MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 10

Let

$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}, b=\hat{i}+2\hat{j}-2\hat{k}$ . Then a unit vector perpendicular to both

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

$\vec{a}+\vec{b}$ is :

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

Explanation

We have,

$\vec { a } =3\hat { i } +2\hat { j } +2\hat { k } ,\, \, \vec { b } =\hat { i } +2\hat { j } -2\hat { k } \\ \vec { a } +\vec { b } =4\hat { i } +4\hat { j } \, \, \, \, \, \, \hat { a } -\hat { b } =2\hat { i } +4\hat { k } \\ \vec { d } =\left( { \vec { a } -\vec { b } } \right) \times \left( { \vec { a } \times \vec { b } } \right)$

Therefore,

$=\left| { \begin{array} { *{ 20 }{ c } }i & { \hat { j } } & { \hat { k } } \\ 2 & 0 & 4 \\ 4 & 4 & 0 \end{array} } \right| =-16\hat { i } +16\hat { j } +8\hat { k }$

Unit vector perpendicular to

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

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

$=\dfrac { { \left( { 16\hat { i } +16\hat { j } +8\hat { k } } \right) } }{ { \sqrt { 576 } } } =\dfrac { { -8\left( { -2\hat { i } +2\hat { j } +\hat { k } } \right) } }{ { 24 } } \\ =\dfrac { { -1 } }{ 3 } \left( { -2\hat { i } +2\hat { j } +\hat { k } } \right)$

Hence, this is the answer.

$\vec a,\vec b, \vec c$ are three coplanar vectors, then

$v\left[ 2\vec { a } +3\vec { b } ,2\vec { b- } 5\vec { c } ,2\vec { c } +3\vec { a } \right]$ is

0%
$0$
0%
$1$
0%
$-\sqrt 3$
0%
$\sqrt 3$

$\vec { x }$ and

$\vec { y }$ be unit vectors and

$\left| \vec { z } \right| =\frac { 2 }{ \sqrt { 7 } }$ such that

$\vec { z } +\vec { z } \times \vec { x } =\vec { y } ,$ then the angle

$\theta$ between

$\vec { x }$ and

$\vec { z }$ can be

0%
${ 30 }^{ \circ }$
0%
${ 60 }^{ \circ }$
0%
${ 90 }^{ \circ }$
0%
None of these

Explanation

We have,

$\vec { z } +\vec { z } \times \vec { x } =\vec { y } \\ \left| { \vec { z } +\vec { z } \times \vec { x } } \right| =\left| { \vec { y } } \right| =1 \\ { \left| { \vec { z } +\vec { z } \times \vec { x } } \right| ^{ 2 } }=1 \\ \left( { \vec { z } +\vec { z } \times \vec { x } } \right) .\left( { \vec { z } +\vec { z } \times \vec { x } } \right) =1 \\ \Rightarrow { \left| { \vec { z } } \right| ^{ 2 } }+\left| { \vec { z } \times \vec { x } } \right| =1$

$\\ \dfrac { 4 }{ 7 } +\dfrac { 4 }{ 7 } { \sin ^{ 2 } }\theta =1 \\ { \sin ^{ 2 } }\theta =\dfrac { 3 }{ 4 } \\ \sin \theta =\dfrac { { \sqrt { 3 } } }{ 2 } \\ \therefore \theta ={ 60^{ o } }$

$\\ Hence,\, the\, option\, B\, is\, the\, correct\, answer.$

The position vector of point C with respect to B is

$\bar { i } +\bar { j }$ . and that of B with respect to A is

$\bar { i } +\bar { j }$ .The position vector of C with respect to A is ____________.

0%
$2\bar { i }$
0%
$-2\bar { i }$
0%
$2\bar { j }$
0%
$-2\bar { j }$

If the vectors

$3 \overline { p } + \overline { q } : 5 \overline { p } - 3 \overline { q }$ and

$2 \overline { p } + \overline { q } ; 4 \overline { p } - 2 \overline { q }$ are pairs of mutually perpendicular vectors then

$\sin ( \overline { p } \overline { q } )$ is:

0%
$\frac { \sqrt { 55 } } { 4 }$
0%
$\frac { \sqrt { 55 } } { 8 }$
0%
$\frac { 3 } { 16 }$
0%
$\frac {{ \sqrt 247 }} { 16 }$

Let

$\hat {a}$ and

$\hat {b}$ two unit vector such that

${ \left( \hat { a } .\hat { b } \right) }^{ 2 }-\left| \hat { a } \times \hat { b } \right|$ is maximum then

$\left| \hat { a } .\hat { b } \right|$ is equal to

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

Explanation

We have,

$\begin{array}{l} { \left( { \vec { a } .\vec { b } } \right) ^{ 2 } }-\left| { \vec { a } \times \vec { b } } \right| \\ ={ \cos ^{ 2 } }\theta -\sin \theta \\ =1-{ \sin ^{ 2 } }\theta -\sin \theta \\ =1-\left[ { { { \sin }^{ 2 } }\theta +\sin \theta +\dfrac { 1 }{ 4 } -\dfrac { 1 }{ 4 } } \right] \\ =\dfrac { 5 }{ 4 } -{ \left( { \sin \theta +\dfrac { 1 }{ 2 } } \right) ^{ 2 } } \\ Max\, value\, =\dfrac { 5 }{ 4 } -\dfrac { 1 }{ 4 } \, \, at\, \, \sin \theta =0 \\ =1 \\ \left| { \vec { a } .\vec { b } } \right| =\left| { \cos \theta } \right| =1 \\ Hence,\, the\, option\, A\, is\, correct\, answer. \end{array}$

The cartesian equation of the plane perpendicular to vector

$3\bar {i}-2\bar {j}-2\bar {k}$ and passing through the point

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

0%
$3x+2y+2z=2$
0%
$3x-2y+2z=2$
0%
$3x+2y-2z=2$
0%
$3x-2y-2z=2$

Explanation

$\begin{array}{l} \left( { \overrightarrow { r } -\overrightarrow { a } } \right) .\overrightarrow { N } =0 \\ \Rightarrow \overrightarrow { a } =2\hat { i } +3\hat { j } -\hat { k } \\ \Rightarrow \overrightarrow { N } =3\hat { i } -2\hat { j } -2\hat { k } \\ \Rightarrow \left( { \overrightarrow { r } -2\hat { i } -3\hat { j } +\hat { k } } \right) .\left( { 3\hat { i } -2\hat { j } -2\hat { k } } \right) =0 \\ \Rightarrow \overrightarrow { r } =x\hat { i } +y\hat { j } +\hat { k } \\ \Rightarrow \left\{ { \left( { x-2 } \right) i+\left( { y-3 } \right) \hat { j } +\left( { z+1 } \right) \hat { k } } \right\} .\left( { 3i-2\hat { j } -2\hat { k } } \right) =0 \\ \Rightarrow 3\left( { x-2 } \right) -2\left( { y-3 } \right) -2\left( { z+1 } \right) =0 \\ \Rightarrow 3x-6-2y+6-2z-2=0 \\ 3x-2y-2z=2\, \, \, \, Ans. \end{array}$

The position vectors of two vertices and the centroid of a triangle are

$\overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } ,\overset { \rightarrow }{ 2i } -\overset { \rightarrow }{ j } +\overset { \rightarrow }{ k }$ and

$\overset { \rightarrow }{ k }$ respectively. The position vector of the third vertex of the triangle is :

0%
$\overset { \rightarrow }{ -3i } +\overset { \rightarrow }{ 2k }$
0%
$\overset { \rightarrow }{ 3i } -\overset { \rightarrow }{ 2k }$
0%
$\overset { \rightarrow }{ i } +\frac { 2 }{ 3 } \overset { \rightarrow }{ k }$
0%
none of these

Explanation

Let the third vertex be

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

and

$A=(1,1,0), B=(2,-1,1), G=(0,0,1)$

Now,

$(0,0,1)=(\dfrac{1+2+x}{3}, \dfrac{1-1+y}{3}, \dfrac{0+1+z}{3})$

Equating this, we get,

$x=-3,y=0,z=2$

Hence,

$C=-3\widehat{i}+2\widehat{k}$

Unit vector perpendicular to the plane of the triangle

$ABC$ with position vectors of the vertices

$A , B , C ,$ is

$($ where

$\Delta$ is the area of the triangle

$A B C$ ) .

0%
$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { \Delta }$
0%
$\dfrac { ( \vec { a } \times \vec { b } + \overline { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 2 \Delta }$
0%
$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 3\Delta }$
0%
$\dfrac { ( \vec { a } \times \vec { b } + \vec { b } \times \vec { c } + \vec { c } \times \vec { a } ) } { 4\Delta }$

Explanation

$\begin{array}{l} Vector\, perpendicular\, to\, the\, plane\, of\, \Delta ABC \\ is\, \, \overrightarrow { AB } \times \, \overrightarrow { AC } \\ =\left( { \, \overrightarrow { b } -\, \overrightarrow { a } } \right) \times \left( { \, \overrightarrow { c } -\, \overrightarrow { a } } \right) \\ =\, \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } \\ Area\, of\, triangle\, =\frac { 1 }{ 2 } \left| { \, \overrightarrow { AB } \times \, \overrightarrow { BC } } \right| \\ 2\Delta =\left| { \, \overrightarrow { AB } \times \, \overrightarrow { AC } } \right| \\ Unit\, vector=\frac { { \left( { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right) } }{ { \left| { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right| } } \\ =\frac { { \left( { \overrightarrow { b } \times \, \overrightarrow { c } +\, \overrightarrow { a } \times \, \overrightarrow { b } +\, \overrightarrow { c } \times \, \overrightarrow { a } } \right) } }{ { 2\Delta } } \\ Hence,\, the\, option\, B\, is\, the\, correct\, answer. \end{array}$

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

$\bar { c }$ are unit vector such that

$\bar { a } +\bar { b } -\bar { c } =0$ . then the angle between

$\bar { a }$ and

$\bar { b }$ is :-

0%
$\dfrac { \pi }{ 6 }$
0%
$\dfrac { \pi }{ 3 }$
0%
$\dfrac { \pi }{ 2 }$
0%
$\dfrac { 2\pi }{ 3 }$

Explanation

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

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

$|\bar{a}+\bar{b}| = |\bar{c}| = 1$

$a^{2} +b^{2}+2ab cos\theta =1$

$1+1+2cos\theta = 1$

$2cos\theta = -1$

$cos\theta = -1/2$

$\theta = 2\pi /3$

1240378_1303716_ans_5fbec7a883e246069119c6e7ea5a09b9.png

Let

$a=(1,-2,3)$ and

$b=(2,7,4)$ then

0%
$a.b=0$
0%
$a.b=-9$
0%
$a.b=4$
0%
$a.b=-4$

Explanation

Given,

$a=(1,-2,3)$ and

$b=(2,7,4)$ .

Now,

$a.b$

$=(1,-2,3).(2,7,4)$

$=2-14+12$

$=0$ .

$u,\ v,\ w$ are non-coplanar vector and

$p,\ q$ are real numbers, then the equality

$[3u\ pv\ pw]-[pv\ w\ qw]-[2w\ qv\ qu]=0$ holds for

0%
Exactly two values of $(p,\ q)$
0%
More than but not all values of $(p,\ q)$
0%
All values of $(p,\ q)$
0%
Exactly one values of $(p,\ q)$

Explanation

$\begin{array}{l} Since, \\ \left[ { 3u\, \, pv\, \, pw } \right] -\left[ { pv\, \, wq\, \, qv } \right] =0 \\ \therefore 3{ p^{ 2 } }\left[ { u.\left( { v\times w } \right) } \right] -pq\, \left[ { v.\left( { w\times u } \right) } \right] \\ -2{ q^{ 2 } }\left[ { w.\left( { v\times u } \right) } \right] =0 \\ \Rightarrow \left( { 3{ p^{ 2 } }-pq+2{ q^{ 2 } } } \right) \left[ { u.\left( { v\times w } \right) } \right] =0 \\ But\, \, \left[ { u\, \, \, v\, \, \, w } \right] \ne 0 \\ \Rightarrow 3{ p^{ 2 } }-pq+2{ q^{ 2 } }=0 \\ \therefore p=q=0 \end{array}$

If the vectors

$\bar { AB } =3\hat { i } +4\hat { k }$ and

$\bar { AC } =5\hat { i } -2\hat j+4\hat k$ are the sides of a triangle ABC, then the length of the median through A is:

0%
$\sqrt { 18 }$
0%
$\sqrt { 72 }$
0%
$\sqrt { 33 }$
0%
$\sqrt { 45 }$

Explanation

We have,

$AB=3\overrightarrow{i}+4\overrightarrow{k}$

$AC=5\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}$

So, the coordinate of

$B\,is\,\left( 3,0,4 \right)$

$C\,is\,\left( 5,-2,4 \right)$

The mid point of $BC$ can be easily found by midpoint formula of two points as

$$\begin{align}

$D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( \dfrac{3+5}{2},\dfrac{0-2}{2},\dfrac{4+4}{2} \right)$

$D\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=\left( 5,-2,4 \right)$

So, the length of median through A is

$AD=\sqrt{{{\left( 4-0 \right)}^{2}}+{{\left( -1-0 \right)}^{2}}+{{\left( 4-0 \right)}^{2}}}$

$AD=\sqrt{33}$

Hence, this is the answer

In the vectors

$\bar { AB } =3\hat { i } +4\hat { k }$ and

$\bar { AC } =5\hat { i } -2\hat { j } +4\hat { k }$ are the series of a triangle ABC, then the length of the median through A is

0%
$\sqrt { 18 }$
0%
$\sqrt { 72 }$
0%
$\sqrt { 33 }$
0%
$\sqrt { 45 }$

Explanation

Take

$A$ as the origin

$(0,0)$

So vector

$AB = 3i + 4k$

$AC= 5i -2j+ 4k$ will become position vector

So cordinate of

$B$ is

$(3,0,4)$ and

$C$ is

$(5, -2, 4)$

The midpoint of

$BC$ can be easily found by midpoint formula of two points as

$D(4,-1,4)$

So length of median

$A$ is

$AD = \sqrt{(4-0)^2 + (-1-0)^2 + (4-0)^2}$

$= \sqrt{33}$

Let

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

$\overrightarrow { c }$ be three non-zero vectors such that no two of them are collinear and

$(\overrightarrow { a } \times \overrightarrow { b } )\times \overrightarrow { c } =\frac { 1 }{ 3 } \left| \overrightarrow { b } \right| \left| \overrightarrow { c } \right| \overrightarrow { a }$ . If

$\theta$ is the angle between vectors

$\overrightarrow { b }$ and

$\overrightarrow { c }$ , then a value of

$\sin { \theta }$ is

0%
$\frac 23$
0%
$\frac { -2\sqrt { 3 } }{ 3 }$
0%
$\frac { 2\sqrt { 2 } }{ 3 }$
0%
$\frac { -\sqrt { 2} }{ 3 }$

Explanation

${\textbf{Step -1: Find the value of sin}}\mathbf{\theta} {\textbf{ using the vector formula}}{\textbf{.}}$

$\Rightarrow\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = \dfrac{1}{3}\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|.\left| {\overrightarrow a } \right|$

$\Rightarrow\left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = \dfrac{1}{3}\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|.\left| {\overrightarrow a } \right|$

$\because \overrightarrow a .\overrightarrow c = 0$

$\Rightarrow - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = \dfrac{1}{3}\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|.\left| {\overrightarrow a } \right|$

$\Rightarrow\overrightarrow b .\overrightarrow c = \dfrac{1}{3}\left| {\overrightarrow b } \right|.\left| {\overrightarrow c } \right|$

$\Rightarrow\cos \theta = - \dfrac{1}{3}$

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

$= \sqrt {1 - \dfrac{1}{9}}$

$= \sqrt {\dfrac{8}{9}}$

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

${\textbf{ Hence, the value of sin}} \mathbf{\theta} {\textbf{ is }}\mathbf{\dfrac{{2\sqrt 2 }}{3}.}$

A unit vector

$d$ is equally inclined at an angle

$\alpha$ with the vectors

$a=\cos \theta. i+ \sin \theta. j , b=-\sin \theta.i+\cos =\theta. j$ and

$c=k$ . Then

$\alpha$ is equal to

0%
$\cos^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$
0%
$\cos^{-1} \left(\dfrac{1}{\sqrt{3}}\right)$
0%
$\cos^{-1} \dfrac{1}{3}$
0%
$\dfrac{\pi}{2}$

Explanation

$\begin{array}{l} a=\cos \theta \, \, i+\sin \theta \, \, j \\ b=-\sin \theta \, \, i+\cos \theta \, \, j \\ c=\hat { k } \\ a.b=\left( { \cos \theta \, \, i+\sin \theta \, \, j } \right) \left( { -\sin \theta \, \, i+\cos \theta \, \, j } \right) \\ \, \, \, \, \, \, \, \, \, =-\cos \theta .\sin \theta +\sin \theta .\cos \theta \\ a.b=\left| a \right| \left| b \right| \, \, \cos \alpha \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ \begin{array}{l} 0=1\times 1\, \cos \alpha \\ 0=\cos \alpha \\ \alpha =\frac { \pi }{ 2 } \end{array} \right] \\ 0=\sqrt { \cos^2 \theta +\sin^2 \theta } .\sqrt { \sin^2 \theta +\cos^2 \theta } \, \, \cos \alpha \end{array}$

The foot of the perpendicular drawn from a point with position vector

$\hat { i } +4\hat { k }$ on the line joining the points

$\hat { j } +3\hat { k }$ ,

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

0%
$4\hat { i } +5\hat { j } +5\hat { k }$
0%
$\frac { 1 }{ 3 } (\hat { i }+\hat { j } +\hat { 8k } )$
0%
$4\hat { i } +4\hat { j } -5\hat { k }$
0%
$4\hat { i } -5\hat { j } +5\hat { k }$

Explanation

$\begin{array}{l} \frac { x }{ 2 } =\frac { { y-1 } }{ { -4 } } =\frac { { z-3 } }{ { -2 } } \\ \frac { x }{ 1 } =\frac { { y-1 } }{ { -2 } } =\frac { { z-3 } }{ { -1 } } \\ Direction\, ratio\, of\, \overrightarrow { AB } :\, r-1,\, -2r+1,\, -r+1 \\ \left( { r-1 } \right) -2\left( { -2r+1 } \right) -\left( { -r-1 } \right) =0 \\ r=\frac { 1 }{ 3 } \\ foot\, of\, perpendicular\left( { \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } ,\frac { 8 }{ 3 } } \right) \\ Now, \\ \frac { 1 }{ 3 } \left( { \hat { i } +\hat { j } +8\hat { k } } \right) \\ Hence, \\ option\, \, B\, \, is\, \, correct\, \, answer. \end{array}$

Let

$ABCD$ is a triangular pyramid with base vectors

$\vec {AB}= 2\bar {i}+3\bar {j}-\bar {k}$ and

$\vec {AC}=\bar {i}-2\bar {k}$ , If volume of the triangular pyramid is

$\sqrt{150}$ unit then its height is

0%
$10$
0%
$20$
0%
$18$
0%
$23$

$\left | \vec{c} \right |$ = 60 and

$\vec{c}$ (

$\hat{i}$ + 2

$\hat{j}$ + 5

$\hat{k}$ ) = 0 , then a value of

$\vec{c}$ .(7

$\hat{i}$ + 2

$\hat{i}$ + 3

$\hat{k}$ ) is :

0%
$4\sqrt{2}$
0%
12
0%
24
0%
$12\sqrt{2}$

For any vectors

$\vec{a}$ , the value of

$(\vec{a}\times \hat{i})^2+(\vec{a}\times \hat{j})^2+(a\times \hat{k})^2$ is equal to?

0%
$3\vec{a^2}$
0%
$\vec{a^2}$
0%
$2\vec{a^2}$
0%
None of these

Explanation

Let any vector such that

$\vec{a}=x \hat{\imath}+y\hat j +z \hat k$

Given that

$(a \times \hat{\imath})^{2}+(a \times \hat{\jmath})^{2}+(a \times \hat{k})^{2}$

$\Rightarrow[(x \hat{\imath}+y j+z \hat{k}) \times \hat{\imath}]^{2}+[(x \hat{\imath}+y \hat{j}+z \hat{k}) x \hat{\jmath}]^{2}+[(x \hat{\imath}+y \hat{j}+z \hat{k}) \times \hat{k}]^{2}$

$\Rightarrow[-y \hat{k}+z \hat{\jmath}]^{2}+[x \hat{k}-z\hat{\imath}]^{2}+[-x \hat{\jmath}+y \hat{\imath}]^{2}$

$\text { As we know that }[-y \hat{k}+z \hat{\jmath}]^{2}=\left[\sqrt{(-y)^{2}+z^{2}}\right]^{2}=y^{2}+z^{2}$

Similarly

$\Rightarrow y^{2}+z^{2}+x^{2}+z^{2}+x^{2}+y^{2}$

$\Rightarrow 2\left[x^{2}+y^{2}+z^{2}\right]$

also

$\left(x^{2}+y^{2}+z^{2}\right)=|a|^{2}=a^{2}$

hence,

$2\left[x^{2}+y^{2}+z^{2}\right]=2 a^{2}$

$\therefore$ (c) is the correct option.

The distance of the point

$\text{P}$ with position vector

$3\hat{i}+6 \hat{j}+8\hat{k}$ from

$y$ - axis

0%
$\sqrt{62}$
0%
$10$
0%
$3\sqrt{5}$
0%
$\sqrt{73}$

The adjacent sides of a parallelogram are

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

$\vec{B}=-2 \hat{i}+4 \hat{j}-\hat{k}$ What is the area of the parallelogram?

0%
$4$ units
0%
$7$ units
0%
$\sqrt{5}$ units
0%
$\sqrt{8}$ units

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

$\vec { a } + \vec { b } + \vec { c } = 0 ,$ the value of

$\vec { a } \cdot \vec { b } + \vec { b } \cdot \vec { c } + \vec { c } \cdot \vec { a }$ is

0%
$1$
0%
$3$
0%
$\dfrac{-3}{2}$
0%
None of these

$\left( \bar { a } -\bar { b } \right) =\bar { \left( a \right) } =\bar { \left( b \right) }$ where

$\bar { a }$ and

$\bar { b }$ are non zero vectors then the angle between

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

0%
${ 120 }^{ 0 }$
0%
${ 45 }^{ 0 }$
0%
${ 60 }^{ 0 }$
0%
${ 90 }^{ 0 }$

If the vectors

$2\hat i + 3 \hat j , 5 \hat i + 6 \hat j , 8 \hat i +$

$\lambda$ j have their initial point at

$(1 , 1)$

$then the value of$

$\lambda$ $ so that the vectors terminated on one line is

0%
5
0%
9
0%
4
0%
0

Explanation

Since initial point of

$2i + 3j, 5i + 6j$ and

$8i + \lambda j$ is

$i + j$ ,

their terminal points will be

$3i + 4j, 6i + 7j,$ and

$9i + (\lambda + 1)j$ .

Now given all the vectors terminate on one straight line.

Hence,

$3i + 3j = \lambda_1(3i + (\lambda + 1 - 7)j)\Rightarrow \lambda_1 = 1$ and

$\lambda = 9$

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

$\bar { OQ } =3\hat { i } -4\hat { j } -2\hat { k }$ then the modulus

$\bar { PQ }$ is

0%
$\sqrt { 13 }$
0%
$\sqrt { 51 }$
0%
$\sqrt { 39 }$
0%
$\sqrt { 67 }$

Explanation

$\vec{PQ}=\vec{OQ}-\vec{OP}$

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

$=\hat{i}-7\hat{j}-\hat{k}$

$\left|\vec{PQ}\right|=\sqrt{{1}^{2}+{\left(-7\right)}^{2}+{\left(-1\right)}^{2}}$

$=\sqrt{1+49+1}=\sqrt{51}$

From the figure the correct relation is :

0%
$\bar { A } +\bar { B } +\bar { E } =\bar { 0 }$
0%
$\bar { C } +\bar { D } =\bar { -A }$
0%
$\bar { B } +\bar { E } +\bar { C } =\bar { -D }$
0%
All of these

Line passing through

$(3,4,5)$ and

$(4,5,6)$ has direction ratios

$\ldots$

0%
$1,1,1$
0%
$\sqrt{3}, \sqrt{3}, \sqrt{3}$
0%
$\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$
0%
$7,9,11$

Explanation

Given points

$(3,4,5)$ and

$(4,5,6)$

The drs are given as

$(4-3,5-4,6-5)=(1,1,1)$

let

$\bar { a } ,\bar { b } ,\bar { c }$ are three mutually perpendicular unit vectors and a unit vector

$\bar { r }$ satisfying the equation

$\left( \bar { b } -\bar { c } \right) \times \left( \bar { r } \times \bar { a } \right) +\left( \bar { c } -\bar { a } \right) \times \left( \bar { r } \times \bar { b } \right) +\left( \bar { a } -\bar { b } \right) \times \left( \bar { r } \times \bar { c } \right) =0$ then

$\bar { r }$ is __________________.

0%
$\dfrac { 1 }{ \sqrt { 3 } } \left( \bar { a } +\bar { b } +\bar { c } \right)$
0%
$\dfrac { 1 }{ \sqrt { 14 } } \left( 2\bar { a } +3\bar { b } +\bar { c } \right)$
0%
$-\dfrac { 1 }{ \sqrt { 14 } } \left( 2\bar { a } +3\bar { b } +\bar { c } \right)$
0%
$-\dfrac { 1 }{ \sqrt { 3 } } \left( \bar { a } +\bar { b } +\bar { c } \right)$

$If \bar { A } = 2 \hat {i} + \hat {j} + \hat {k}\ and\ \bar { B } = \hat {i} + \hat {j} + \hat {k}$ two vectors,then the unit vector is

0%
Perpendicular to $\bar { A }\ is\ \dfrac { -\hat { j } +\hat { k } }{ \sqrt { 2 } }$
0%
Parallel to $\bar { A }\ is\ \dfrac { 2 +\hat { j } +\hat { k } }{ \sqrt { 6 } }$
0%
Parallel to $\bar { B }\ is\ \dfrac { -\hat { j } +\hat { k } }{ \sqrt { 2 } }$
0%
Parallel to $\dfrac { \hat { i } +\hat { j } +\hat { k } }{ 3 }$

$\vec { x }$ is a vector in the direction of

$(2,-2,1)$ of magnitude

$6$ and

$\vec { y }$ is a vector in the direction of

$(1,1,-1)$ of magnitude

$\sqrt{3}$ , then

$\left| \vec { x } +2\vec { y } \right| =...$

0%
$40$
0%
$\sqrt { 35 }$
0%
$\sqrt { 17 }$
0%
$2\sqrt { 10 }$

Explanation

they given x direction

we need to find unit vector in that direction and multiply with the magnitude of x

they given y direction

we need to find unit vector in that direction and multiply with the magnitude of y

$\vec { x } =\cfrac { 6\left( 2\hat { i } -2\hat { j } +\hat { k } \right) }{ 3 }$

$,\vec { y } =\cfrac { \sqrt { 3 } \left( \hat { i } +\hat { j } -\hat { k } \right) }{ \sqrt { 3 } }$
so

$\left| \vec { x } +2\vec { y } \right|$

$=\left| 6\hat { i } -2\hat { j } \right| =\sqrt { 40 }$

$=2\sqrt { 10 }$

The position vector of a point P is

$\overrightarrow r=x \overrightarrow i + y \overrightarrow j+x \overrightarrow k$ , Where

$x,y,z,\epsilon N$ and

$\overrightarrow a= \overrightarrow i+ \overrightarrow j+\overrightarrow k$ . If

$\overrightarrow r. \overrightarrow a=10$ , then the number of possible positions of P is ___________.

0%
$30$
0%
$72$
0%
$66$
0%
$36$

Explanation

The number of positive integral solution of the equation

$x_1+x_2+x_3+...x_r=n$ is

$^{n-1}C_{r-1}$

We have

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

Also,

$\vec{r}.\vec{a}=10$

$\Rightarrow\,x+y+z=10$

Thus number of positive integral solution is

$= ^{10-1}C_{3-1}= ^{9}C_{2}=\dfrac{9!}{7!2!}=\dfrac{9\times 8}{2}=36$

Option

$(d)$ is correct.

Let

$\overline { a } =4\hat { i } +3\hat { j } -\hat { k }$ and

$\overline { b } =2\hat { i } -6\hat { j } -3\hat { k } .$ Then a unit vector

$\bot$ to both

$\overline { a }$ and

$\overline { b }$ is.

0%
$\dfrac { 1 }{ 7 } \left( -3\hat { i } -2\hat { j } +3\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( 3\hat { i } +2\hat { j } -6\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right)$
0%
$\dfrac { 1 }{ 7 } \left( -3\hat { i } +2\hat { j } -6\hat { k } \right)$

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

$\vec{b}=2\hat{i}+\hat{j}+2\hat{k}$ . Find the projection vector of

$\vec{b}$ on

$\vec{a}$ .

0%
$\displaystyle\frac{8}{9}\left(\hat{i}+2\hat{j}+2\hat{k}\right)$
0%
$\displaystyle\frac{8}{9}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$
0%
$\displaystyle\frac{9}{8}\left(\hat{i}+2\hat{j}+2\hat{k}\right)$
0%
$\displaystyle\frac{9}{8}\left(2\hat{i}+\hat{j}+2\hat{k}\right)$

Unit vector perpendicular to vector

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

$B=2\hat { { i } } +4\hat { { j } } +6\hat { { k } }$ both is

0%
$\dfrac { 3\hat { { j } } -2\hat { { k } } }{ \sqrt { 13 } }$
0%
$\dfrac { 3\hat { { k } } -2\hat { { j } } }{ \sqrt { 13 } }$
0%
$\dfrac { - \hat { { j } } +2\hat { { k } } }{ \sqrt { 13 } }$
0%
$\dfrac { \hat { { i } } + 3\hat { { j } } -\hat { { k } } }{ \sqrt { 13 } }$

If the position vector

$\vec{a}$ of point

$(12, n)$ is such that

$\left | \vec{a} \right | = 13$ , then find the value (s) of

$n$ .

0%
$\pm 6$
0%
$\pm 4$
0%
$\pm 5$
0%
$\pm 7$

Express

$\vec{AB}$ in terms of unit vectors

$\hat{i}$ and

$\hat{j}$ , when the points are:
A(4,-1), B(1,3)
Find

$\left | \vec{AB} \right |$ in each case.

0%
$\vec{AB} = -3\hat{i}-4\hat{j}, \left | \vec{AB} \right | = 5$
0%
$\vec{AB} = +3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5$
0%
$\vec{AB} = -3\hat{i}+4\hat{j}, \left | \vec{AB} \right | = 5$
0%
none of these

Explanation

$A(4, -1)$

$B(1, 3)$

$\vec{OA}=4\hat{i}-\hat{j}$ ;

$\vec{OB}=\hat{i}+3\hat{j}$

$\vec{AD}=\vec{OB}-\vec{OA}=(\hat{i}+3\hat{j})-(4\hat{i}-\hat{j})=-3\hat{i}+4\hat{j}$

$|\vec{AB}|=\sqrt{(-3)^2+4^2}=5$ .

What is the scalar projection of

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

$\vec{b}=4\hat{i}+4\hat{j}+7\hat{k}$ ?

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

Explanation

Formula,

scalar projection of a on b

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

Given,

$a=i+2j+k,b=4i+4j+7k$

$=\dfrac{(i+2i+k)\cdot (4i+4j+7k)}{|4i-4j+7k|}$

$=\dfrac{4+8+7}{\sqrt{4^2+(4)^2+7^2}}$

$=\dfrac{19}{9}$

$a, b, c$ are vectors such that

$a+b+c = 0$ and

$|a| = 7, |b| = 5, |c| = 3$ , then the angle between

$c$ and

$b$ is

0%
$\dfrac{\pi}{3}$
0%
$\dfrac{\pi}{6}$
0%
$\dfrac{\pi}{4}$
0%
$\pi$

Explanation

We have,

$a+b+c = 0$

$\Rightarrow b+c = -a$

$\Rightarrow |b+c| = |-a|$

$\Rightarrow |b+c| = |a|$

$\Rightarrow |b+c|^2 = |a|^2$

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

$\Rightarrow |b|^2+|c|^2+2|b||c| \cos \theta = |a|^2$

$\Rightarrow (5)^2 + (3)^2 + 2\times 5\times 3 \cos \theta = (7)^2$

$\Rightarrow 25 + 9 + 30 \cos \theta = 49$

$\Rightarrow 30 \cos \theta = 15$

$\Rightarrow \cos \theta = \dfrac{1}{2}$

$\Rightarrow \theta = 60^o$ or

$\dfrac{\pi}{3}$

$\therefore$ Angle betweenn

$b$ and

$c$ is

$\dfrac{\pi}{3}$ .

If a unit vector

$\vec{a}$ makes an angle

$\dfrac{\pi }{3}$ with

$\hat{i},\dfrac{\pi }{4}$ with

$\hat{j}$ and an accute angle

$\theta$ with

$\hat{k},$ then find

$\theta$ and hence, the components of

$\vec{a}$ .

0%
$\dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}-\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k}$
0%
$\dfrac{\pi }{3};\,\vec{a}=\dfrac{-1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k}$
0%
$\dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k}$
0%
$\dfrac{\pi }{3};\,\vec{a}=\dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}-\dfrac{1}{2}\hat{k}$

Explanation

The direction cosines of vector

$\vec{a}$ are

$l = cos \dfrac{\pi }{3} = \dfrac{1}{2},m = cos \dfrac{\pi }{4} = \dfrac{1}{\sqrt{2}}$ and

$n = cos \,\theta$ .

$\therefore l^{2}+m^{2}+n = 1 \Rightarrow \dfrac{1}{4}+\dfrac{1}{2}+n^{2} = 1 \Rightarrow n^{2} = \dfrac{1}{4} \Rightarrow n = \dfrac{1}{2}\Rightarrow cos\,\theta = \dfrac{1}{2}\Rightarrow \theta = \dfrac{\pi }{3}$
Since

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

$\therefore \vec{a} = l\hat{i}+m\hat{j}+n\hat{k}$

$\Rightarrow \vec{a} = \dfrac{1}{2}\hat{i}+\dfrac{1}{\sqrt{2}}\hat{j}+\dfrac{1}{2}\hat{k}$
Hence, components of

$\vec{a}$ are

$\dfrac{1}{2}\hat{i},\dfrac{1}{\sqrt{2}}\hat{j},\dfrac{1}{2}\hat{k}.$

The adjacent sides of a parallelogram are represented by the vectors

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

$\vec{b} = 2\hat{i}+\hat{j}+2\hat{k}.$ Find unit vectors parallel to the diagonals of the parallelogram.

0%
$\dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}+\hat{k})$
0%
$\dfrac{1}{\sqrt{2}}(-\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k})$
0%
$\dfrac{1}{\sqrt{22}}(3\hat{i}+2\hat{j}+3\hat{k}),\dfrac{1}{\sqrt{2}}(\hat{i}+\hat{k})$
0%
$\dfrac{1}{\sqrt{2}}(+\hat{i}+2\hat{j}+\hat{k}),\dfrac{1}{\sqrt{6}}(\hat{i}-\hat{k})$

Explanation

Gven

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

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

From the figure

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

and

$\vec a+\vec d=\vec b$

$\Rightarrow\vec d=\vec b-\vec a=2\hat i+\hat j+2\hat k-(\hat i+\hat j+\hat k)=\hat i+\hat k$

The unit vector parallel to

$\vec c$ is

$\dfrac{\vec c}{|\vec c|}=\dfrac{3\vec i+2\vec j+3\vec k}{\sqrt{3^2+2^2+3^2}}=\dfrac{1}{\sqrt{22}}(3\vec i+2\vec j+3\vec k)$

The unit vector parallel to

$\vec d$ is

$\dfrac{\vec d}{|\vec d|}=\dfrac{\vec i+\vec k}{\sqrt{(1)^2+(1)^2}}=\dfrac{1}{\sqrt{2}}(\vec i+\vec k)$

1615939_1662342_ans_f87054d78fba4c41ad58852c889f4207.png

The unit vector normal to the plane containing

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

$\vec{b}=(\hat{i}+\hat{j}+\hat{k})$ is?

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

Let

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

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

$\vec{b}$ and

$\vec{c}$ whose projection on

$\vec{a}$ is of magnitude

$\sqrt{(2/3)}$ is

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

$\bar{a}$ and

$\bar{b} = 3 \hat{i} + 6 \hat{j} + 6 \hat{k}$ are collinear and

$\bar{a} . \bar{b} = 27$ , then

$\bar{a}$ is equal to

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

Explanation

Since,

$\vec { a }$ and

$\vec { b }$ are collinear vector. Therefore,

$\vec { a }=\lambda\vec { b }$ ............(i)

$\because \vec { a }.\vec { b }=27$

$\Rightarrow |\vec { a }||\vec { b }|\cos 0^0=27$

$\Rightarrow |\vec { b }|.\sqrt{9+36+39}=27$

$\Rightarrow |\vec {a }|=\dfrac{27}{9}=3$

By Equation (i),

$\vec { }=\lambda \vec { b }$

$\Rightarrow |\vec { a}|=|\lambda||\vec { b }|$

$3=|\lambda|.9$

$\Rightarrow |\lambda|=\pm\dfrac{1}{3}$

$\therefore \vec { a }=\pm \dfrac{1}{3}(3\hat{i}+6\hat{j}+6\hat{k})$

$\vec { }=\pm (\hat{i}+2\hat{j}+2\hat{k})$

Let

$O$ be the circumcentre,

$G$ be the centroid and

$O$ be the orthocentre of a

$\triangle ABC$ . Three vectors are taken through

$O$ and are represented by

$\vec{a}=\vec{OA}, \vec{b}=\vec{OB}$ and

$\vec{c}=\vec{OC}$ then

$\vec{a}+\vec{b}+\vec{c}$ is

0%
$\vec{OG}$
0%
$2\vec{OG}$
0%
$\vec{OO}$
0%
None of them

$(\vec a\times \vec b)^2 +(\vec a. \vec b)^2 =144$ and

$|\vec a|=4$ , then

$|\vec b|=$

0%
$16$
0%
$8$
0%
$3$
0%
$12$

A parallelogram is constructed on the vectors

$\vec{a}=3\vec{\alpha}-\vec{\beta}, \vec{b}=\vec{\alpha}+3\vec{\beta}$ if

$|\vec{\alpha}|=|\vec{\beta}|=2$ and angle between

$\vec{\alpha}$ and

$\vec{\beta}$ is

$\pi/3$ then the length of a diagonal of the parallelogram is

0%
$4\sqrt{5}$
0%
$4\sqrt{3}$
0%
$4\sqrt{7}$
0%
$None\ of\ these$

$A, B, C$ and

$D$ have position vectors

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

$\vec{d}$ respectively, such that

$\vec{a} - \vec{b} = 2 (\vec{d} - \vec{c})$ . Then

0%
$AB$ and $CD$ bisect each other
0%
$BD$ and $AC$ bisect each other
0%
$AB$ and $CD$ trisect each other
0%
$BD$ and $AC$ trisect each other

Explanation

$\vec{a} - \vec{b} = 2 (\vec{d} - \vec{c})$

$\therefore \frac{\vec{a} + 2 \vec{c}}{2 + 1} = \frac{\vec{b} + 2 \vec{d}}{2 + 1}$

$\Rightarrow$

$AC$ and

$BD$ trisect each other as L.H.S. is the position vector of a point trisecting

$A$ and

$C$ , and R.H.S. that of

$B$ and

$D$ .

$p\hat{i}+3\hat{j}+4\hat{k}$ and

$\sqrt{q}\hat{i}+4\hat{k}$ are two vectors, where

$p,q>0$ are two scalars, then the length of the vectors is equal to

0%
All value of $(p,q)$
0%
Only finite number of values of $(p,q)$
0%
Infinite number of values of $(p,q)$
0%
No value fo $(p,q)$

$(\vec r. \hat i)(\vec r \times \hat i)+ (\vec r. \hat j)(\vec r \times \hat j) +(\vec r. \hat k)(\vec r \times \hat k)$ is equal to

0%
$3\ \vec r$
0%
$\vec r$
0%
$\vec 0$
0%
$None\ of\ these$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.