quad magnitude=1$

$C)Ans.$

$\overline {OA}=i+j+k, \overline {AB}=3i-2j+k,\overline {BC}=i+2j-2k$ and

$\overline {CD}=2i+j+3k$ then find the vector

$\overline{OD}$ .

0%
$7i-2j-6k$
0%
$7i+2j+3k$
0%
$7i+2j+5k$
0%
None of these

Explanation

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

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

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

$\Rightarrow \overrightarrow {OB} = 3i - 2\hat j + \hat k + \left( {\hat i + \hat j + \hat k} \right)$

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

Now,

$\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB}$

$\Rightarrow \overrightarrow {OC} = \overrightarrow {BC} + \overrightarrow {OB}$

$\Rightarrow \overrightarrow {OC} = \left( {i + 2j - 2k} \right) + \left( {4\hat i - \hat j + 2\hat k} \right)$

$\Rightarrow \overrightarrow {OC} = 5\hat i + \hat j$

similarly,

$\overrightarrow {CD} = \overrightarrow {OD} - \overrightarrow {OC}$

$\Rightarrow \overrightarrow {OD} = \overrightarrow {CD} + \overrightarrow {OC}$

$\Rightarrow \overrightarrow {OD} = \left( {2\hat i + \hat j + 3\hat k} \right) + \left( {5\hat i + \hat j} \right)$

$\Rightarrow \overrightarrow {OD} = 7\hat i + 2\hat j + 3\hat k$

Let

$\vec{a}$ ,

$\vec{b}$ ,

$\vec{c}$ and

$\vec{d}$ are four distinct vectors satisfying the conditions

$\vec{a} \times \vec{b}$ =

$\vec{c} \times \vec{d}$ &

$\vec{a} \times \vec{c}$ =

$\vec{b} \times \vec{d}$ then

$\vec{a}.\vec{b} + \vec{c}.\vec{d} \neq \vec{a}.\vec{c} + \vec{b}.\vec{d}$ .

0%
True
0%
False

Four forces act on a point object. The object will be in equilibrium, if:

0%
all of them are in the same plane
0%
they are opposite to each other in pairs
0%
the sum of x, y and z - components of forces zero separately
0%
they form a closed figure of 4 sides when added as Polygon law

The value of

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

0%
$0$
0%
$-1$
0%
$2\ \bar {a}\times (\bar {b}+\bar {c})$
0%
$[\bar {a}\bar {b}\bar {c}]$

Explanation

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

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

$= \bar {a} \times \bar{b} + \bar{a} \times \bar{c} + \bar{b} \times \bar{c} - \bar{a} \times \bar{b} - \bar{a} \times \bar{c} - \bar{b} \times \bar{c}$

$= 0$ (Option A)

Formula:

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

Which of the following is the unit vector perpendicular to $\vec{A}$ and $\vec{B}$ ?

0%
${{\vec{A}\times \vec{B}} \over {AB\sin \theta }}$
0%
$\left|{{\vec{A}\times \vec{B}} \over {AB\sin \theta }}\right|$
0%
${{\vec{A}\times \vec{B}} \over {AB\cos \theta }}$
0%
$\left|{{\vec{A}\times \vec{B}} \over {AB\cos \theta }}\right|$

The vector that must be added to the vector

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

$3\hat{i}+6\hat{j}-7\hat{k}$ so that the resultant vector is a unit vector along the y-axis is:

0%
$4\hat{i}+2\hat{j}+5\hat{k}$
0%
$-4\hat{i}-2\hat{j}+5\hat{k}$
0%
$3\hat{i}+4\hat{j}+5\hat{k}$
0%
Null vector

Explanation

$\textbf{Step 1: Equation formation as per given data}$

Let vector

$\vec{C}$ is added to vector

$\vec{A}$ and

$\vec{B}$ to give a resultant unit vector along

$y$ -axis.

$\therefore \vec{A} + \vec{B} + \vec{C} = \hat{j}$

$\textbf{Step 2: Solving equation}$

$(\hat{i} - 3\hat{j} + 2 \hat{k}) + (3 \hat{i} + 6 \hat{j} - 7 \hat{k}) + \vec{C} = \hat{j}$

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

Hence the vector is

$\vec{C} = -4 \hat{i} - 2\hat{j} + 5 \hat{k}$ .

Option

$B$ is correct.

The vector equation of the plane containing the line

$\vec {r}=(-2\hat {i}-3\hat {j}+4\hat {k})+\lambda(3\hat {i}-2\hat {j}-\hat {k})$ and the point

$\hat {i}+2\hat {j}+3\hat {k}$ is:

0%
$\vec {r}\cdot(\hat {i}+3\hat {k})=10$
0%
$\vec {r}\cdot(\hat {i}-3\hat {k})=10$
0%
$\vec {r}\cdot(3\hat {i}+\hat {k})=10$
0%
$none\ of\ these$

Let

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

$\vec{c}$ be three non-zero vectors such that no two of these are collinear. If the vector

$\vec{a}+2\vec{b}$ is collinear with

$\vec{c}$ and

$\vec{b}+3\vec{c}$ is collinear with

$\vec{a}(\lambda$ being some non-zero scalar), then

$\vec{a}+2\vec{ b}+6\vec{c}$ equals

0%
$\lambda\vec{a}$
0%
$\lambda\vec{b}$
0%
$\lambda\vec{c}$
0%
$\vec{0}$

Explanation

$\vec { a } +2\vec { b } =\lambda\vec { c } \quad -(i)$

$\vec { b } +3\vec { c } =\mu\vec { a } \quad -(ii)$

$Substituting\quad \vec { a } \quad from\quad (i)\quad in\quad (ii)$

$\implies\quad \vec { b } +3\vec { c } =\mu(\lambda\vec { c } -2\vec { b } )$

$\implies\quad (1+2\mu)\vec { b } =(\lambda\mu-3)\vec { c }$

$\implies\quad But\quad since\quad \vec { b } \quad and\quad \vec { c } \quad are\quad not\quad collinear.$

$\implies\quad Either\quad 1+2\mu=0,\quad \lambda\mu-3=0$

$\implies\quad \mu=\cfrac { -1 }{ 2 }$

$and\quad \cfrac { -\lambda}{ 2 } =3$

$\implies\quad \lambda=-6$

$Substituting\quad in\quad (i)$

$\implies\quad \vec { a } +2\vec { b } +6\vec { c } =\vec { 0 }$

$\therefore D)Answer.$

Component of

$\vec{a}=\hat{i}-\hat{j}-\hat{k}$ perpendicular to the vector

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

0%
$\dfrac{1}{3}(\hat{i}+2\hat{j}+2\hat{k})$
0%
$\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$
0%
$\dfrac{1}{3}(\hat{i}+4\hat{j}+2\hat{k})$
0%
$\dfrac{1}{3}(\hat{i}+2\hat{j}+\hat{k})$

Explanation

$\bar{a}=\hat{i}-\hat{j}-\hat{k}$

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

$a_{\bot}=a-\dfrac{a.b}{b.b}(\bar{b})$

$a_{\bot}=(\hat{i}-\hat{j}-\hat{k})-\dfrac{(\hat{i}-\hat{j}-\hat{k}).(2\hat{i}+\hat{j}-\hat{k})}{(\sqrt{(2)^{2}+(1)^{2}+(1)^{2}})^{2}} (2\hat{i}+\hat{j}-\hat{k})$

$a_{\bot}=(\hat{i}-\hat{j}-\hat{k})-\dfrac{(2-1+1)}{6}(2\hat{i}+\hat{j}-\hat{k})$

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

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

$a_{\bot}=\dfrac{\hat{i}-4\hat{j}-2\hat{k}}{3}=\dfrac{1}{3}(\hat{i}-4\hat{j}-2\hat{k})$

$\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar vectors such that

$\vec{d}\cdot \vec{a}=\vec{d}\cdot \vec{b}=\vec{d}\cdot \vec{c}=0$ , then

$\vec{d}$ is :-

0%
a scalar
0%
a null vector
0%
has non-zero magnitude
0%
an imaginary number

Explanation

$\because \overline { a } ,\overline { b }\ and\quad \overline { c }$ are not coplanar vectors.

Hence,

$\overline { d }$ can't be perpendicular to

$\overline { a } ,\overline { b }\ and\quad \overline { c }$ at the same time.

$\therefore$ The only way for

$\overline {d} .\overline{a} = 0,\ \overline { d } .\overline { b } =0$ and

$\overline { d } .\overline { c } =0$ to be true is if

$\overline { d }$ is a null vector.

$\therefore B)Ans.$

$\overline a = \overline i - \overline k ,\overline b = x\overline i + \overline j + (1 - x)\overline k$ and

$\overline c = y\overline i + \lambda \overline j + (1 + x - y)\overline k$ then

$\left[ {\overline a \overline b \overline c } \right]$ depends on:-

0%
neither x nor y
0%
both x and y
0%
only x
0%
only y

Explanation

$\textbf{Step1-Find scalar triple product of vcectors}$

$\overline { a } .[\overline { b } \times \overline { c } ]$

$\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & \lambda & 1+x-y \end{matrix} \right|$

$\implies\quad 1(1+x-y-\lambda +\lambda x)-1(\lambda x-y)$

$\implies\quad 1+x-y-\lambda +\lambda x-\lambda x+y$

$\therefore [\begin{matrix} a & b & c \end{matrix}]$

$\text{depends only on x,not y.}$

$\textbf{Hence , option C is correct}$

$A(1, -1, -3)$ ,

$B(2, 1, -2)$ &

$C(-5, 2, -6)$ are the position vectors of the vertices of a triangle ABC. The length of the bisector of its internal angle at A is?

0%
$\sqrt{10}/4$
0%
$3\sqrt{10}/4$
0%
$\sqrt{10}$
0%
None

Explanation

$AB=\sqrt { { (2-1) }^{ 2 }+{ (1+1) }^{ 2 }+{ (-2+3) }^{ 2 } }$

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

$=\sqrt { 6 }$

$AC=\sqrt { { (-5-1) }^{ 2 }+{ (2+1) }^{ 2 }+{ (-6+3) }^{ 2 } }$

$=\sqrt { 36+9+9 }$

$=\sqrt { 54 }$

$=3\sqrt { 6 }$

$\cfrac { AB }{ AC } =\cfrac { BE }{ EC } (\because \triangle AEB\sim \triangle AEC\quad \therefore \angle BAE=\angle EAC)$

$\implies\quad \cfrac { BE }{ EC } =\cfrac { AB }{ AC } =\cfrac { \sqrt { 6 } }{ 3\sqrt { 6 } } =\cfrac { 1 }{ 3 }$

Using section formula,

$E=(\cfrac { -5+2(3) }{ 1+3 } ,\cfrac { 2+3(1) }{ 1+3 } ,\cfrac { -6+3(-2) }{ 4 } )$

$\implies\quad E=(\cfrac { -5+6 }{ 4 } ,\cfrac { 2+3 }{ 4 } ,\cfrac { -6-6 }{ 4 } )$

$=(\cfrac { 1 }{ 4 } ,\cfrac { 5 }{ 4 } ,{ -3 })$

$\therefore Length\quad AE=\sqrt { { (1-\cfrac { 1 }{ 4 } ) }^{ 2 }+{ (-1-\cfrac { 5 }{ 4 } ) }^{ 2 }+{ (-3+3) }^{ 2 } }$

$=\sqrt { \cfrac { 9 }{ 16 } +\cfrac { 81 }{ 16 } +0 }$

$=\sqrt { \cfrac { 90 }{ 16 } }$

$=\sqrt { \cfrac { 9\times 10 }{ 16 } }$

$=\cfrac { 3\sqrt { 10 } }{ 4 }$

1025403_1059013_ans_53fa8ff0139c453c8ba2f29845e77aab.png

$\hat {i},\hat {j},\hat {k}$ are positive vectors of

$A,B,C$ and

$\vec {AB}=\vec {CX}$ , then positive vector of

$X$ is

0%
$-\hat {i}+\hat {j}+\hat {K}$
0%
$\hat {i}-\hat {j}+\hat {K}$
0%
$\hat {i}+\hat {j}-\hat {K}$
0%
$\hat {i}+\hat {j}+\hat {K}$

The

$P.V.'s$ of the vertices of a

$\triangle ABC$ are

$\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+\bar {j}+\bar {k}, 4\bar {i}+5\bar {j}+\bar {k}$ . The

$P.V.$ of the circumcentre of

$\triangle ABC$ is

0%
$\dfrac{5}{2}\bar {i}+3\bar {j}+\bar {k}$
0%
$5\bar {i}+\dfrac{3}{2}\bar {j}+\bar {k}$
0%
$5\bar {i}+3\bar {j}+\dfrac{1}{2}\bar {k}$
0%
$\bar {i}+\bar {j}+\bar {k}$

Explanation

We have,

$A(1,1,1)$

$B(4,1,1)$

$C(4,5,1)$

$AB=3$

$BC=4$

$CA=5$

Hence,

Triangle is right angle at

$B$ .

$\therefore$ CIrcumcentre is mid point of

$AC = \left( {\frac{5}{2},3,1} \right)$

$= \left( {\frac{5}{2}\overline i ,3\overline j ,\overline k } \right)$

Then,

Option

$A$ is correct answer.

The position vector of point P, is?

0%
$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$
0%
$\dfrac{|\vec{a}||\vec{c}|}{3|\vec{c}+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$
0%
$\dfrac{2|\vec{a}||\vec{c}|}{3|\vec{c}|+2|\vec{a}|}\left\{\dfrac{\vec{a}}{|\bar{a}|}+\dfrac{\vec{c}}{|\vec{c}|}\right\}$
0%
$\dfrac{3|\vec{a}||\vec{c}|}{3|\vec{c}|+2\vec{a}|}\left\{\dfrac{\vec{a}}{|\vec{a}|}-\dfrac{\vec{c}}{|\vec{c}|}\right\}$

Explanation

According to question:

$\begin{array}{l} B=\overrightarrow { a } +\overrightarrow { c } \\ \overrightarrow { E } =\dfrac { { 2\overrightarrow { c } +\overrightarrow { a } +\overrightarrow { c } } }{ 3 } =\dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } \\ then, \\ Position\, vector\, is: \\ \, \, \, \, \, \overrightarrow { OP } =\lambda \, (\overrightarrow { a } +\overrightarrow { c } )-----(i) \\ \, \, \, \, Now, \\ \, \, \, \, \, \, \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \, \, \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } +\overrightarrow { a } } }{ 3 } -\overrightarrow { a } } \right) \\ \, \, \, \overrightarrow { p } =\overrightarrow { a } +\mu \left( { \dfrac { { 3\overrightarrow { c } -2\overrightarrow { a } } }{ 3 } } \right) -----(ii) \\ Now,\, equating\, \, eqn.\, \, (i)\, \, \& \, (ii) \\ \lambda \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) =\overrightarrow { a } +\dfrac { \mu }{ 3 } \left( { 3\overrightarrow { c } -2\overrightarrow { a } } \right) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { { 2\mu } }{ 3 } \, \, \, \, \, \, \, (cofficient\, compare) \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =\dfrac { { 3\mu } }{ 3 } ,\, \, \, \, \, \, \, \, \, \mu =\dfrac { \lambda }{ { \left| { \overline { a } } \right| } } \\ And, \\ \dfrac { \lambda }{ { \left| { \overline { a } } \right| } } =1-\dfrac { 2 }{ 3 } .\, \dfrac { \lambda }{ { \left| { \overline { c } } \right| } } \\ \lambda \left( { \dfrac { 1 }{ { \left| { \overline { a } } \right| } } +\dfrac { 2 }{ 3 } .\, \dfrac { 1 }{ { \left| { \overline { c } } \right| } } } \right) =1 \\ \lambda =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \\ \, positon\, \, vector\, of\overrightarrow { \, p } : \\ \overrightarrow { p } =\dfrac { { 3\left| { \overline { a } } \right| \, \left| { \overline { c } } \right| } }{ { 3\left| { \overline { c } } \right| +2\left| { \overline { a } } \right| } } \, \, \, \left( { \dfrac { { \overrightarrow { a } } }{ { \left| { \overline { a } } \right| } } +\dfrac { { \overrightarrow { c } } }{ { \left| { \overline { c } } \right| } } } \right) \\ so\, ,\, the\, correct\, \, option\, is\, A. \end{array}$

$a^b=b^c=ab$ , then

$b+c$ always equals?

0%
$\dfrac{1}{bc}$
0%
$\dfrac{1}{2}bc$
0%
$1$
0%
$bc$

Explanation

${ a }^{ b }={ b }^{ c }=ab=p(let)$

$\implies\quad { a }^{ b }=p$

taking log

$\implies\quad b\log { a } =\log { p } \quad -(i)$

${ b }^{ c }=p$

Taking log

$\implies\quad c\log { b } =\log { p } \quad -(ii)$

$ab=p$

Taking log

$\log { a } +\log { b } =\log { p }$

Using

$(i)\& (ii),$ we can say:

$\cfrac { \log { p } }{ b } +\cfrac { \log { p } }{ c } =\log { p }$

$\implies\quad \cfrac { 1 }{ b } +\cfrac { 1 }{ c } =1$

$\implies\quad b+c=bc$

$\overline { a } =\cfrac { 1 }{ \sqrt { 10 } } \left( 3\overline { i } +\overline { k } \right) ;\overline { a } =\cfrac { 1 }{ 7 } \left( 2\overline { i } +3\overline { j } -6\overline { k } \right)$ then the value of

$(2\overline { a } -\overline { b } ).[(\overline { a } \times \overline { b } )\times (\overline { a } +2\overline { b } )]$

0%
$5$
0%
$3$
0%
$-5$
0%
$-3$

Let P, Q, R and S be the points on the plane with position vectors

$-2\hat{i}-\hat{j}, 4\hat{i}, 3\hat{i}+3\hat{j}$ and

$-3\hat{i}+2\hat{j}$ respectively. the quadrilateral PQRS must be a.

0%
Parallelogram, which is neither a rhombus nor a rectangle
0%
Square
0%
Rectangle, but not a square
0%
Rhombus, but not a square

Explanation

Consider the problem

As we given

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

On converting above vectors into Cartesian coordinates

Then , consider,

$P\left( { - 2, - 1} \right),Q\left( {4,0} \right),R\left( {3,3} \right),S\left( { - 3,2} \right)$

By distance formula

$=\sqrt{(x_1-x_2)^2+(y_1+y_2)^2}$

$d(PQ)=\sqrt{(4+2)^2+(0+1)^2}=\sqrt{37}$

Similarly,

$d(QR)=\sqrt{10}$

$d(RS)={\sqrt{37}}$

$d(SP)=\sqrt{10}$

Therefore,

$d(PQ)=d(RS)$ and

$d(QR)=d(SP)$

Hence,

$PQRS$ is parallelogram

Now,

Slope

$=\frac{y_2-y_1}{x_2-x_1}$

Therefore, let slope is

$S$

then ,

$S(PQ)=\frac{0-(-1)}{4-(-2)}=\frac{0+1}{4+2}=\frac{1}{6}$

Similarly,

$S(QR)=\frac{3}{-1}=-3$

$S(RS)=\frac{-1}{-6}=\frac{1}{6}$

$S(PS)=-3$

$S(PQ)=S(RS)$ this implies

$PQ \parallel RS$

And

$S(QR)=S(SP)$ this implies

$QR \parallel SP$

$S(PQ) \ne S(QR)$ i.e. they aren't parallel

$S(PQ) \times S(QR) \ne -1$ i.e. they aren't perpendicular.

Therefore

$PQRS$ is a parallelogram but it is nor a rhombus and not rectangle.

Hence option

$A$ is the correct answer.

$\vec a,\vec b,\vec c$ non-zero vectors such that

$\vec a$ is perpendicular to

$\vec b$ and

$\vec c$ and non-zero vector coplanar with

$\vec a + \vec b$ and

$2\vec b - \vec c$ and

$\vec d.\vec a = 1$ , then the minimum value of

$\left| {\vec d} \right|$

0%
$\frac{2}{{\sqrt {13} }}$
0%
$\frac{1}{{\sqrt {13} }}$
0%
$\frac{3}{{\sqrt {13} }}$
0%
$\frac{4}{{\sqrt {13} }}$

$a,b,c \in N$ , the number of points having position vectors

$a\hat i + b\hat j + c\hat k$ such that

$6 \le a + b + c \le 10$ is

0%
110
0%
116
0%
120
0%
127

Explanation

Given:

$6\le\,a+b+c\le \,10$

$\because\,a,b,c,\in N$

$\therefore\,a+b+c=6\,or\,7\,or\,8\,or\,9\,or\,10$

and

$a,b,c\ge 1$

$\therefore\,x+y+x=3\,or\,4\,or\,5\,or\,6\,or\,7$ where

$x,y,z\ge 1$

Let

$x=a-1,\,y=b-1,\,$ and

$z=c-1$

$\therefore\,$ Required number of points

$=^{3-1+3}C_{3}+^{3-1+4}C_{4}+^{3-1+5}C_{5}+^{3-1+6}C_{6}+^{3-1+7}C_{7}$

$=^{5}C_{3}+^{6}C_{4}+^{7}C_{5}+^{8}C_{6}+^{9}C_{7}$

$=\dfrac{5!}{3!2!}+\dfrac{6!}{2!4!}+\dfrac{7!}{5!2!}+\dfrac{8!}{6!2!}+\dfrac{9!}{7!2!}$

$=\dfrac{20}{2}+\dfrac{30}{2}+\dfrac{42}{2}+\dfrac{56}{2}+\dfrac{72}{2}$

$=10+15+21+28+36=110$

$\bar a + \bar b$ is perpendicular to

$\bar b$ and

$\bar a + 2\bar b$ is perpendicular to

$\bar a$ then.

0%
$\left| {\bar a} \right| = \left| {\bar b} \right|$
0%
$\left| {\bar a} \right| = \sqrt 2 \left| {\bar b} \right|$
0%
$\left| {\bar b} \right| = \sqrt 2 \left| {\bar a} \right|$
0%
$\left| {\bar a} \right| = \left| {\bar b} \right|\sqrt 3$

Explanation

Since

$\vec a+\vec b$ is

$\bot$ to

$\vec b$

So,

$\vec b\cdot (\vec a+\vec b)=0$

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

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

$\vec a.\vec b=-|\vec b|^2$ ----

$(1)$

Also

$\vec a+2\vec b$ is perpendicular to

$\vec a$

So,

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

$\vec a.\vec a+2\vec a.\vec b=0$

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

$|\vec a|^2+2 \times -|\vec b|^2=0$ from

$(1)$

$|\vec a|=\sqrt 2|\vec b|$

The length of the projection of the line segment joining points

$(5, -1, 4)$ and

$(4, -1, 8)$ on the plane

$x+y+z=7$ .

0%
$\dfrac{2}{\sqrt{3}}$
0%
$\dfrac{2}{3}$
0%
$\sqrt{5}$
0%
$\sqrt{\dfrac{2}{3}}$

Explanation

Let

$A\left(5,-1,4\right)$ and

$B\left(4,-1,3\right)$ be a line

then

$\vec{AB}=\left(4-5\right)\hat{i}+\left(-1+1\right)\hat{j}+\left(3-4\right)\hat{k}$

$=-\hat{i}-\hat{j}$

$\left|AB\right|=\sqrt{{1}^{2}+{1}^{2}}=\sqrt{2}$ units

Given:Equation of the plane is $$x+y+z=7$

Normal vector

$\vec{n}=\hat{i}+\hat{j}+\hat{k}$

Equation of plane is

$\vec{r}.\hat{n}=d$

$\left(x\hat{i}+y\hat{j}+z\hat{k}\right).\left(\hat{i}+\hat{j}+\hat{k}\right)=7$

$\left|n\right|=\sqrt{1+1+1}=\sqrt{3}$

Length of Projection

$AB$ along the plane is

$P=\left|AB\right|\cos{\left(90-\theta\right)}=\left|AB\right|\sin{\theta}$

Here

$\cos{\theta}=\dfrac{\vec{AB}.\vec{n}}{\left|AB\right|.\left|\vec{n}\right|}$

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

$=\dfrac{-1-1+0}{\sqrt{6}}=-\dfrac{2}{\sqrt{6}}\times\dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{2\sqrt{3}}=\dfrac{\sqrt{2}}{\sqrt{3}}$

$\Rightarrow\,\sin{\theta}=\sqrt{1-{\cos}^{2}{\theta}}=\sqrt{1-{\left(\dfrac{\sqrt{2}}{\sqrt{3}}\right)}^{2}}=\sqrt{1-\dfrac{2}{3}}=\dfrac{1}{\sqrt{3}}$

Now,

$P=\left|AB\right|\sin{\theta}=\sqrt{2}\times\dfrac{1}{\sqrt{3}}=\sqrt{\dfrac{2}{3}}$

$\therefore\,$ Length of projection of line segment on the plane is

$\sqrt{\dfrac{2}{3}}$

1491504_1124573_ans_48d3fe915cbe40b09b5e3dbf59383a0e.png

A unit vector

${\vec a}$ in the plane of

$\vec = 2\hat i + \hat j$ and

$\vec c = \hat i - \hat j + \hat k$ is such that angle between

${\vec a}$ and

${\vec b}$ is the same angle between

${\vec a}$ and

${\vec d}$ where

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

0%
$\frac{{\hat i + \hat j + \hat k}}{{\sqrt 3 }}$
0%
$\frac{{\hat i - \hat j + \hat k}}{{\sqrt 3 }}$
0%
$\frac{{2\hat i + \hat j}}{{\sqrt 5 }}$
0%
$\frac{{2\hat i - \hat j}}{{\sqrt 5 }}$

Explanation

Solution -

Let

$\vec{a}=\dfrac{1}{\sqrt{x^{2}+y^{2}+z^{2}}}(x\hat{i}+y\hat{j}+z\hat{k})$

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

$\vec{a}\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 0 \\ 1 & -1 & 1 \end{vmatrix}=\vec{a}.(\hat{i}-2\hat{j}-3\hat{k})$

$x-2y-3z=0$ . ____ (1)

Also

$\dfrac{\vec{a}.\vec{b}}{\left | \vec{a} \right |\left | \vec{b} \right |}=\dfrac{\vec{a}.\vec{d}}{\left | \vec{a} \right ||\vec{d}|}$

$\dfrac{2x+y}{\sqrt{5}}=\dfrac{y+2z}{\sqrt{5}}$

$x=z$ ____ (2)

$y=-x=-z$ (from (1))

From options

$\vec{a}=\dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{1^{2}+1^{2}+1^{2}}}=\dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$

B is correct.

1130805_1101680_ans_663224c40f76407695a188f98f9ed6fe.jpg

If the unit vectors

$\vec{e}_1 \, and \, \vec{e}_2$ are inclined at an angle

$2 \theta \, and \, |\vec{e}_1 - \vec{e}_2| < 1$ , then for

$\theta \in [0, \pi] , \theta$ may lie in the interval

0%
$\left[0, \dfrac{\pi}{6} \right]$
0%
$\left[ \dfrac{\pi}{6} , \dfrac{\pi}{2} \right]$
0%
$\left[ \dfrac{5 \pi}{6} , \pi \right]$
0%
$\left[\dfrac{\pi}{2}, \dfrac{5 \pi}{6} \right]$

Explanation

$|\hat a-\hat b|<1$

$|\hat a|^2+|\hat b|^2-2|\hat a||\hat b|\cos 2\theta <1$

$2(1-\cos 2\theta)<1$

$2\times 2\sin^2\theta <1$

$\sin^2\theta <\dfrac 14$

$\dfrac{-1}{2}< \sin \theta <\dfrac 12$

$\therefore \theta \in (0, \dfrac \pi6 )\cup (\dfrac{5\pi}{6}, \pi)$

1457530_1103391_ans_ceddf8d2599941d58370a273f2026849.png

Let

$\vec{a},\vec{b},\vec{c}$ be three non-zero vectors such that

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

$\lambda \vec{b}\times \vec{a}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=0$ , then

$\lambda$ is

0%
$1$
0%
$2$
0%
$-1$
0%
$-2$

$[\vec{a}\vec{b}\vec{c}]=2$ , then find the value of

$[(\vec{a}+2\vec{b}-\vec{c}(\vec{a}-\vec{b})(\vec{a}-\vec{b}-\vec{c})]$ .

0%
$6$
0%
$8$
0%
$12$
0%
$11$

Explanation

$[(\vec{a}+2\vec{b}-\vec{c})(\vec{a}-\vec{b})(\vec{a}-\vec{b}-\vec{c})]$

$=\begin{vmatrix} 1 & 2 & -1\\ 1 & -1 & 0\\ 1 & -1 & -1\end{vmatrix}[\vec{a}\vec{b}\vec{c}]$

$=3[\vec{a}\vec{b}\vec{c}]=6$

A non-zero vectors

$\overrightarrow{a}$ is such that its projections along the vectors

$\dfrac{\hat{i}+\hat{j}}{\sqrt{2}}$ and

$\dfrac{-\hat{i}+\hat{j}}{\sqrt{2}}$ and

$\hat{k}$ are equal then unit vector along

$\overrightarrow{a}$ is

0%
$\dfrac{\sqrt{2}\hat{j}-\hat{k}}{\sqrt{3}}$
0%
$\dfrac{\hat{j}-\sqrt{2}\hat{k}}{\sqrt{3}}$
0%
$\dfrac{\sqrt{2}}{\sqrt{3}}\hat{j}+\dfrac{\hat{k}}{\sqrt{3}}$
0%
$\dfrac{\hat{j}-\hat{k}}{\sqrt{2}}$

$a$ and

$b$ are unit vectors along

$OA, OB$ and

$OC$ bisects the angle

$AOB$ . The unit vector along

$OC$ is

0%
$\dfrac{a+b}{2}$
0%
$\dfrac{b-b}{2}$
0%
$\dfrac{a+b}{|a+b|}$
0%
$\dfrac{a-b}{|a-b|}$

Explanation

$\overline{OC} = k\left(\dfrac{\overline{OA}+\overline{OB}}{2}\right)$

$\overline{OC}$ unit vector

$=\dfrac{\dfrac{k}{2} (\overline{OA} + \overline{OB})}{\dfrac{k}{2} |\overline{OA} + \overline{OB}|}$

$=\dfrac{\overline{a} + \overline{b}}{|\overline{a }+\overline{b}| }$
1049319_1129083_ans_aafbae1cb2b541a7a4fa880ecf84b5dc.png

$\overrightarrow {a}$ ,

$\overrightarrow {b}$ and

$\overrightarrow {c}$ be three non-zero vectors, non-coplanar and if

$\overrightarrow {d}$ is such that

$\bar { a } =\dfrac { 1 }{ y } \left( \overrightarrow { b } +\overrightarrow { c } +\overrightarrow { d} \right)$ and where

$x$ and

$y$ are non-zero real numbers, then

$\dfrac { 1 }{ xy } \left( \vec { a } +\vec { b } +\vec { c } +\vec { d } \right) =$

0%
$-\vec {a}$
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$\vec {0}$
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$-2\vec {a}$
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$-3\vec {c}$

If the vector

$6\hat { i } -3\hat { j } -6\hat { k }$ is decomposed into vectors parallel and perpendicular to the vector

$\hat { i } +\hat { j } +\hat { k }$ then the vectors are :

0%
$-\left( \hat { i } +\hat { j } +\hat { k } \right) +\hat { 7i } -\hat { 2j } -\hat { 5k }$
0%
$-2\left( \hat { i } +\hat { j } +\hat { k } \right) +\hat { 8i } -\hat { j } -\hat { 4k }$
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$\left( \hat { i } +\hat { j } +\hat { k } \right)+ \hat { 4i } -\hat { 5j } -\hat { 8k }$
0%
$none$

Explanation

Given vector :

$6\widehat i - 3\widehat j - 6\widehat k$

Let

$6\widehat i - 3\widehat j - 6\widehat k$ =

$c+d$ , where c is parallel to $$\widehat i + \widehat j + \widehat k$$ and d is perpendicular to $$\widehat i + \widehat j + \widehat k$$

hence $$c = m\left( {\widehat i + \widehat j + \widehat k} \right)$$ where m is a scalar

$d = {a_1}\widehat i + {a_2}\widehat j + {a_3}\widehat k$ where

$\begin{array} { *{ 20 }{ l } }{ d\cdot \left( { \hat { i } +\hat { j } +\hat { k } } \right) =0 } \\so\, { { a_{ 1 } }+{ a_{ 2 } }+{ a_{ 3 } }=0 } \\Now\,\, { 6\hat { i } -3\hat { j } -6\hat { k } =m\left( { \hat { i } +\hat { j } +\hat { k } } \right) +\left( { { a_{ 1 } }\hat { i } +{ a_{ 2 } }\hat { j } +{ a_{ 3 } }\hat { k } } \right) } \\so\, equation \,coefficients,\,\, { 6=m+{ a_{ 1 } } } \\ { 3=m+{ a_{ 2 } } } \\ { 6=m+{ a_{ 3 } } } \end{array}$

But

${{a_1} + {a_2} + {a_3} = 0}$ so

$m =1$

then

$\begin{array}{l} { a_{ 1 } }=7;{ a_{ 2 } }=-2;{ a_{ 3 } }=-5 \\ 6\widehat { i } -3\widehat { j } -6\widehat { k } =c+d=m\left( { \widehat { i } +\widehat { j } +\widehat { k } } \right) +\left( { { a_{ 1 } }\widehat { i } +{ a_{ 2 } }\widehat { j } +{ a_{ 3 } }\widehat { k } } \right) \\ =-1\left( { \widehat { i } +\widehat { j } +\widehat { k } } \right) +\left( { 7\widehat { i } -2\widehat { j } -5\widehat { k } } \right) \end{array}$

The value of

$\left(a.i\right)i+\left(a.j\right)j+\left(a.k\right)k$ in terms of vector

$a$

0%
$\overrightarrow{a}$
0%
$\overrightarrow{a}\hat{i}$
0%
$\overrightarrow{a}\hat{j}$
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$\overrightarrow{a}\hat{k}$

Explanation

Let

$\hat{a}={a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}$

$\therefore \overrightarrow{a}.\hat{i}=\left({a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}\right).\hat{i}={a}_{1}$ ,

$\overrightarrow{a}.\hat{j}={a}_{2},\overrightarrow{a}.\hat{k}={a}_{3}$

$\therefore \left(a.i\right)i+\left(a.j\right)j+\left(a.k\right)k$

$={a}_{1}\times\hat{i}+{a}_{2}\times\hat{j}+{a}_{3}\times\hat{k}=\overrightarrow{a}$

If the position vectors of A, B, C, D are

$\vec{a}, \vec{b}. 2\vec{a}+3\vec{b}, \vec{a}-2\vec{b}$ respectively, then

$\vec{AC}, \vec{DB}, \vec{BA}, \vec{DA}$ are?

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$\vec{a}+3\vec{b}, 3\vec{b}-\vec{a}, \vec{a}-\vec{b}, 2\vec{b}$
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$2\vec{b}, \vec{b}-2\vec{a}, 3\vec{b}+\vec{a}, \vec{b}-\vec{a}$
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$\vec{a}-3\vec{b}, 3\vec{b}-\vec{a}, \vec{a}+\vec{b}, 2\vec{b}$
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$-2\vec{b}, \vec{b}-2\vec{a}, 3\vec{b}-\vec{a}, \vec{b}-\vec{a}$

Explanation

Given,

$\vec {OA}=\vec a$

$\vec {OB}=\vec b$

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

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

Then,

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

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

$\vec {BA}=\vec {OA}-\vec {OB}=\vec a-\vec b$

$\vec {DA}=\vec {OA}-\vec {OD}=2\vec b$

Therefore,

$\vec {AC},\vec {DB},\vec {BA}$ and

$\vec {DA}$ are

$\vec a+3\vec b,3\vec b-\vec a,\vec a-\vec b,2\vec b$ respectively.

What vector must be added to the two vectors

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

$2\hat{i}-\hat{j}-\hat{k}$ , so that the resultant may be a unit vector along x-axis

0%
$2\hat{i}+\hat{j}+\hat{k}$
0%
$-2\hat{i}+\hat{j}-\hat{k}$
0%
$2\hat{i}-\hat{j}+\hat{k}$
0%
$-2\hat{i}-\hat{j}-\hat{k}$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 8 - MCQExams.com

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Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 8 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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