MCQ Questions for Class 12 Commerce Maths Vector Algebra Quiz 2

Give the vector from

$(2,-7,0)$ to

$(1,-3,-5)$ .

0%
$(3,4,-5)$
0%
$(-1,4,-5)$
0%
$(3,-10,-5)$
0%
None of these

Explanation

Given points

$A(2,-7,0)$ and

$B(1,-3,-5)$

$\vec{OA}=2\hat{i}-7\hat{j}$

$\vec{OB}=\hat{i}-3\hat{j}-5\hat{k}$

vector from A to B

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

$\vec{AB}=\hat{i}-3\hat{j}-5\hat{k}-(2\hat{i}-7\hat{j})$

$\vec{AB}=\hat{i}-3\hat{j}-5\hat{k}-2\hat{i}+7\hat{j}$

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

$AB=(-1,4,-5)$

Find the vector which joins the point A

$(4,5,6)$ to B

$(10,11,12)$ .

0%
$14\hat i+16\hat j+18\hat k$
0%
$-6\hat i-6\hat j-6\hat k$
0%
$6\hat i+6\hat j+6\hat k$
0%
None of these

Explanation

Given points

$A(4,5,6)$ and

$B(10,11,12)$

$\vec{OA}=4\hat{i}+5\hat{j}+6\hat{k}$

$\vec{OB}=10\hat{i}+11\hat{j}+12\hat{k}$

vector

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

$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-(4\hat{i}+5\hat{j}+6\hat{k})$

$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-4\hat{i}-5\hat{j}-6\hat{k}$

$\vec{AB}=6\hat{i}+6\hat{j}+6\hat{k}$

Give the vector from

$(1,-3,-5)$ to

$(2,-7,0)$ .

0%
$(1,-4,5)$
0%
$(3,-10,-5)$
0%
$(1,4,5)$
0%
None of these

Explanation

Given points

$A(1,-3,-5)$ and

$B(2,-7,-0)$

$\vec{OA}=\hat{i}-3\hat{j}-5\hat{k}$

$\vec{OB}=2\hat{i}-7\hat{j}$

vector from

$A$ to

$B$

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

$\vec{AB}=2\hat{i}-7\hat{j}-(\hat{i}-3\hat{j}-5\hat{k})$

$\vec{AB}=2\hat{i}-7\hat{j}-\hat{i}+3\hat{j}+5\hat{k}$

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

$AB=(1,-4,5)$

Find the vector

$w$ with the initial point

$(4,1,2)$ and final point

$(1,6,5)$ .

0%
$(3,-5,-3)$
0%
$(0,7,3)$
0%
$(-3,5,3)$
0%
None of these

Explanation

Given points

$A(4,1,2)$ and

$B(1,6,5)$

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

$\vec{OB}=\hat{i}+6\hat{j}+5\hat{k}$

vector

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

$\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-(4\hat{i}+\hat{j}+2\hat{k})$

$\vec{AB}=\hat{i}+6\hat{j}+5\hat{k}-4\hat{i}-\hat{j}-2\hat{k}$

$\vec{AB}=-3\hat{i}+5\hat{j}+3\hat{k}$

$AB=(-3,5,3)$

Find

$2v$ , when

$u=(3,4,-2)$ and

$v=(0,-4,0)$ .

0%
$(0,-8,0)$
0%
$(3,8,0)$
0%
$(3,-8,-2)$
0%
None of these

Explanation

Given points

$u(3,4,-2)$ and

$v(0,-4,0)$

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

$\vec{v}=-4\hat{j}$

vector

$2\vec{v}=2\times -4\vec{j}$

$2\vec{v}=-8\vec{j}=(0,-8,0)$

Find the magnitude of the vector which joins the point

$A$

$(4,5,6)$ to

$B$

$(10,11,12)$ .

0%
$12.29$
0%
$10.39$
0%
$10.29$
0%
None of these

Explanation

Given points

$A(4,5,6)$ and

$B(10,11,12)$

$\vec{OA}=4\hat{i}+5\hat{j}+6\hat{k}$

$\vec{OB}=10\hat{i}+11\hat{j}+12\hat{k}$

vector

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

$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-(4\hat{i}+5\hat{j}+6\hat{k})$

$\vec{AB}=10\hat{i}+11\hat{j}+12\hat{k}-4\hat{i}-5\hat{j}-6\hat{k}$

$\vec{AB}=6\hat{i}+6\hat{j}+6\hat{k}$

magnitude of AB

$\left | \vec{AB} \right |=\sqrt{6^2+6^2+6^2}$

$\left | \vec{AB} \right |=\sqrt{36+36+36}$

$\left | \vec{AB} \right |=\sqrt{36\times 3}$

$\left | \vec{AB} \right |=6\sqrt{3}$

$\left | \vec{AB} \right |=6\times 1.732$

$\left | \vec{AB} \right |=10.39$

Which of the following is not a unit vector for all values of

$\theta$ ?

0%
$(\cos\theta)i - (\sin\theta)j$
0%
$(\sin\theta)i + (\cos\theta)j$
0%
$(\sin2\theta)i - (\cos\theta)j$
0%
$(\cos2\theta)i - (\sin2\theta)j$

Explanation

Let

$n$ be any vector.

Then

$n$ is a unit vector, if

$||n||=1$

We know that,

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

For

$n=(\sin 2\theta)i - (\cos \theta)j$

$||n||=\sin^{2} 2\theta + \cos^{2} \theta \neq 1$

Hence,

$(\sin^{2} 2\theta)i + (\cos^{2} \theta)j$ is not a unit vector.

Find the vector

$w$ with the initial point

$(8,-10,3)$ and final point

$(1,10,7)$ .

0%
$(7,10,-4)$
0%
$(-4,10,4)$
0%
$(-7,20,4)$
0%
None of these

Explanation

Given points

$A(8,-10,3)$ and

$B(1,10,7)$

$\vec{OA}=8\hat{i}-10\hat{j}+3\hat{k}$

$\vec{OB}=\hat{i}+10\hat{j}+7\hat{k}$

vector w from A to B

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

$\vec{w}=\hat{i}+10\hat{j}+7\hat{k}-(8\hat{i}-10\hat{j}+3\hat{k})$

$\vec{w}=\hat{i}+10\hat{j}+7\hat{k}-8\hat{i}+10\hat{j}-3\hat{k}$

$\vec{AB}=-7\hat{i}+20\hat{j}+4\hat{k}=(-7,20,4)$

The vector joining vector

$\vec{A}$ to

$\vec{B}$ is represented by:

0%
$\vec{A}-\vec{B}$
0%
$\vec{B}-\vec{A}$
0%
$\vec{A}+\vec{B}$
0%
None of these

Explanation

Given vectors

$\vec{A}$ and

$\vec{B}$

$\vec{AB}=\vec{B}-\vec{A}$

Find the magnitude of the vector which joins the point

$A$

$(-1,-3,-1)$ to

$B$

$(0,0,0)$ .

0%
$\sqrt{21}$
0%
$\sqrt{10}$
0%
$\sqrt{11}$
0%
None of these

Explanation

Given points

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

$B(0,0,0)$

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

$\vec{OB}=0$

vector AB from A to B

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

$\vec{AB}=0-(-\hat{i}-3\hat{j}-\hat{k})$

$\vec{AB}=+\hat{i}+3\hat{j}+\hat{k})$

magnitude of AB

$\left | \vec{AB} \right |=\sqrt{1^2+3^2+1^2}$

$\left | \vec{AB} \right |=\sqrt{1+9+1}$

$\left | \vec{AB} \right |=\sqrt{11}$

Find the vector

$w$ with the initial point

$(a,b,c)$ and final point

$(a+1,b+2,c+3)$ .

0%
$(a,b+2,c+3)$
0%
$(a-1,b-2,c-3)$
0%
$(1,2,3)$
0%
None of these

Explanation

Given points

$A(a,b,c)$ and

$B(a+1,b+2,c+3)$

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

$\vec{OB}=(a+1)\hat{i}+(b+2)\hat{j}+(c+3)\hat{j}$

vector

$w$ from

$A$ to

$B$

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

$\vec{w}=(a+1)\hat{i}+(b+2)\hat{j}+(c+3)\hat{j}-(a\hat{i}+b\hat{j}+c\hat{k})$

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

$\vec{w}=\hat{i}+2\hat{j}+3\hat{j}=(1,2,3)$

Find the magnitude of the vector which joins the point

$A$

$(a,2,c)$ to

$B$

$(a+1,5,c+3)$ .

0%
$\sqrt{18}$
0%
$\sqrt{21}$
0%
$\sqrt{20}$
0%
None of these

Explanation

Given points

$A(a,2,c)$ and

$B(a+1,5,c+3)$

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

$\vec{OB}=(a+1)\hat{i}+5\hat{j}+(c+3)\hat{j}$

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

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

$\vec{AB}=a\hat{i}+5\hat{j}+c\hat{k}+\hat{i}+3\hat{j}-a\hat{i}-2\hat{j}-c\hat{k}$

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

magnitude of AB

$\left | \vec{AB} \right |=\sqrt{1^2+3^2+3^2}$

$\left | \vec{AB} \right |=\sqrt{1+9+9}$

$\left | \vec{AB} \right |=\sqrt{19}$

none of option is matched

The coordinates of a point in

$3$ -D space is

$(3,1,2)$ . Then the position vector of the point is:

0%
$3\hat i + 3\hat j + 3\hat k$
0%
$3\hat i + 2\hat j + \hat k$
0%
$3\hat i + \hat j + 2\hat k$
0%
$\hat i + 2\hat j + 3\hat k$

Explanation

A position vector is a Euclidean vector that represents the position of a point P in space in relation to an arbitrary reference.

Given point is

$(3,1,2)$ in

$3$ -D space.

$x,y$ and

$z$ coordinates are represented by vectors

$i,j$ and

$k$ respectively.

Hence, position vector is given by

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

$\vec{a} = (2, 1, -1), \vec{b} = (1,-1,0), \vec{c} = (5, -1, 1)$ , then what is the unit vector parallel to

$\vec{a} + \vec{b} - \vec{c}$ in the opposite direction ?

0%
$\dfrac{\hat{i}+\hat{j}-2 \hat{k}}{3}$
0%
$\dfrac{\hat{i}-2 \hat{j} + 2 \hat{k}}{3}$
0%
$\dfrac{2 \hat{i} - \hat{j} + 2 \hat{k}}{3}$
0%
None of the above.

Explanation

GIven :

$\vec{a} = (2, 1, -1), \vec{b} = (1,-1,0), \vec{c} = (5, -1, 1)$

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

Magnitude of the vector

$=\sqrt { 4+1+4 } =3$

So unit vector in the opposite direction

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

Hence, C is correct.

The position vectors of the points

$A$ and

$B$ are respectively

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

$\hat { i } +\hat { j } -\hat { k }$ . What is the length of

$AB$ ?

0%
$11$
0%
$9$
0%
$7$
0%
$6$

Explanation

Now, for vectors

$A=a_{ 1 }\hat { i } +b_{ 1 }\hat { j } +c_{ 1 }\hat { k }$

$B=a_{ 2 }\hat { i } +b_{ 2 }\hat { j } +c_{ 2 }\hat { k }$

length of

$AB=\sqrt { (a_{ 2 }-a_{ 1 })^{ 2 }+(b_{ 2 }-b_{ 1 })^{ 2 }+(c_{ 2 }-c_{ 1 })^2 }$

So, for given vectors

Length of AB

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

$=\sqrt { 4+36+9 }$

$=7$

The unit vector perpendicular to the vector

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

$\hat{i}+\hat{j}$ forming a right handed system, is

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

Explanation

Let

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

and

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

Let the unknown vector be

$\vec{c}$

Given

$\vec{c}$

$\perp \vec{a}$ and

$\vec{b}$ and

$|c|=1$

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

$\therefore \vec{c}=\dfrac{\vec{a}\times\vec{b}}{|\vec{a}\times\vec{b}|}$

$\Rightarrow \hat{c}=\dfrac{2\hat{k}}{2}$

$\Rightarrow \hat{c}=\hat{k}$

Correct option is A

Point

$(4, 0)$ lies on ________.

0%
$\vec {XO}$
0%
$\vec {YO}$
0%
$\vec {OX}$
0%
$\vec {OY}$

Explanation

$\vec {XO}$ is positive x-axis, so

$(4, 0)$ lies on it.

State the following statement is True or False

If the starting and end points of a vector are collinear, it is known as a unit vector.

0%
True
0%
False

If the points

$A$ and

$B$ are

$\left( 1,2,-1 \right)$ and

$\left( 2,1,-1 \right)$ respectively, then

$\vec { AB }$ is

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

Explanation

$\vec{ AB } =$

$\left < 2-1, 1-2, -1+1 \right >$

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

$\therefore\quad\vec{AB} =$

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

$\vec {a} . \hat {i} = \vec {a} . (\hat {i} + \hat {j}) = \vec {a} (\hat {i} + \hat {j} + \hat {k})$ , thus

$\vec {a}=$

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

Explanation

Let

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

And

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

$\Rightarrow x.1=x+y+z=x+y$

$\Rightarrow y+z=0$ and

$y=0 \Rightarrow z=0$

Hence,

$\vec{a}=k\hat{i}$ where

$k$ can be any real no

from options ,

$\vec{a}=i$

The Polygon Law of Vector Addition is simply an extension of ____________.

0%
Parallelogram Law of Vector Addition
0%
Triangular Law of Vector Addition
0%
Both A and B
0%
None of the above

$\left| {\widehat a - \widehat b} \right| = \sqrt 3$ , then

$\left| {\widehat a + \widehat b} \right|$ may be:-

0%
$1$
0%
${{\sqrt 3 } \over 2}$
0%
Either (1) and (2)
0%
None of these

Explanation

$\left| \overset { \wedge }{ a } -\overset { \wedge }{ b } \right| =\sqrt { 3 }$

$Squaring\quad both\quad sides$

$\implies\quad { \left| \overset { \wedge }{ a } \right| }^{ 2 }+{ \left| \overset { \wedge }{ b } \right| }^{ 2 }-2(\overset { \wedge }{ a } .\overset { \wedge }{ b } )={ (\sqrt { 3 } ) }^{ 2 }$

$\because \overset { \wedge }{ a } \& \overset { \wedge }{ b } \quad represent\quad unit\quad vectors.$

$\implies\quad 1+1-2(\overset { \wedge }{ a } .\overset { \wedge }{ b } )=3$

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

$\implies\quad 2\overset { \wedge }{ a } .\overset { \wedge }{ b } =-1\quad -(i)$

$\left| \overset { \wedge }{ a } +\overset { \wedge }{ b } \right| =\sqrt { { \left| \overset { \wedge }{ a } \right| }^{ 2 }+{ \left| \overset { \wedge }{ b } \right| }^{ 2 }+2\overset { \wedge }{ a } .\overset { \wedge }{ b } }$

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

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

$\left| \overrightarrow { a } \right| =7,\ \left| \overrightarrow { b } \right| =11,\ \left| \overrightarrow { a } +\overrightarrow { b } \right| = 10\sqrt 3$ , then

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

0%
10
0%
$\sqrt 10$
0%
$2\sqrt 10$
0%
20

Explanation

Consider

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

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

Using the values given

$\Rightarrow {\left|\overrightarrow{a}-\overrightarrow{b}\right|}^{2}+{\left(10\sqrt{3}\right)}^{2}=2{7}^{2}+2{11}^{2}$

$\Rightarrow {\left|\overrightarrow{a}-\overrightarrow{b}\right|}^{2}+300=98+248=340$

$\Rightarrow {\left|\overrightarrow{a}-\overrightarrow{b}\right|}^{2}=340-300=40$

$\Rightarrow \left|\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{4\times 10}$

$\Rightarrow \left|\overrightarrow{a}-\overrightarrow{b}\right|=2\sqrt{10}$

For

$A(1, -2, 4), B(5, -1, 7), C(3, 6, -2), D(4, 5, -1)$ , the projection of

$\overline {AB}$ on

$\overline {CD}$ is ________.

0%
$(2\sqrt {3}, -2\sqrt {3}, 2\sqrt {3})$
0%
$\dfrac {3}{13} (4, 1, 3)$
0%
$(1, -1, 1)$
0%
$(2, -2, 2)$

Explanation

$\overline{AB}=\overline{B}-\overline{A}$

$A(1, -2, 4), B(5, -1, 7)$ ....given

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

$\overline{B}=5\overline i-\overline j +7\overline k$

$\therefore \overline{AB}=(5-1)\overline i+(-1+2)\overline k+(7-4)\overline k$

$\therefore \overline {AB}=4\overline i+\overline j+3\overline k$

$\overline {AB} = (4, 1, 3)$

$C(3, 6, -2), D(4, 5, -1)$

$\overline{C}=3\overline i+6\overline j -2\overline k$ and

$\overline{D}=4\overline i+5\overline j -\overline k$

$\therefore \overline{CD}=(4-3)\overline i+(5-6)\overline k+(-1+2)\overline k$

$\therefore \overline {CD}=\overline i+\overline j-\overline k$

$\overline{CD} = (1, -1, 1)$

Now projection of

$\overline {AB}$ on

$\overline CD =$

$\left (\dfrac {\overline {AB} \cdot \overline {CD}}{|CD|}\right ) \cdot \hat {CD}$

$= \left (\dfrac {4 - 1 + 3}{\sqrt {4}}\right ) \cdot \dfrac {(1, -1, 1)}{\sqrt {3}}$

$= (2, -2, 2)$ .

In Polygon Law of Vector Addition, the head of first vector is joined to the tail of last vector.

0%
True
0%
False

$\bar{a}$ is unit vector, then

$|\bar{a}\times \hat{i}|^2+|\bar{a}\times \hat{j}|^2+|\bar{a}\times \hat{k}|^2=$ _____________.

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

Explanation

Let

$x, y$ and

$z$ be the direction angles of the vector

$\overrightarrow{a}$

Then,

$|\overrightarrow{a} \times \widehat{i}| =|\overrightarrow{a}||\widehat{i}|sin(x)=sinx$

Similarly,

$|\overrightarrow{a} \times \widehat{j}| =siny$

$|\overrightarrow{a} \times \widehat{k}| =sinz$

Therefore,

$|\overrightarrow{a} \times \widehat{i}|^2+|\overrightarrow{a} \times \widehat{j}|^2+|\overrightarrow{a} \times \widehat{k}|^2=sin^2x+sin^2y+sin^2z$

$=(1-cos^2x)+(1-cos^2y)+(1-cos^2z)$

$=3-(cos^2x+cos^2y+cos^2z)$

$=3-1$

$=2$

Hence, the answer is option (A).

$\vec { a }$ and

$\vec { b }$ are non-zero non-collinear vectors, then

$\left[ \vec { a } \quad \vec { b } \quad \hat { i } \right] \hat { i } +\left[ \vec { a } \quad \vec { b } \quad \hat { j } \right] \hat { j } +\left[ \vec { a } \quad \vec { b } \quad \hat { k } \right] \hat { k }$ is equal to

0%
$\vec { a } +\vec { b }$
0%
$\vec { a } \times \vec { b }$
0%
$\vec { a } -\vec { b }$
0%
$\vec { b } \times \vec { a }$

Explanation

Let

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

$\vec{b} = b_{1} \hat{i}+ b_2 \hat{j} + b_3 \hat{k}$

$[\vec{a}\,\vec{b} \,\vec{c}]. \hat{i}+ [\vec{a}\, \vec{b}] \hat{j} + [\vec{a}\,\vec{b}\, \hat{k}] \hat{k}$

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

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

1080076_793336_ans_4acb916a451d45199890598f7774e473.jpg

The vector

$z = 3 - 4i$ is turned anticlockwise through an angle of

$180^{\circ}$ and stretched

$\dfrac{5}{2}$ times. The complex number corresponding to the newly obtained vector is ....

0%
$-\dfrac{15}{2}-10i$
0%
$-\dfrac{15}{2}+10i$
0%
$\dfrac{15}{2}+10i$
0%
$\dfrac{15}{2}-10i$

Explanation

$z=3-4i$ lie in fourth quadrant in complex plane, after turned anticlockwise through

$180^0$ this will lie in II quadrant, therefore, the number will be

$-3+4i$ now after stretching it

$2.5$ times i.e., multiplying by

$2.5$ , the required complex number will be

$\Rightarrow 2.5\times (-3+4i)$

$\Rightarrow -\dfrac{15}{2}+10i$

Hence, this is the answer.

The set of values of

$c$ for which the angle between the vectors

$cx\hat{i}-6\hat{j}+3\hat{k}$ and

$x\hat{i}-2\hat{j}+2cx\hat{k}$ is acute for every

$x\in R$ is

0%
$(0,4/3)$
0%
$[0,2/3]$
0%
$(11/9,4/3)$
0%
$[0,1/3)$

Explanation

$cx^{2}+12+6cx>0\implies c(3c-4)<0\implies c\in(0,\dfrac{4}{3})$

The vector

$\vec a + \vec b,\vec a - k\vec b$ where

$k$ scalar are collinear, for

0%
$k=0$
0%
$k=-1$
0%
$k=1$
0%
$k=2$

Explanation

Say

$\vec{A}=\vec{a}+\vec{b}, \vec{B}=\vec{a}-k\vec{b}$

If two vectors are collinear then

$\vec{B}=\lambda\vec{A}$ ......(1)

Now,

$\vec{B}=1(\vec{a}-k\vec{b})$

$\lambda=1$

From equation (1), we get

$\vec{B}=1\times \vec{A}$

$\vec{a}+\vec{b}=\vec{a}-k\vec{b}$

$k=-1$

Hence for

$k=-1$ ,

$\vec{A}$ and

$\vec{B}$ are collinear.

A line passes through the points whose position vectors

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

$\hat { i } -3\hat { j } +\hat { k }$ . Then the position vector of a point on it at a unit distance from the first point is

0%
$\dfrac { 1 }{ 5 } \left( 5\hat { i } +\hat { j } -7\hat { k } \right)$
0%
$\dfrac { 1 }{ 5 } \left( 5\hat { i } +9\hat { j } -13\hat { k } \right)$
0%
$\left( \hat { i } -4\hat { j } +3\hat { k } \right)$
0%
$\left( \hat { i } +4\hat { j } +3\hat { k } \right)$

Explanation

Let points of the position vector

$\hat i + \hat j - 2\hat k$ and

$\hat i - 3\hat j + \hat k$ are

$A(1,1-2)$ and

$(1,-3,1)$

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

Required points which divide

$AB$ in ratio

$1:4$

The required position vector of the point

$=\dfrac{1 \times (\vec i -3 \vec j + \vec k)+4(\vec i+\vec j-2 \vec k)}{1+4}=\dfrac{1}{5}(5 \vec i + \vec j -7 \vec k)$

The position vector of point F, is?

0%
$\dfrac{{3\left| {\overline a } \right|\,\left| {\overline c } \right|}}{{3\left| {\overline c } \right| + 2\left| {\overline a } \right|}}\,\,\,\left( {\dfrac{{\overrightarrow a }}{{\left| {\overline a } \right|}} + \dfrac{{\overrightarrow c }}{{\left| {\overline c } \right|}}} \right)$
0%
$\vec{a}+\left|\dfrac{\vec{a}}{\vec{c}}\right|\vec{c}$
0%
$\vec{a}+\dfrac{2|\vec{a}|}{|\vec{c}|}\vec{c}$
0%
$\vec{a}-\left|\dfrac{\vec{a}}{\vec{c}}\right|\vec{c}$

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}$

$\bar{a}=2\bar{i}+\bar{j}+\bar{k}, \bar{b}=\bar{i}+5\bar{j}, \bar{c}=4\bar{i}+4\bar{j}-2\bar{k}$ then the length of the projection of

$(3\bar{a}-2\bar{b})$ in the direction of

$\bar{c}$ .

0%
$3$
0%
$-3$
0%
$33$
0%
$-33$

Explanation

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

$\implies 4\hat{i} - 7\hat{j} +3\hat{k}$

Scalar projection =

$\cfrac{(3\vec{a} -2\vec{b}).\vec{c}}{|\vec{c}|}$

$\implies \cfrac{16-28-6}{\sqrt{16+16+4}}$

$\implies \cfrac{18}{\sqrt{36}}$

$\implies 3$

In a triangle ABC, if

$2\vec { AC } =3\vec { CB }$ , then

$2\vec { OA } +3\vec { OB }$ equals ?

0%
$5\vec { OC }$
0%
$-\vec { OC }$
0%
$\vec { OC }$
0%
None of these

Explanation

$\begin{array}{l} Given\, in\, \, \, \Delta ABC, \\ \, \, \, \, \, \, \, \, \, \, \, \, 2\, \overrightarrow { AB } =3\overrightarrow { CB, } \\ \, \, \, \, \, \, \, \, \, \, \, \, 2\overrightarrow { OA } =3\overrightarrow { OB } , \\ We\, know\, that, \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { AC } =\overrightarrow { OC } \, -\overrightarrow { OA } \\ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \overrightarrow { CB } =\overrightarrow { OB } -\overrightarrow { OC } \\ then, \\ \, \, \, 2(\overrightarrow { OC } -\overrightarrow { OA } )=3(\overrightarrow { OB } -\overrightarrow { OC } ) \\ 2\, \overrightarrow { OC } +2\overrightarrow { \, OA } =3\overrightarrow { OB } -3\overrightarrow { OC } \\ 2\, \, \overrightarrow { OC } -3\, \overrightarrow { \, OC } =3\, \overrightarrow { \, OB } +2\overrightarrow { OA } \\ 5\, \overrightarrow { OC } =\, \, 2\overrightarrow { OA } \, +3\, \overrightarrow { \, OB } \end{array}$

$\,So,\,\,\,\,2\overrightarrow {OA} \, + 3\,\overrightarrow {\,OB} = 5\,\overrightarrow {OC}$

$|\overrightarrow{a}| = 5, |\overrightarrow{a} - \overrightarrow{b}|=8$ and

$|\overrightarrow{a} + \overrightarrow{b}| = 10$ , then

$|\overrightarrow{b}|$ is equal to:

0%
$1$
0%
$\sqrt{57}$
0%
$3$
0%
$57$

Explanation

Given,

$\left| {\overline a } \right| = 5\,$

$\,\left| {\overline a - \overline b } \right| = 8\, \, equation - - - (1$

$\,\left| {\overline a + \overline b } \right| = 10\,\, equation - - - (2$

Squaring and adding both equation (1) & (2)

$\underline { \begin{array}{l} { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }-2\left| a \right| \left| b \right| \cos \theta =64 \\ { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }+2\left| a \right| \left| b \right| \cos \theta =100 \end{array} }$

$\begin{array}{l} 2\left( { { { \left| a \right| }^{ 2 } }+{ { \left| b \right| }^{ 2 } } } \right) =164 \\ { \left| a \right| ^{ 2 } }+{ \left| b \right| ^{ 2 } }=82 \\ we\, have, \\ \left| a \right| =5 \\ then, \\ { \left( 5 \right) ^{ 2 } }+{ \left| b \right| ^{ 2 } }=82 \\ { \left| b \right| ^{ 2 } }=82-25 \\ { \left| b \right| ^{ 2 } }=57 \\ \left| b \right| =\sqrt { 57 } \end{array}$

So the correct Ans is option B.

If the vector

$OP$ in

$XY$ plane whose magnitude is

$\sqrt3$ makes an angle

$60^o$ with

$Y-$ axis, the length of the component of the vector in direction of

$X-$ axis is :

0%
$1$
0%
$\sqrt3$
0%
$\dfrac {1}{2}$
0%
$\dfrac {3}{2}$

Explanation

Vector

$V=x \uparrow + y \uparrow$

$\tan 30^0=\dfrac{y}{x} \Rightarrow \dfrac{1}{\sqrt{3}}\dfrac{y}{x} \Rightarrow \boxed{x=\sqrt{3}y}$

given

$|v|=\sqrt{3}$

$\Rightarrow \sqrt{x^2+y^2}=\sqrt{3}$

$\Rightarrow \sqrt{3y^2+y^2}=\sqrt{3}$ (using (1)

$\Rightarrow 4y^2=3$

$\Rightarrow \boxed{y=\dfrac{\sqrt{3}}{2}}$

$x=\sqrt{3}y=\sqrt{3}\times \dfrac{\sqrt{3}}{2}$

$\boxed{x=\dfrac{3}{2}}$

$\therefore$ component of vector in

$x-axis$ direction is

$\dfrac{3}{2}$ .

1048735_1110116_ans_342bd4c8cf644983972b109ae980c294.png

$| \overline{a} | = 1 , | \overline{b} | = 2, | \overline{a} - \overline{b} |^2 + | \overline{a} + 2 \overline{b} |^2 = 20,$ then

$( \overline{a} , \overline{b} ) =$

0%
$\dfrac {\pi}{3}$
0%
$\dfrac {\pi}{4}$
0%
$\dfrac {\pi}{6}$
0%
$\dfrac {2 \pi}{3}$

Explanation

Given

$|\bar{a}|=1 , |\bar{b}|=2, |\bar{a}-\bar{b}|^{2}+|\bar{a}+2\bar{b}|^{2}=20$

$|\bar{a}|^{2}+|\bar{b}|^{2}-2|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})+|\bar{a}|^{2}+4|\bar{b}|^{2}+4|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})=20$

$\Rightarrow 2|\bar{a}|^{2}+5|\bar{b}|^{2}+2|\bar{a}||\bar{b}| \cos (\bar{a},\bar{b})=20$

$\Rightarrow 2(1)^{2}+5(2)^{2}+2\times 1\times 2\times \cos (\bar{a},\bar{b})=20$

$\Rightarrow 22+4 \cos (\bar{a},\bar{b})=20$

$\cos (\bar{a},\bar{b})=\dfrac{-1}{2}$

$\Rightarrow \pi -(\bar{a},\bar{b})=\dfrac{\pi }{3}$

$\Rightarrow (\bar{a},\bar{b})=(\pi -\dfrac{\pi }{3})=\dfrac{2\pi }{3}$

Which of the following expressions are meaningful?

0%
$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$
0%
$\left(\overrightarrow{u}.\overrightarrow{v}\right).\overrightarrow{w}$
0%
$\left(\overrightarrow{u}.\overrightarrow{v}\right)\overrightarrow{w}$
0%
$\overrightarrow{u}\times \left(\overrightarrow{v}.\overrightarrow{w}\right)$

Explanation

$\left( A \right)\,\,\overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$ is meaningful as

${\overrightarrow v \times \overrightarrow w }$ is a vector. and dot product of two vectors is meaningful.

$\left( B \right)\,\left( {\overrightarrow u \times \overrightarrow v } \right).\overrightarrow w$ is no;t meaningful as

${\overrightarrow u \times \overrightarrow v }$ is scalar and dot product of scalar & vector is not meaningful.

$\left( C \right)\,\left( {\overrightarrow u \times \overrightarrow v } \right).\overrightarrow w$ is meaningful as

${\overrightarrow u \times \overrightarrow v }$ is scalar and expression is of the form

$.\lambda \overrightarrow w$ .

$\left( D \right)\,\,\,\overrightarrow u \left( {\overrightarrow v \times \overrightarrow w } \right)$ is not meaningfulllll as

${\overrightarrow v \times \overrightarrow w }$ is scalar and cross product of vector & scalar is not defined.

Then, We get

Option

$A$ &

$C$ is correct answer.

For three vectors

$\overrightarrow{u},\overrightarrow{v},\overrightarrow{w}$ which of the following expressions is not equal to any of remaining is

0%
$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$
0%
$\left(\overrightarrow{v}\times \overrightarrow{w}\right).\overrightarrow{u}$
0%
$\overrightarrow{v}.\left(\overrightarrow{u}\times \overrightarrow{w}\right)$
0%
$\left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$

Explanation

In a scalar triple product dot and cross can be interchanged.

$\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right) =\left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$
the vectors can be changed cyclically.

$\therefore \left(\overrightarrow{v}\times \overrightarrow{w}\right).\overrightarrow{u}= \left(\overrightarrow{u}\times \overrightarrow{v}\right).\overrightarrow{w}$

$\Rightarrow \overrightarrow{v}.\left(\overrightarrow{u}\times \overrightarrow{w}\right) =-\overrightarrow{u}.\left(\overrightarrow{v}\times \overrightarrow{w}\right)$

The ratio in which the line joining

$(2,-4,3)$ and

$(-4,5,-6)$ is divided by the plane

$3x+2y+z-4=0$ is

0%
$2 : 1$
0%
$4 : 3$
0%
$1 : 4$
0%
$2 : 3$

Explanation

We have given points as shown in the figure

$Eq{u^n}\,of\,line\,\,AB:\,\frac{{x - 2}}{6} = \frac{{y + 4}}{{ - 9}} = \frac{{z - 3}}{9}$

Now

Points

$P$ is

$\left( {6r + 2,\, - 9r - 4,\,9r + 3} \right)$

$3\left( {6r + 2} \right) + 2\left( { - 9r - 4} \right) + 9r + 3 = 4$

$9r + 1 = 4$

$\therefore r = \dfrac{1}{3}$

So,

$P\left( {4, - 7,6} \right)$

Now,

$\frac{{AP}}{{BP}} = \sqrt {\frac{{{{\left( { - 2} \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2}}}{{{{\left( 8 \right)}^2} + {{\left( {12} \right)}^2} + {{\left( 12 \right)}^2}}}}$

$= \sqrt {\frac{{22}}{{352}}}$

$= \sqrt {\frac{{11}}{{176}}}$

$= \sqrt {\frac{1}{{16}}} = \frac{1}{4}$

$P$ divides externally in the ratio of

$1:4$ .

Hence, The option

$C$ is the correct answer.

1205109_1184656_ans_e420a5f7aef340c5bd21a1f278bcaefd.JPG

State true or false.

The vectors

$\vec { a } = - 4 \hat { i } - \hat { j } , \vec { b } = \hat { i } - 4 \hat { j } \text { and } \vec { c } = 3 \hat { i } + 5 \hat { j }$ form a right angled-triangle.

0%
True
0%
False

Explanation

$\vec{AB}=\vec{b}-\vec{a}=\hat{i}-4\hat{j}+4\hat{i}+\hat{j}=5\hat{i}-3\hat{j}$

$\vec{BC}=\vec{c}-\vec{b}=3\hat{i}+5\hat{j}-\hat{i}+4\hat{j}=2\hat{i}+9\hat{j}$

$\vec{CA}=\vec{a}-\vec{c}=-4\hat{i}-\hat{j}-3\hat{i}-5\hat{j}=-7\hat{i}-6\hat{j}$

Now,

$\vec{AB}.\vec{CA}=\left(5\hat{i}-3\hat{j}\right)\left(-7\hat{i}-6\hat{j}\right)=-35+18\neq 0$

Now,

$\vec{BC}.\vec{CA}=\left(2\hat{i}+9\hat{j}\right)\left(-7\hat{i}-6\hat{j}\right)=-14-54\neq 0$

Thus,

$\vec{BC}$ and

$\vec{CA}$ are not perpendicular to each other.

Hence,

$ABC$ is not a right angled triangle.

A vector

$\overrightarrow { A }$ points vertically downward(south)and

$\overrightarrow { B }$ points towards east, then the vector product

$\overrightarrow { A } \times \overrightarrow { B }$ is:

0%
Along west
0%
Along east
0%
Zero
0%
Outwards or inwards

Explanation

$\vec{A}\times \vec{B}$ is perpendicular to the plane of paper and acting outward. It is denoted by

$\otimes$

1233850_1187059_ans_093aba5e1fa34db68873a6396371d675.png

$\bar{a}$ and

$\bar{b}$ are the position vectors of A and B respectively. Points P and Q divide AB, internally and externally in the same ratio 2:1 , then PQ is equal to

0%
$\frac{3}{2}(\bar{b}-\bar{a})$
0%
$\frac{4}{3}(\bar{a}-\bar{b})$
0%
$\frac{5}{6}(\bar{b}-\bar{a})$
0%
$\frac{4}{3}(\bar{b}-\bar{a})$

Explanation

Position vector of P will be

$\vec{OP}=\left(\dfrac{2\bar{b}+\bar{a}}{3}\right)$ (P divides AB internally in the ratio

$2:1$ )

Position vector of Q will be

$\vec{OQ}=\left(\dfrac{2\bar{b}-\bar{a}}{2-1}\right)=\left(\dfrac{2\bar{b}-\bar{a}}{1}\right)$

(Q divides AB externally in the ratio

$2:1$ )

$\vec{PQ}=\vec{OQ}-\vec{OP}=\left(2\vec{b}-\vec{a}\right)-\left(\dfrac{2\bar{b}+\bar{a}}{3}\right)=\dfrac{4\vec{b}}{3}-\dfrac{4\vec{a}}{3}$

$=\dfrac{4}{3}(\vec{b}-\vec{a})$ .

1195569_1195681_ans_6e9713bb8c944ac481ccd799b7dc1b6d.jpg

Two vector

$A$ and

$B$ have equal magnitudes. Then the vector

$A+B$ is perpendicular to

0%
$A\times{B}$
0%
$A-B$
0%
$3A-3B$
0%
All of these

Explanation

$\vec{A}\times \vec{B}$ is a vector perpendicular to plane

$\vec{A}+\vec{B}$ and hence perpendicular to

$\vec{A}+\vec{B}$

The work done by the force

$\vec { F } = 2 \hat { i } - \hat { j } - \hat { \mathbf { k } }$ in moung an object along the vector

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

0%
$- 9$ units
0%
$9$ units
0%
$-15$ units
0%
None of these

Explanation

$\vec { F } = 2 \hat { i } - \hat { j } - \hat { k }$

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

$Work=\vec { F }.\vec{r}$

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

$Work=6-2+5$

$Work=11-2=9units$

Hence the correct option is (B).

Let

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

$\vec{b}$ is a vector such that

$\vec{a}.\vec{b}=0$ and

$\vec{a}\times \vec{b}=0$ . Then which of following is correct?

0%
$\vec{b}=0$
0%
$\vec{b} \perp\vec{a}$
0%
$\vec{b}$ is non-zero vector
0%
$\vec{b}$ is parallel to $\vec{a}$

Explanation

$\begin{array}{l} \vec { a } .\vec { b } =0 \\ \vec { a } \times \vec { b } =0 \\ \left| { \vec { a } } \right| \left| { \vec { b } } \right| \cos \theta =0 \\ \left| { \vec { b } } \right| \cos \theta =0 \\ or, \\ \vec { a } =\hat { i } +\hat { j } +\hat { k } \\ \left| { \vec { a } } \right| =\sqrt { 3 } \\ \left| { \vec { a } } \right| \left| { \vec { b } } \right| \sin \theta =0 \\ \left| { \vec { b } } \right| \sin \theta =0 \\ N0w, \\ \left| { \vec { b } } \right| =0 \\ \vec { b } =0 \\ Hence, \\ option\, \, A\, \, is\, \, correct\, \, answer. \end{array}$

Given that

$\vec{ A } \times \vec{ B } =\vec{ B } \times \vec { C } =\vec { 0 }$ if

$\vec{ A } \vec { B } \vec { C }$ are not null vectors, Find the value of

$\vec{ A } \times \vec{ C }$

0%
$\vec { A } \times \vec { B }$
0%
$\vec { 0 }$
0%
$\vec{ C } \times \vec { B }$
0%
$\vec { C } \times \vec { A }$

Explanation

We have,

$\begin{array}{l} \vec { A } \times \vec { B } =\vec { B } \times \vec { C } =\vec { 0 } \\ \vec { A } \times \vec { B } +\vec { C } \times \vec { B } =\vec { 0 } \\ \left( { \vec { A } +\vec { C } } \right) \times \vec { B } =\vec { 0 } \\ \vec { A } +\vec { C } =\lambda \vec { B } \\ \vec { A } \times \vec { C } +0=\lambda \left( { \vec { B } \times \vec { C } } \right) \\ \vec { A } \times \vec { C } =\vec { 0 } \\ Hence,\, the\, option\, B\, is\, correct.\, \end{array}$

$\vec{r} = \vec{x}\hat{i}+\vec{y}\hat{j}$ is the equation of:

0%
$yoz$ plane
0%
a straight line joining the points $\vec{i}$ and $\vec{j}$
0%
$zox$ plane
0%
$xoy$ plane

Explanation

A vector is defined by its magnitude and its orientation with respect to a set of coordinates. It is often useful in analyzing vectors to break them into their component parts. For two-dimensional vectors, these components are horizontal and vertical.

Vector can always be represented in two dimension and three dimension. In two dimension, vector is represented in the form of

$x\hat{i}+y\hat{j}$ . In three dimension, it is represented by

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

Thus,

$\overrightarrow{r}=x\hat{i}+y\hat{j}$ is vector in

$xy$ plane.

The unit vector in the direction of

$\overrightarrow{a}$ is

0%
$\dfrac{\vec{a}}{|\vec{a}|}$
0%
$\vec{a}|\vec{a}|$
0%
$a^2$
0%
$\hat{i}$

Explanation

Consider the given vector

$\vec a$ .

Unit vector

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

Hence, this is the answer.

$\vec{a}$ be the position vector whose tip is (5,-3), find the coordinates of a point B such that

$\vec{AB} = \vec{a},$ the coordinates of A being (4,-1).

0%
(9, -4)
0%
(-9, -4)
0%
(9, 4)
0%
none of these

Explanation

$\vec{a}$ is position vector of

$(5, -3)$

$\vec{a}=5\hat{i}-3\hat{j}$

Coordinates of

$A=(4, -1)$

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

$\vec{AB}=\vec{a}=5\hat{i}-3\hat{j}$

$\vec{OB}-\vec{OA}=5\hat{i}-3\hat{j}$

$\Rightarrow \vec{OB}=5\hat{i}-3\hat{j}+\vec{OA}$

$=5\hat{i}-3\hat{j}+4\hat{i}-3\hat{j}$

$=a\hat{i}-4\hat{j}$

$\vec{OB}=9\hat{i}-4\hat{j}$

Coordinates of point B

$=(9, -4)$ .

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 2 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

CBSE Questions for Class 12 Commerce Maths Vector Algebra Quiz 2 - MCQExams.com

0:0:2

Practice Class 12 Commerce Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.