MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 13

$\vec { u } =\vec { a } -\vec { b } ~;\vec{ v } =\vec { a } +\vec { b } ~~\&~ |\vec { a }| =|\vec { b }| =2,$ then

$\left| \vec { u } \times \vec { \upsilon } \right|$ is equal to:

0%
$\sqrt { 2\left( 4-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) }$
0%
${ 2 }\sqrt { \left( 16+\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) }$
0%
${ 2 }\sqrt { \left( 4-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) }$
0%
${ 2 }\sqrt { \left( 16-\left( \vec { a } .\vec { b } \right) ^{ 2 } \right) }$

$\vec{r}.\hat{i}=2\vec{r}.\hat{j}=4\vec{r}.\hat{k}$ and

$\left|\vec{r}\right|=\sqrt{84}$ , then

$\left|\vec{r}.\left(2\hat{i}-3\hat{j}+\hat{k}\right)\right|$ is equal to

0%
$0$
0%
$2$
0%
$4$
0%
$6$

$\bar{a} = \hat{i} + \hat{j} - 2 \hat{k}, \bar{b} = 2 \hat{i} - 0 \hat{j} + \hat{k}, \bar{c} = 3 \hat{i} - \hat{k}$ and

$\bar{c} = m \bar{a} + n \bar{b}$ then m + n = ....

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

If the position vectors of

$A, B, C, D$ are

$3\hat{i} + 2\hat{j} + \hat{k}, 4\hat{i} + 5\hat{j} + 5\hat{k}, 4\hat{i} + 2\hat{j} - 2\hat{k}, 6\hat{i} + 5\hat{j} - \hat{k}$ respectively then the position vector of the point of intersection of

$\bar{AB}$ and

$\bar{CD}$ is

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

Explanation

$\begin{aligned} \overrightarrow{A B} &=\vec{b}-\vec{a} \\ &=(4 i+5 \hat{\jmath}+5 \hat{k})-(3 \hat{i}+2 \hat{\jmath}+\hat{k}) \\ &=\hat{\imath}+3 \hat{\jmath}+4 \hat{k} , and \end{aligned}$

$\begin{aligned} \overrightarrow{C D} &=\vec{d}-\vec{c} \\ &=(6 \hat{\imath}+5 \hat{\jmath}-\hat{k})-(4 \hat{\imath}+2 \hat{\jmath}-2 \hat{k}) \\ &=\overrightarrow{2} \hat{\imath}+3 \hat{\jmath}+\hat{k} \\ \text { Let } \overrightarrow{A B} & \text { and } \overrightarrow{C D} \text { intersect at } \vec{P} . \\ \text { Line } A B=(3 i+2 j+\hat{k})+\lambda(i+3 \hat{j}+4 \hat{k}) & \text { and } \end{aligned}$

Line

$C D=(4 i+2 \hat{\imath}-2 \hat{k})+\mu(2 \hat{i}+3 \hat{j}+\hat{k})$

W.r.t. line

$A_{0}$ ,

$P=(3+\lambda, 2+3 \lambda, 1+4 \lambda)$

and

$w \cdot r \cdot t \cdot$ line

$C D$ ,

$P=(4+2 \mu, 2+3 \mu,-2+\mu)$

Comparing both points,

$\begin{aligned} \Rightarrow & 3+\lambda^{\prime}=4+2 \mu, \\ & 2+3 \lambda=2+3 \mu \\ & 1+4 \lambda=-2+\mu \end{aligned}$

Solving these equs.,

$\therefore \vec{p}=2 \hat{\imath}-\hat{\jmath}-3 \hat{k}$

$\therefore$ option

$D$ is correct.

Let

$\vec{a}=\hat{i}-\hat{j},\vec{b}=\hat{i}-\hat{j}=\vec{c}=\hat{i}-\hat{j}$ , if

$\vec{d}$ is a unit vector such that

$\vec{a}.\vec{d}=0=|\vec{b}\vec{c}\vec{d}|$ then

$\vec{d}$ equals:

0%
$\pm\dfrac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}$
0%
$\pm\dfrac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{3}}$
0%
$\pm\dfrac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{3}}$
0%
$\pm\hat{k}$

$a=i-j,b=i+j,c=i+3j+5k$ and

$n$ is a unit vector such that

$b,n=0,a,n=0$ then the value of

$|c,n|$ is equal to

0%
$1$
0%
$3$
0%
$5$
0%
$2$

Explanation

$\begin{array}{l} { b_{ n } }=0 \\ \left( { \hat { i } +\hat { j } } \right) \left( { a\hat { i } +\hat { i } e } \right) =0 \\ a+b=c \\ \left( { -\hat { j } } \right) \left( { a\hat { i } +b\hat { j } } \right) =0 \\ a-b=0\, \, \, \, \, \, \, a=0=b \\ \left( { \bar { c } \cdot \bar { n } } \right) \\ \left( { \hat { i } +3\hat { j } +5\hat { b } } \right) \cdot \left( { c\hat { k } } \right) \\ =5c\, =\sqrt { { a^{ 2 } }+{ b^{ 2 } }+{ c^{ 2 } } } =1 \\ =5 \end{array}$

The value of

$|\vec{a} \times \hat{i} |^2 + |\vec{a} \times \hat{j} |^2 + |\vec{a} \times \hat{k} |^2$ is

0%
$a^2$
0%
$2a^2$
0%
$3a^2$
0%
none of these

The values of

$\lambda$ such that

$(x, y, z) \neq (0, 0, 0)$ and

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

0%
$0, 1$
0%
$0, -1$
0%
$1, -1$
0%
$0, 1, -1$

Explanation

$\Rightarrow(x+3 y-4 z-\lambda x)\hat{i}+(x-3 y+5 z-\lambda y) \hat{j}+(3 x+y-\lambda z) \hat{k}=0$

$\Rightarrow x+3 y-4 z-\lambda x=0$

$x-3 y+5 z-\lambda y=0$

$3 x+y-\lambda z=0$

$\left[\begin{array}{ccc}1-\lambda & 3 & -4 \\ 1 & -3-\lambda & 5 \\ 3 & 1 & -\lambda\end{array}\right]\left[\begin{array}{c}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$

$\Rightarrow\left|\begin{array}{ccc}1-\lambda & 3 & -4 \\ 1 & -3-\lambda & 5 \\ 3 & 1 & -\lambda\end{array}\right|=0 \quad\left\{\begin{array}{c}\because \text { for non-trivial } \\ \text { soution of equation } \\ A x=y \\ |A|=0\end{array}\right\}$

$\Rightarrow(1-\lambda)((-3-\lambda)(-\lambda)-5)-1(-3 \lambda+4)+3(15+4(-3-\lambda))=0$

$=(1-\lambda)\left(3 \lambda+\lambda^{2}-5\right)-4+3 \lambda+3(3-4 \lambda)=0$

$=(1-\lambda)\left(\lambda^{2}+3 \lambda-5\right)-4+3 \lambda+9-12 \lambda=0$

$=\lambda^{2}+3 \lambda-5-\lambda^{3}-3 \lambda^{2}+5 \lambda-4+3 \lambda+9-12 \lambda=0$

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

$\Rightarrow \lambda^{3}+2 \lambda^{2}+\lambda^{2}=0$

$\lambda\left(\lambda^{2}+2 \lambda+1\right)=0$

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

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

$\lambda=-1$

$\Rightarrow \lambda=0,-1$

$\therefore$ option

$B$ is correct.

Let

$\vec{u}$ ,

$\vec{v}$ ,

$\vec{w}$ be such that

$\left | \vec{u} \right |$ = 1,

$\left | \vec{v} \right |$ = 2,

$\left | \vec{w} \right |$ = 3 . If the projection

$\vec{v}$ along

$\vec{u}$ is equal to that of

$\vec{w}$ along

$\vec{u}$ and

$\vec{v}$ ,

$\vec{w}$ are perpendicular to each other , then

$\left | \vec{u} \vec{v} + \vec{w} \right |$ equals

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

Explanation

Given,

$|\vec{u}|=1$ and

$|\vec{v}|=2,|\vec{w}|=3$

and,

$\quad \vec{v} \cdot \vec{u}=\vec{w} \cdot \vec{u}$

$\vec{v} \cdot \vec{u}=\vec{\omega} \cdot \vec{u}$

or,

$\quad \vec{v} \cdot \vec{\omega}=0$

Now,

$\quad|\vec{u}-\vec{v}+\vec{w}|$

Squaring,

$|\vec{u}|^{2}+\left.\vec{v}\right|^{2}+|\vec{w}|^{2}-2 \vec{u} \cdot \vec{v}-2 \vec{v} \cdot \vec{\omega}$

$+2 \vec{u} \cdot \vec{w}$

$\Rightarrow 1+4+9-2\vec{u} \cdot\vec{v}+2 \overrightarrow{u} \cdot \vec{v}$ -0

$\Rightarrow \quad 14$

$|\vec{u}-\vec{v}+\vec{w}|=\sqrt{14}$

option

$A=\sqrt{14}$

Vector

$\vec { x }$ satisfying the relation

$\vec { A } . \overline { x } = c$ and

$\vec { A } \times \vec { x } = \vec { B }$ is

0%
$\frac { c \vec { A } - ( \overline { A } \times \vec { B } ) } { | \vec { A } | }$
0%
$\frac { c \vec { A } - ( \vec { A } \times \vec { B } ) } { | \vec { A } | ^ { 2 } }$
0%
$\frac { c \vec { A } + ( \vec { A } \times \vec { B } ) } { | \vec { A } | ^ { 2 } }$
0%
None

For any vector

$\vec a$ , the value of

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

0%
$4{ |\vec a| }^{ 2 }$
0%
$2{ |\vec a| }^{ 2 }$
0%
${|\vec a |}^{ 2 }$
0%
$3{ |\vec a| }^{ 2 }$

let

$\bar { a }$ be a unit vector and

$\bar { b }$ be a non-zero vector not parallel to

$\bar { a }$ if two sides of a triangle are represented by the vectors

$\sqrt { 3 } \left( \bar { a } \times \bar { b } \right) \quad and\quad \bar { b } - \left( \bar { a } .\bar { b } \right) \bar { a }$ then the angles of triangle are

0%
$90^{ \circ },{ 60 }^{ \circ },{ 30 }^{ \circ }$
0%
$45^{ \circ },{ 45 }^{ \circ },{ 90 }^{ \circ }$
0%
$60^{ \circ },{ 60 }^{ \circ },{ 60 }^{ \circ }$
0%
$75^{ \circ },{ 45 }^{ \circ },{ 60 }^{ \circ }$

In a triangle ABC, if

$A=(0, 0), B=(3, 3\sqrt{3}), C=(-3\sqrt{3}, 3)$ then the vector of magnitude

$2\sqrt{2}$ units directed along

$\overline{AO}$ , where O is the circumcentre of triangle ABC is?

0%
$(1-\sqrt{3})\bar{i}+(1+\sqrt{3})\bar{j}$
0%
$\sqrt{3}\bar{i}+2\bar{j}$
0%
$\bar{i}-\sqrt{3}\bar{j}$
0%
$\bar{i}+2\bar{j}$

In a parallelogram ABD,

$|\overset { \_ }{ A\overset { \rightarrow }{ B } } |=a,|\overset { \_ }{ A\overset { \rightarrow }{ D } } |=b$ and

$|\overset { \_ \\ \quad \quad \rightarrow }{ AC } |=c,$ ,

$\overset { \_ \rightarrow }{ AB } .\overset { \_ \rightarrow }{ DB }$ has the value :

0%
$\frac { 1 }{ 2 } ({ 3a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 })$
0%
$\frac { 1 }{ 2 } ({ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 })$
0%
$\frac { 1 }{ 2 } ({ a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 })$
0%
$\frac { 1 }{ 3 } ({ b }^{ 2 }+{ c }^{ 2 }-{ a }^{ 2 })$

If the position vectors of the vertices

$A,B$ and

$C$ of a

$\Delta ABC$ are respectively

$4\hat{i}+7\hat{j}+8\hat{k},2\hat{i}+3\hat{j}+4\hat{k}$ and

$2\hat{i}+5\hat{j}+7\hat{k}$ , then the position vector of the point, where the bisector of

$\angle A$ meets

$BC$ is:

0%
$\frac{1}{2}(4\hat{i}+8\hat{j}+11\hat{k})$
0%
$\frac{1}{3}(6\hat{i}+13\hat{j}+18\hat{k})$
0%
$\frac{1}{4}(8\hat{i}+14\hat{j}+19\hat{k})$
0%
$\frac{1}{3}(6\hat{i}+8\hat{j}+15\hat{k})$

$If\quad |\overrightarrow { a } |$ =2 and

$\quad |\overrightarrow { b } |$ =3 and

$\quad |\overrightarrow { a } |$ .

$\quad |\overrightarrow { b } |$ =Then (a x (a x (a x (a x b)))) is equal to

0%
$4\hat { b }$
0%
$-4\hat { b }$
0%
$4\hat { a }$
0%
$-4\hat { a }$

If C is the mid-point of AB and P is any point outside AB , then

0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ + $\overrightarrow{PC}$ = 0
0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ + 2 $\overrightarrow{PC}$ = $\overrightarrow{0}$
0%
$\overrightarrow{PA}$ $\overrightarrow{PB}$ = $\overrightarrow{PC}$
0%
$\overrightarrow{PA}$ $\overrightarrow{PB}$ = 2 $\overrightarrow{PC}$

$\overset { \rightarrow }{ a }$ and

$\overset { \rightarrow }{ b }$ are vectors such that

$|\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } |=\sqrt { 29 }$ and

$\overset { \rightarrow }{ a } \times (2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )=(2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )\times \overset { \rightarrow }{ b }$ , then a possible value of

$(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } )(-7\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )$ is

0%
0
0%
3
0%
4
0%
8

$\hat{u}$ and

$\hat{v}$ are unit vectors and

$\theta$ is the acute angle between them , then 2

$\hat{u}$ 3

$\hat{v}$ is a unit vector for

0%
No value of $\theta$
0%
Exactly one value of $\theta$
0%
Exactly two values of $\theta$
0%
More than two values of $\theta$

Explanation

Given,

$\bar{u}$ and

$\vec{v}$ are unit vector

$|\vec{u}|=1$

$|\vec{v}|=1$

Now,

$\Rightarrow|2 \vec{u}||\overrightarrow{3 v}| \sin \theta=1$

$|\overrightarrow{2 u} \times \rightarrow{\vec{3 v}}|$

$[\because|\vec{a} \times \vec{b}|=|\vec a|\vec b|| \sin \theta]$

$\Rightarrow 2 \cdot 3|\bar{u}| \nabla \mid \sin \theta=1 \\$

$\sin \theta=\frac{1}{6}$

$\text { There is only one value of } \theta \\$

$\text { for which }|\vec 24 \times \vec 3V| \text { is a } \\$

$\text { unit vector. }$

$\text { option } B$

Let

$\vec{a}$ =

$\hat{i}$

$\hat{j}$ ,

$\vec{b}$ =

$\hat{j}$

$\hat{k}$ ,

$\vec{c}$ =

$\hat{k}$

$\hat{i}$ . If

$\vec{d}$ is a unit vector such that

$\vec{a}.\vec{d}$ = 0 =

$\left | \vec{b}\vec{c}\vec{d} \right |$ then

$\vec{d}$ equals :

0%
$\dfrac{\hat{i} + \hat{j} 2\hat{k}}{\sqrt{6}}$
0%
$\dfrac{\hat{i} + \hat{j} \hat{k}}{\sqrt{3}}$
0%
$\dfrac{\hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$
0%
$\hat{k}$

The value of

$\lambda$ for

$\left( x,y,z \right) \neq \left( 0,0,0 \right)$ and

$\left( i+j+3k \right) x+\left( 3i-3j+k \right) y+\left( -4i+5j \right) z=\lambda \left( xi+yj+zk \right)$ are

0%
0, -1
0%
0, 1
0%
-2, 0
0%
0, 2

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

$\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0$ then the value of

$\overset { \rightarrow }{ a. } \overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a. }$ is

0%
1
0%
-1
0%
-3/2
0%
none of these

$\overline { a } =-2\overline { i } +3\overline { j } +4\overline { k }$ and

$\overline { b } =-2\overline { i } -2\overline { j } +3\overline { k }$ then

$\overline { a } .\overline { b }$ is

0%
2
0%
-2
0%
6
0%
none of these

Let position vector of the orthocentre of

$\triangle ABC$ be

$\overrightarrow{r}$ . then, which of the following statement(s) is\are correct (Given position vector of points

$a\hat{i},b\hat{j},c\hat{k}$ and

$abc=0$ )

0%
$\displaystyle \overrightarrow { r } .\bar { i } =\frac{a}{\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}}}$
0%
$\displaystyle \overrightarrow { r } .\bar { i } =\frac{1}{a\left(\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}+\frac{1}{{c}^{2}} \right) }$
0%
$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$
0%
$\displaystyle \frac {\overrightarrow { r } .\bar { i }}{\overrightarrow { r } .\bar { j }}+\frac {\overrightarrow { r } .\bar { j }}{\overrightarrow { r } .\bar { k }}+\frac {\overrightarrow { r } .\bar { k }}{\overrightarrow { r } .\bar { i }}=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$

If C is the mid point of AB and P is any point outside AB , then

0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ + $\overrightarrow{PC}$ = 0
0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ + $2\overrightarrow{PC}$ = $\overrightarrow{0}$
0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ = $\overrightarrow{PC}$
0%
$\overrightarrow{PA}$ + $\overrightarrow{PB}$ = $2\overrightarrow{PC}$

A particle in a plane from A to E along the shown path. It is given that AB=BC=DE=10 metre. Then the magnitude of net displacement of particle is :

0%
10 m
0%
15 m
0%
5 m
0%
20 m

Let

$a=\displaystyle\sum_{i < j}\left(\dfrac{1}{^{n}C_i}+\dfrac{1}{^{n}C_j}\right)$ and

$b=\displaystyle\sum_{i < j}\left(\dfrac{i}{^{n}C_i}+\dfrac{j}{^{n}C_j}\right)$ , then?

0%
$b=(n-1)a$
0%
$b=(n+1)a$
0%
$b=\dfrac{n}{2}a$
0%
$b=na$

In parallelogram ABCD,

$|\overline { AB } |=a,|\overline { AD } |=b$ and

$|\overline { AC } |=c$ then

$\overline { DB } ,\overline { AB }$ has the value

0%
$\frac {3a^2+b^2-c^2}{2}$
0%
$\frac {3b^2+c^2-a^2}{2}$
0%
$\frac {3c^2+b^2-a^2}{2}$
0%
$\frac {a^2+b^2+c^2}{2}$

$\sum_{i=1}^{n} \vec{a}_{i}=\vec{0}$ where

$\left|\vec{a}_{i}\right|=1 \forall i,$ then the value of

$\sum_{1 \leq i<j \leq n} \vec{a}_{i} \cdot \vec{a}_{j}$ is

0%
$-n/2$
0%
$-n$
0%
$n/2$
0%
$n$

If the vectors

$\vec{a}$ ,

$\vec{b}$ ,

$\vec{c}$ satisfying

$\vec{a}$ +

$\vec{b}$ + 2

$\vec{c}$ = 0 . If

$\left | \vec{a} \right |$ = 1 ,

$\left | \vec{b} \right |$ = 4 ,

$\left | \vec{c} \right |$ = 2 , then

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

$\vec{b}.\vec{c}$ +

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

0%
$-\dfrac{7}{2}$
0%
$-\dfrac{17}{2}$
0%
$\dfrac{17}{2}$
0%
$\dfrac{7}{2}$

The projection of the join of the point (3, 4, 2), (5, 1, 8) on the line whose d.c.'s are

$\left( \dfrac { 2 }{ 7 } ,-\dfrac { 3 }{ 7 } ,\dfrac { 6 }{ 7 } \right)$ is

0%
7
0%
$\left( \dfrac { 46 }{ 13 } \right)$
0%
$\left( \dfrac { 42 }{ 13 } \right)$
0%
$\left( \dfrac { 38 }{ 13 } \right)$

$a$ ,

$b$ and care three mutually perpendicular vectors, then the projection of the vector

$\left|\frac{a}{|a|}+m \frac{b}{|b|}+n \frac{(a \times b)}{|a \times b|}\right.$ along the angle bisector of the vector

$a$ and

$b$ is

0%
$\frac{l^{2}+m^{2}}{\sqrt{l^{2}+m^{2}-n^{2}}}$
0%
$\sqrt{1^{2}+m^{2}+n^{2}}$
0%
$\frac{\sqrt{1^{2}+m^{2}}}{\sqrt{l^{2}-m^{2}+n^{2}}}$
0%
$\frac{1+m}{\sqrt{2}}$

Let OAB be a regular triangle with side unity (o being otogin). Also M, N are the points of intersection of AB, M being closer to A and N closer to B. Position vectors of A, B, M and N are

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

$\vec { n }$ respectively. Which of the following hold (s) good ?

0%
$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow \dfrac { 2 }{ 3 }$ and $y=\dfrac { 1 }{ 3 }$
0%
$\vec { m } =x\vec { a } +y\vec { b } \Rightarrow x=\dfrac { 5 }{ 6 }$ and $y=\dfrac { 1 }{ 6 }$
0%
$\vec { m } .\vec { n }$ equals $\dfrac { 13 }{ 18 }$
0%
$\vec { m } .\vec { n }$ equals $\dfrac { 15 }{ 18 }$

The position vector of A is

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

$\vec { AB } =5\vec { i } +7\vec { j } +6\vec { k }$ , then the position vector of B is

0%
$-7\vec { i } -10\vec { j } -10\vec { k }$
0%
$7\vec { i } -10\vec { j } +10\vec { k }$
0%
$7\vec { i } +10\vec { j } -10\vec { k }$
0%
$7\vec { i } +10\vec { j } +10\vec { k }$

The position vector of a point lying on the joining the points whose position vectors are

$\overline i + \overline j -\overline k$ and

$\overline i - \overline j +\overline k$ is

0%
$\overline j$
0%
$\overline i$
0%
$\overline k$
0%
$\overline 0$

Area of diagonals is, ..., where diagonals are

$a = 2 \hat {i} - 3 \hat { j } + 5 \hat { k }$ , and

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

0%
$\sqrt { 21.5 }$
0%
$\sqrt { 31.5 }$
0%
$\sqrt { 28.5 }$
0%
$\sqrt{ 38.5 }$

$\bar { a } ,\bar { b } ,\bar { c }$ are position vectors of the points A,B,C respectively such that

$9\bar { a } -7\bar { b } -2\bar { c } =\bar { 0 }$ then point B divides AC in the ratio.....

0%
Internally 7:2
0%
Externally 9:2
0%
Internally 9:7
0%
Externally 2:7

A vector

$A=\overrightarrow { l } =\overrightarrow { xj } =3\overrightarrow { k }$ is rotated through an angle and is also doubled in magnitude resulting in

$\overrightarrow { B } =4\overrightarrow { l } +\left( 4x-2 \right) \overrightarrow { j } +2\overrightarrow { k }$ . An acceptable value of x is

0%
1
0%
2
0%
3
0%
4/3

A stone projected vertically upwards raises 's' feets in 't' seconds where

$_{ S }=112t-{ 16t }^{ 2 }$ then the maximum height it reached is

0%
195 ft
0%
194 ft
0%
196 ft
0%
216 ft

Explanation

$s=112 t-16 t^{2}$

for maximum height

$\dfrac{d s}{d t}=0$

$\dfrac{d\left(112 t-16 t^{2}\right)}{d t}=0$

$\dfrac{d(112 t)}{d t}-\dfrac{d\left(16 t^{2}\right)}{d t}=0$

$112-16(2) t^{2-1}=0$

$112-32 t=0$

$32 t=112$

$t=\dfrac{112}{32}=\dfrac{28}{8}=\frac{7}{2}$

$\therefore$ maximum height

$=112\left(\dfrac{7}{2}\right)-16\left(\dfrac{7}{2}\right)^{2}$

$=56 \times 7-4 \times 49$

$=392-196$

$=196 \mathrm{ft}$

Given

$\overline { a } = x \hat { i } + y \hat { j } + 2 \hat { k } , \overline { b } = \hat { i } - \hat { j } + \hat { k } , \overline { c } = \hat { i } + 2 \hat { j } ;( \overline { a } \hat { } \overline { b } ) = \pi / 2 , \overline { a } .\overline { c } = 4$ then

0%
$[ \overline { a } \overline { b } \overline { c } ] ^ { 2 } = \left| \overline { a } \right|$
0%
$[ \overline { a } \overline { b } \overline { c } ]= \left| \overline { a } \right|$
0%
$[ \overline { a } \overline { b } \overline { c } ]= 0$
0%
none of these

$\vec { a }$ and

$\vec { b }$ are vectors such that

$| \vec { a } + \vec { b } | = \sqrt { 29 }$ and

$\vec { a } \times ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) = ( 2 \hat { i } + 3 \hat { j } + 4 \hat { k } ) \times \vec { b } ,$ then
a possible value of

$( \vec { a } + \vec { b } ) \cdot ( - 7 \hat { i } + 2 \hat { j } + 3 \hat { k } )$ is

0%
0
0%
3
0%
4
0%
8

For any three

$\vec { a } , \vec { b } , \vec { c } ( \vec { a } - \vec { b } ) ( \vec { b } - \vec { c } ) \times ( \vec { c } - \vec { a } )$ is equal to

0%
$\vec { b } \cdot ( \vec { c } \times \vec { a } )$
0%
$2 \vec { a } \cdot ( \vec { b } \times \vec { c } )$
0%
$0$
0%
none of these

The point D,E,F divide BC, CA and Ab of the triangle ABC in the ratio 1 : 4, 3 : 2 and 3 : 7 respectively and the point K divides AB in the ratio 1 : 3 then

$(\overrightarrow { AD } + \overrightarrow { BE } +\overrightarrow { CF }) : \overrightarrow { CK }$ is equal to

0%
5 : 2
0%
2:5
0%
1:1
0%
none of these

$\left( \vec { r } .\vec { i } \right) \left( \vec { r } \times \vec { i } \right) +\left( \vec { r } .\vec { j } \right) \left( \vec { r } \times \vec { j } + \right) \left( \vec { r } .\vec { k } \right) \left( \vec { r } \times \vec { k } \right)$ is equal to

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

If u =

$\hat{j} + 4 \hat{K}, V = \hat{i} - 3 \hat{K} w = cos \theta \hat{i} + sin \theta \hat{j}$ are vectors in 3- dimensional space, then the maximum possible value of

$|u \times v.w|$ is

0%
$\sqrt{13}$
0%
$\sqrt{14}$
0%
5
0%
7

If u =

$\hat{j} + 4\hat{k}, V = \hat{i} =- 3K and W = cos \theta i + sin \theta \hat{i}$ are vectors in 3-dimension space, then the maximum possible value of

$|u \times v. w|$ is

0%
$\sqrt 13$
0%
$\sqrt 14$
0%
5
0%
7

$\hat{i}\times (\vec{a}\times \hat{i})+\hat{j}\times (\vec{a}\times \hat{j})+\hat{k}\times (\vec{a}\times \hat{k})=.....\left\{(\vec{a}.\hat{i})\hat{i}+(\vec{a}.\hat{j})\hat{j}+(\vec{a}.\hat{k})\hat{k}\right\}$

0%
$-1$
0%
$0$
0%
$2$
0%
$None\ of\ these$

The magnitude of two vectors which can be represented in the form i+j+(2x)k is

$\sqrt{18}$ . Then the unit vector that is perpendicular to these two vectors is

0%
$\frac{-i+j}{\sqrt{2}}$
0%
$\frac{i-j}{8\sqrt{2}}$
0%
$\frac{-i+j}{8}$
0%
$\frac{-i+j}{2\sqrt{2}}$

The length of vector

$\overrightarrow{A G}$ is

0%
$\sqrt{17}$
0%
$\sqrt{51} / 3$
0%
$3 / \sqrt{6}$
0%
$\sqrt{59} / 4$

Explanation

Point

$G$ is

$\left(\dfrac{4}{3}, \dfrac{1}{3}, \dfrac{8}{3}\right) \cdot$ Therefore

$\begin{array}{l}|\overrightarrow{A G}|^{2}=\left(\dfrac{5}{3}\right)^{2}+\dfrac{1}{9}+\left(\dfrac{5}{3}\right)^{2}=\dfrac{51}{9} \\\Rightarrow|\overrightarrow{A G}|=\dfrac{\sqrt{51}}{3} \\\overrightarrow{A B}=-4 \hat{i}+4 \hat{j}+0 \hat{k} \\\overrightarrow{A C}=2 \hat{i}+2 \hat{j}+2 \hat{k}\end{array}$

$\therefore \overrightarrow{A B} \times \overrightarrow{A C}=-8\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 0 \\ 1 & 1 & 1\end{array}\right|$

$=8(\hat{i}+\hat{j}-2 \hat{k})$
Area of

$\Delta A B C=\dfrac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=4 \sqrt{6}$

$\overrightarrow{A D}=-3 \hat{i}-5 \hat{j}+3 \hat{k}$
The length of the perpendicular from the vertex

$D$ on the opposite face

$=\mid$ projection of

$\overrightarrow{A D}$ on

$\overrightarrow{A B} \times \overrightarrow{A C} \mid$

$\begin{array}{l}=\left|\dfrac{(-3 \hat{i}-5 \hat{j}+3 \hat{k})(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}\right| \\ \\=\left|\dfrac{-3-5-6}{\sqrt{6}}\right|=\dfrac{14}{\sqrt{6}}\end{array}$

1656314_1784899_ans_ea98f53b591b4639a861d7272b7ca73f.PNG

1656314_1784899_ans_ea98f53b591b4639a861d7272b7ca73f.PNG

For non-zero vectors

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

$\vec{c},|(\vec{a} \times \vec{b}) \cdot \vec{c}|=|\vec{a}||\vec{b}||\vec{c}|$ holds if and only if

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

Explanation

$\textbf{Step 1: Simplifying given expression}$

$\text{Given that }$

$|(\vec a \times \vec b).\vec c|=|\vec a||\vec b||\vec c|$

$\implies ||\vec a||\vec b|sin\theta \hat n.\vec c|=|\vec a||\vec b||\vec c|$

$\text{Here }\hat n \text{ is the resultant of vector } a\text{ and }b$

$\text{and } \hat n.\vec c=|\hat n||\vec c|cos\alpha$

$\implies ||\vec a||\vec b|sin\theta |\hat n||\vec c|cos\alpha |=|\vec a||\vec b||\vec c|$

$\text{Since }\hat n \text{ is a unit vector }|\hat n|=1$

$\textbf{Step 2: Calculating}$

$\implies ||\vec a||\vec b||\vec c|sin\theta.cos\alpha|=|\vec a ||\vec b||\vec c|$

$\implies |sin\theta ||cos\alpha|=1$

$\textbf{Step 3: Calculating result}$

$\text{If we assume that }\theta = \dfrac {\pi}2 , \alpha =0$

$\implies \vec a \text{ and } \vec b \text{ are perpendicular}$

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

$\text{As }\alpha=0, \vec c \text{ is along the resultant of vector }a\text{ and }b$

$\implies \vec c \text{ is perpendicular to }\vec a \text { and }\vec b$

$\implies \vec c.\vec a=0 \text{ and } \vec c.\vec b=0$

$\textbf{Hence, }\mathbf{\vec a.\vec b= \vec a.\vec c= \vec b.\vec c=0}$

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 13 - 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 13 - MCQExams.com

0:0:1

Practice Class 12 Commerce Maths Quiz Questions and Answers

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