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Multiplication of vectors - class-XI

Description: multiplication of vectors
Number of Questions: 48
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Tags: systems of particles and rotational motion basic mathematical concepts work, energy and power mathematical methods physics motion in a plane
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Three vectors satisfy the relation $\displaystyle \overrightarrow { A } .\overrightarrow { B } =0$ and $\displaystyle \overrightarrow { A } .\overrightarrow { C } =0$, then $\displaystyle \overrightarrow { A } $ is parallel to:

  1. $\displaystyle \overrightarrow { C } $

  2. $\displaystyle \overrightarrow { B } $

  3. $\displaystyle \overrightarrow { B } \times \overrightarrow { C } $

  4. $\displaystyle \overrightarrow { B } .\overrightarrow { C } $


Correct Option: C
Explanation:

Using : $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (\vec{A} . \vec{C})  - \vec{C} (\vec{A}.\vec{B})$

Given : $\vec{A}.\vec{C}  = 0$  and  $\vec{A}.\vec{B}  = 0$ 
$\therefore$         $\vec{A} \times (\vec{B} \times \vec{C})  = \vec{B}  (0)  - \vec{C} ( 0)   = 0$
Thus, $\vec{A}$ is parallel to  $(\vec{B} \times \vec{C})$.

Vectors $\bar { A }$, $\bar { B }$ and $\bar { C }$ are such that $ \bar { A } \bullet \bar { B } =0$ and $ \bar { A } \bullet \bar { C } =0$. Then the vector parallel to $\bar { A }$ is

  1. $\bar { A } \times \bar { B }$

  2. $\bar { A }+ \bar { B }$

  3. $\bar { B} \times \bar { C }$

  4. $\bar { B}$ and $\bar { B}$


Correct Option: C

The vector $\overrightarrow { B } = 5\hat { i } + 2\hat {  j}-S \hat {  k} $ is perpendicular to the vector $\overrightarrow {  A}= 3\hat {  i} +\hat { j } + 2\hat {k  } $ if S=

  1. $1$

  2. $4.7$

  3. $6.3$

  4. $10.5$


Correct Option: D

$\vec {A}$ and $\vec {B}$ are vectors expressed as $\vec {A} =2\hat {i}+\hat {j}$ and $\vec {B} =\hat {i}-\hat {j}$. Unit vector perpendicular to $\vec {A}$ and $\vec {B}$ is

  1. $\dfrac{\hat {i}-\hat {j}+\hat {k}}{\sqrt{3}}$

  2. $\dfrac{\hat {i}+\hat {j}-\hat {k}}{\sqrt{3}}$

  3. $\dfrac{\hat {i}+\hat {j}+\hat {k}}{\sqrt{3}}$

  4. $\hat {k}$


Correct Option: D

Two particles are simultaneously projected in opposite direction horizontally from a given point in space where gravity g is uniform.If $u _1 and u _2$ be their initial speeds, then the time t after which their velocities are mutually perpendicular is given by

  1. $\dfrac{\sqrt{u _1 u _2}}{g}$

  2. $\dfrac{\sqrt{u^2 _1 + u^2 _2}}{g}$

  3. $\dfrac{\sqrt{u _1(u _1 + u _2)}}{g}$

  4. $\dfrac{\sqrt{u _2(u _1 + u _2)}}{g}$


Correct Option: A
Explanation:

${v _1} = {u _1}\hat i - gt\hat j$

${v _2} =  - {u _2}\hat i - gt\hat j$

$for\,\,\,\,\,\,{v _1} \bot {v _2}$

${{\bar v} _1}.{{\bar v} _2} = 0$

$({u _1}\hat i - gt\hat j).( - {u _2}\hat i - gt\hat j) = 0$

$ - {u _1}{u _2} + {g^2}{t^2} = 0$

${g^2}{t^2} = {u _1}{u _2}$

$t = {{\sqrt {{u _1}{u _2}} } \over g}$

If the magnitude of two vectors are $8$ unit and $5$ and their scalar product is zero, the angle between the two vectors is

  1. Zero

  2. ${ 30 }^{ o }$

  3. ${ 60 }^{ o }$

  4. ${ 90 }^{ o }$


Correct Option: D

If $\overrightarrow { A } +\overrightarrow { B } =\overrightarrow { R }$ and $\left( \overrightarrow { A } +2\overrightarrow { B }  \right)$ is perpendicular to $\overrightarrow { A }$, then

  1. $R=2B$

  2. $R=B/2$

  3. $R=B$

  4. $R=B/\sqrt { 2 }$


Correct Option: A

The angle between the vectors $(\overline{\mathrm{A}}$ x $\overline{\mathrm{B}})$ and $(\overline{\mathrm{B}}\times\overline{\mathrm{A}})$ is:

  1. $0^{0}$

  2. $180^{0}$

  3. $45^{0}$

  4. $90^{0}$


Correct Option: B
Explanation:

($A$$\times $$B$) = - ($B$$\times $$A$) which are equal and opposite in direction.
Hence it will have angle in between $180^0$

In a clockwise system, which of the following is true?

  1. $\hat { j } \times \hat { k } =\hat { i } $

  2. $\hat { i }\ .\hat { i } =0$

  3. $\hat { j } \times \hat { j } =1$

  4. $\hat { k } \ .\hat { i } =1$


Correct Option: A
Explanation:

In clockwise system:
$\hat{i} \times\hat{ j} = \hat{k}$
$\hat{ j} \times \hat{k} = \hat{i}$
$\hat{k} \times \hat{i} = \hat{j}$

The value of $ (\bar { A } +\bar { B } )\times (\bar { A } -\bar { B } )$ is 

  1. $0$

  2. ${A}^{2}-{B}^{2}$

  3. $\bar { B } \times \bar { A }$

  4. $2(\bar { B } \times \bar { A })$


Correct Option: D
Explanation:
$(\overrightarrow A+\overrightarrow B)\times(\overrightarrow A-\overrightarrow B)=\overrightarrow A\times \overrightarrow A-\overrightarrow A\times \overrightarrow B+\overrightarrow B\times \overrightarrow A-\overrightarrow B\times \overrightarrow B=2(\overrightarrow B\times \overrightarrow A)$


The velocity of a particle is $\vec{v}=6\hat{i}+2\hat{j}-2\hat{k}.$ The component of the velocity parallel to vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ is :- 

  1. $6\hat{i}+2\hat{j}+2\hat{k}$

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

  3. $\hat{i}+\hat{j}+\hat{k}$

  4. $6\hat{i}+2\hat{j}-2\hat{k}$


Correct Option: B
Explanation:

$\vec{v}=6\hat{i}+2\hat{j}-2\hat{k}$


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

component $\vec {a  }$ is $(\vec { v } \cdot\hat {  a})$
$\vec {  v}.\hat{a}=\vec{v}.\dfrac{\vec{a}}{|a|}=\vec{v}.\dfrac{(\hat { i }+\hat { j }+\hat { k })}{\sqrt{1+1+1}}$

$\Rightarrow \vec{v}.\hat{a}=\dfrac{(6\hat { i }+2\hat { j }-2\hat { k }).(\hat { i }+\hat { j }+\hat { k })}{\sqrt{3}}$

$\Rightarrow \vec{v}.\hat{a}=\dfrac{6+2-2}{\sqrt{3}}=2\sqrt{3}$

$\Rightarrow$ component $=(\vec{v}\hat{j})\hat{j}=2\sqrt{3}\dfrac{(\hat {  i}+\hat { j }+\hat { k })}{\sqrt{3}}$

$\boxed{component=2\hat { i }+2\hat { i }+2\hat { k }}$

If $\overline {A} \times\overline {B} =\overline {C}$ which of the following statement is not correct?

  1. $\overline {C} \top \overline {A}$

  2. $\overline {C} \top \overline {B}$

  3. $\overline {C} \top \overline {A} \times \overline {B}$

  4. $\overline {C} \top \overline {A} + \overline {B}$


Correct Option: C
Explanation:

The cross product $\vec{A}\times\vec{B}$  is defined as a vector $\vec{C}$ that is perpendicular (orthogonal) to both A and B, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

In C option $\vec{C}$ is not perpendicular to  $\vec{A}\times\vec{B}$  
Hence C option is correct 

Two forces of magnitude 20N and 20N act along the adjacent sides of the parallelogram and the magnitude of the resultant force of these two forces is $20\sqrt{3}$. Then the angle between these forces is:

  1. $30^0$

  2. $45^0$

  3. $90^0$

  4. $60^0$


Correct Option: A

What is the unit vector perpendicular to the following vectors $ 2\hat{i} + 2\hat{j}- k$ and $6\hat{i}-3\hat{j}+2k$ 

  1. $\frac{\hat{i}+10\hat{j}-18k}{5\sqrt{17}}$

  2. $\frac{\hat{i}-10\hat{j}+18k}{5\sqrt{17}}$

  3. $\frac{\hat{i}-10\hat{j}-18k}{5\sqrt{17}}$

  4. $\frac{\hat{i}+10\hat{j}+18k}{5\sqrt{17}}$


Correct Option: C

$(\overline{A} + \overline{B} )\times ( \overline{A} - \overline{B} )$ is

  1. $(\overline{A} ^2- \overline{B} ^2)$

  2. $2\overline{A} \overline{B} $

  3. $(\overline{A} \times \overline{B} )$

  4. $( \overline{B} \times \overline{A} $


Correct Option: B
Explanation:

$(\bar A+\bar B)\times (\bar A-\bar B)$

$=\bar A \bar A-\bar A\bar B+\bar A\bar B-\bar B \bar B$
$=2\bar A\bar B$ .

The vectors $\vec{A}=4\hat{i}+3\hat{j}+\hat{k}$ and $\vec{B}=12\hat{i}+9\hat{j}+3\hat{k}$ are parallel to each other.

  1. True

  2. False


Correct Option: A

The momentum of a particle is $\vec { P } =\vec { A } +\vec { B } { t }^{ 2 }$, where $\vec { A }$ and $\vec { B }$ are constant perpendicular vectors. The force acting on the particle when its acceleration is at ${45}^{o}$ with its velocity is

  1. $2\sqrt \frac {A}{B}\vec {B}$

  2. $2\vec {B}$

  3. $zero$

  4. $2 \vec A$


Correct Option: A

Find the projection of $ \vec A =2\hat { i } -\hat { j } +\hat { k } \quad on\quad \vec  B  =\quad \hat { i } -2\hat { j } +\hat { k }  $

  1. $ \frac { 5 }{ \sqrt { 6 } } $

  2. $ \frac { 7 }{ 10 } $

  3. $ \frac { 6 }{ \sqrt { 5 } } $

  4. $ \frac { 5 }{ \sqrt { 3 } }


Correct Option: A
Explanation:

Given that,

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

 $ \vec{B}=\hat{i}-2\hat{j}+\hat{k} $

Now, the projection  $\vec{A}$ on $\vec{B}$

  $ =\dfrac{\vec{A}\centerdot \vec{B}}{|\vec{B}|} $

 $ =\dfrac{5}{\sqrt{6}} $

Hence, this is the required solution

The resultant of the two vector is having magnitude 2 and 3 is 1. What is their cross product 

  1. $6$

  2. $3$

  3. $1$

  4. $0$


Correct Option: D

The vector of  magnitude 18 which is perpendicular to both vectors $4\hat i-\hat j+3\hat k \,and  -2\hat i+\hat j-2\hat k$  is

  1. $12\hat i+ 12 \hat j-6\hat k$

  2. $6\hat i-12\hat j-12\hat k$

  3. $12\hat i+6\hat j+12 \hat k$

  4. $-6\hat i+12\hat j+12\hat k$


Correct Option: A

The component of vector $2\ \hat {i}+3\hat {j}$ along vector $-\hat {j}+5\hat {i}$ is:

  1. $\dfrac{7}{\sqrt{13}}$

  2. $\dfrac{7}{\sqrt{26}}$

  3. $\dfrac{13}{\sqrt{13}}$

  4. $none\ of\ these$.


Correct Option: B

If $\vec {u},\vec {v}$ and $\vec {w}$ are three non-coplanar vectors, then
$(\vec {u}+\vec {v}-\vec {w}).(\vec {u}-\vec {v})\times (\vec {v}-\vec {w})$ equals

  1. $3\vec {u}.\vec {v} \times \vec {w}$

  2. $0$

  3. $\vec {u}.\vec {v}\times \vec {w}$

  4. $\vec {u}.\vec {w} \times \vec {v}$


Correct Option: B

Let $\vec {a}$ and $\vec {b}$ to two unit vectors. If the vectors $\vec {c}=\hat {a}+2\hat {b}$ and $\vec {d}=5\hat {a}-4\hat {b}$ are perpendicular to each other, then teh angle between $\vec {a}$ and $\vec {b}$ is

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

  2. $\dfrac {\pi}{4}$

  3. $\dfrac {\pi}{6}$

  4. $\dfrac {\pi}{2}$


Correct Option: D

Which of the following vector is perpendicular to the vector $A=2\hat{i}+3\hat{j}+4\hat{k}$?

  1. $\hat{i}+\hat{j}+\hat{k}$

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

  3. $\hat{i}-3\hat{j}+\hat{k}$

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


Correct Option: D

Which of the following vector is perpendicular to the vector $\vec { A } =\hat { 2i } +\hat { 3j } +\hat { 4k } $?

  1. $\hat { i } +\hat { j } +\hat { k } $

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

  3. $\hat { i } -\hat {3 j } +\hat { k } $

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


Correct Option: D

Find a vector $\vec {x}$ which is perpendicular to both $\vec {A}$ and $\vec {B}$ but has magnitude equal to that of $\vec {B}$. Vector $\vec {A}=3\hat{i}-2 \hat {j} +\hat {k}$ and $\vec {B}=4\hat{i}+3 \hat {j} -2\hat {k}$

  1. $\displaystyle \frac{1}{\sqrt{10}}(\hat{i}+10\hat{j}+17\hat{k})$

  2. $\displaystyle \frac{1}{\sqrt{10}}(\hat{i}-10\hat{j}+17\hat{k})$

  3. $\sqrt {\displaystyle \frac{29}{390}}(\hat{i}-10\hat{j}+17\hat{k})$

  4. $\sqrt {\displaystyle \frac{29}{390}}(\hat{i}+10\hat{j}+17\hat{k})$


Correct Option: D
Explanation:

$\vec {A} \times \vec {B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 1\ 4 & 3 & -2\end{vmatrix}$

$\hat {n}=\displaystyle \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\displaystyle \frac{\hat{i}+10\hat{j}+17\hat{k}}{\sqrt{390}}$

$\vec{x}=|\vec{B}|\hat{n}=\displaystyle \frac{\sqrt{29}(\hat{i}+10\hat{j}+17\hat{k})}{\sqrt{390}}$

Three vectors $\vec A, \vec B$ and $\vec C$ satisfy the relation $\vec {A}\cdot \vec {B}=0$ and $\vec{A}\cdot \vec{C}=0$. The vector $A$ is parallel to :

  1. $\vec {B}. \vec {C}$

  2. $\vec {B}$

  3. $\vec {C}$

  4. $\vec {B} \times \vec {C}$


Correct Option: D
Explanation:

$\vec A$ is perpendicular to $\vec B$ and $\vec A$ is perpendicular to $\vec C$. Thus $\vec A$ must lie along the direction of the cross product of $\vec B$ and $\vec C$.

Alternatively:
Given,
$\vec {A}.\vec {B}=0$
$\vec{A}.\vec{C}=0$
$ \Rightarrow \vec {A}.\vec {B} - \vec {A}.\vec {C}=\vec{A}( \vec{B} -\vec{C}) =0$
$ \Rightarrow \vec{A} \perp (\vec{B} -\vec{C})$
$\Rightarrow \vec{A} \parallel (\vec{B} \times \vec{C})$

A vector $\vec{A}$ is along +ve x-axis. Another vector $\vec{B}$ such that $\vec{A} \times \vec{B}=\vec{0}$ could be

  1. $4 \hat {j}$

  2. $-4 \hat {i}$

  3. $-(\hat {i}+\hat {j})$

  4. $(\hat {j}+\hat {k})$


Correct Option: B

If $\vec{A}=5 \hat {i}+7 \hat{j}-3 \hat {k}$ and $\vec{B}=15 \hat {i}+21 \hat{j}+a \hat {k}$ are parallel vectors then the value of $a$ is:

  1. -3

  2. 9

  3. -9

  4. 3


Correct Option: C
Explanation:

$\left( \vec { A } \times \vec { B }  \right) =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 5 & 7 & -3 \ 15 & 21 & a \end{matrix} \right| =0\quad or\quad 5a+45=0\quad or\quad or\quad 7a+63=0\quad \Rightarrow a=-9$

If $\vec{A}\times\vec{B}=\vec{C}$, then choose the incorrect option : [$\vec{A}$ and $\vec{B}$ are non zero vectors]

  1. $\vec{C}$ is prependicular to $(\vec{A} + \vec{B})$

  2. $\vec{C}$ is prependicular to $(\vec{A} - \vec{B})$

  3. $\vec{C}$ is prependicular to $(\vec{A} \times \vec{B})$

  4. $\vec{C}$ is prependicular to $\vec{A}$ and $\vec{B}$


Correct Option: C

If $\overrightarrow a  + b + \overrightarrow c  = 0$ The angle between $\overrightarrow a \,\,and\,\,\overrightarrow b \,,b\,and\,\overrightarrow c \,and\,{150^0}\,\,and\,\,{120^0}$ respectively.The the magnitude of vectors $\overrightarrow a ,\overrightarrow b \,\,and\,\,\overrightarrow c $ are in ratio of .

  1. $1:2:2$

  2. $1:2$:$\sqrt 3 $

  3. $\sqrt 3 $:2:1

  4. $2$:$\sqrt 3 $:1


Correct Option: B

If  $\vec { A } = 4 \vec { i } + 5 \vec { j } - 6 \vec { k }$  and  $\vec { B } = 2 \vec { i }  - 3 \vec { j } + 4 \vec { k }$  then  $( \vec { A } + \vec { B } ) \cdot (\vec { A } - \vec { B } )$  is

  1. $6$

  2. $48$

  3. $67$

  4. $13$


Correct Option: B

If the two given vectors $ 2 \hat i + 3 \hat j + 4 \hat k $ and $ 6 \hat i +  \alpha \hat j + \beta \hat k $ are parallel , the value of $ \alpha $ and $ \beta $ will be

  1. $9$ and $12$

  2. $3$ and $14$

  3. $6$ and $8$

  4. $4$ and $12$


Correct Option: A

If $\overrightarrow{A}=4\widehat{i}+6\widehat{j} $  and  $\overrightarrow{B}=2\widehat{i}+3\widehat{j}$ .Then :

  1. $\overrightarrow{A}.\overrightarrow{B} =29$

  2. $\overrightarrow{A}\times \overrightarrow{B}=0$

  3. $\dfrac{|\overrightarrow{A}|}{|\overrightarrow{B}|}=\dfrac{2}{1} $

  4. angles between $ \overrightarrow{A}$ and $\overrightarrow{B} $ is $ 30^{\circ}$


Correct Option: B
Explanation:
(A) $\vec A.\vec B=(4\hat i+6\hat j).(2\hat i+3\hat j)=8+18=26$

(B) $\overrightarrow{A} \times \overrightarrow{B}= \begin{vmatrix} \widehat{i}& \widehat{j} &\widehat{k}\\  4& 6  & 0\\ 2 & 3 & 0\end{vmatrix} = \widehat{i}(0-0)-\widehat{j}(0-0)+\widehat{k}(12-12)=0 $

(C) $|\vec A|=\sqrt{4^2+6^2}=7.21$
     $|\vec B|=\sqrt{2^2+3^2}=3.60$
     $\dfrac{|\vec A|}{|\vec B|}=2$

(D) Angle between $\vec A$ and $\vec B$ $= cos ^{-1}\dfrac{\vec A.\vec B}{|\vec A|.|\vec B|}=cos^{-1}1=0^0$

If $  \overrightarrow{A} \times \overrightarrow{B}=0,$ $  \overrightarrow{B} \times \overrightarrow{C}=0, $then $  \overrightarrow{A} \times \overrightarrow{C}= $

  1. $AC$

  2. $\dfrac{AB^2}{C} $

  3. $Zero$

  4. $None  of  these$


Correct Option: C
Explanation:

$  \overrightarrow{A} \times \overrightarrow{B}=0   \overrightarrow{A}  \ and\  \overrightarrow{B} $ are parallel.
$  \overrightarrow{B} \times \overrightarrow{C}=0  \Rightarrow  \overrightarrow{B}  \ and \  \overrightarrow{C} $ are parallel 
 $\Rightarrow \vec{A} \parallel \vec{C}$

Consider a vector $F=4\hat{i}-3\hat{j} $. Another vector which is perpendicular to $\vec F$ is:

  1. $ 4\hat{i}+3\hat{j}$

  2. $ 6\hat{i}$

  3. $7\hat{k} $

  4. $ 3\hat{i}-4\hat{j}$


Correct Option: C
Explanation:

The vector perpendicular to the $i$ & $j$ plane would be along the unit vector $k$.
Another vector in $i$ & $j$ plane can be perpendicular to $\vec{F}$
$(3\widehat{i} +4\widehat{j}) \perp (4\widehat{i} -3\widehat{j}) $
But,  from the options only $7\widehat{k} \perp (4\widehat{i} -3\widehat{j})$

Show that the vector is parallel to a vector $\displaystyle \vec{A}=\hat{i}-\hat{j}+2\hat{k}$ is parallel to a vector $\displaystyle \vec{B}=3\hat{i}-3\hat{j}+6\hat{k}.$

  1. $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

  2. $\displaystyle \frac{1}{4}$ times the magnitude of $\displaystyle \vec{B}.$

  3. $\displaystyle \frac{1}{2}$ times the magnitude of $\displaystyle \vec{B}.$

  4. None of these


Correct Option: A
Explanation:

A vector $\displaystyle \vec{A}$ is parallel to an another vector $\displaystyle \vec{B}$ if it can be written as
$\displaystyle \vec{A}= m\vec{B}$ where $m$ is a constant.
Here, $\displaystyle \vec{A}=\left ( \hat{i}-\hat{j}+2\hat{k} \right )=\frac{1}{3}\left ( 3\hat{i}-3\hat{j}+6\hat{k} \right )$
or $\displaystyle \vec{A}=\frac{1}{3}\vec{B}$
This implies that $\vec A || \displaystyle \vec{B}$ and magnitude of $\displaystyle \vec{A}$ is $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

If $\vec{a}=x _1\hat {i}+y _1\hat {j}$ and $\vec{b}=x _2\hat {i}+y _2\hat {j}$. The condition that would make $\vec{a}$ and $\vec{b}$ parallel to each other is........... .

  1. $x _1y _2=x _2y _1$

  2. $x _1/y _1=x2y2$

  3. $x _1y _1=x _2/y _2$

  4. $x _1y _1=y _2/x _2$


Correct Option: A
Explanation:

$\vec{a}\times\vec{b}=\begin{pmatrix}x _1\hat {i}+y _1\hat {j}\end{pmatrix}\times\begin{pmatrix}x _2\hat {i}+y _2\hat {j}\end{pmatrix}$

$\;\;\;\;\;\;\;\;\;\;\;=x _1y _2\hat {k}-x _2y _1\hat {k}=\vec{0}\Rightarrow x _1y _2=x _2y _1$

A vector $\bar{P} _{1}$ is along the positive x- axis. If its cross product with another vector $\bar{P} _{2}$ is zero, then $\bar{P} _{2}$ could be:

  1. $4\hat{j}$

  2. $-4\hat{i}$

  3. $(\hat{i}+\hat{k})$

  4. $-(\hat{i}+\hat{j})$


Correct Option: B
Explanation:

The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B} = \hat{n} |A| |B| \sin x $, where  $\hat{n}$ is the unit vector perpendicular to both A and B vectors and x is the angle between them.
Here in this question the vector product of $\vec{P1}$ and $\vec{P2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2) the $\sin$ of the angle between them is zero.
From the given options, the vector parallel to the given vector  $\hat{i}$ is  $-4\hat{i}$. 

If three vectors satisfy the relation $ \overrightarrow A . \overrightarrow B = 0 $ and $ \overrightarrow A . \overrightarrow C = 0 $ , then $ \overrightarrow A $ can be parallel to

  1. $ \overrightarrow C $

  2. $ \overrightarrow B $

  3. $ \overrightarrow B \times \overrightarrow C $

  4. $ \overrightarrow B . \overrightarrow C $


Correct Option: C
Explanation:

$\displaystyle \displaystyle \overrightarrow A .  \overrightarrow B = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow B$
and, $\overrightarrow A .  \overrightarrow C = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow C$
Also, $\overrightarrow B \times \overrightarrow C $ is perpendicular to both $\vec { B } \ and  \ \vec { C }$
$Thus, \overrightarrow { A } ||\, (\overrightarrow { B } \times \overrightarrow { C } )$

Consider the following statements A and B given below and identify the correct answer:
A) lf $\vec{\mathrm{A}}$ is a vector, then the magnitude of the vector is given by $\sqrt{\vec{A}\times \vec{A}}$
B) lf $\vec{a}=m\vec{b}$ where 'm' is a scalar, the value of 'm' is equal to $\frac{\vec{a} \cdot  \vec{b}}{b^{2}}$

  1. both A & B are correct

  2. A is correct but B is wrong

  3. A is wrong but B is correct

  4. both A and B are wrong


Correct Option: C
Explanation:

Magnitude of a vector A can be given as $ \sqrt{\overrightarrow{A}.\overrightarrow{A}} $. 
Hence, statement A is wrong .
Also, if $ \overrightarrow{a} = m \overrightarrow{b}$, taking dot product with b vector on both sides,
$ \overrightarrow{a}.\overrightarrow{b} = m \  b^2 \Rightarrow m = \dfrac {\overrightarrow{a}.\overrightarrow{b}}{b^2} $
Hence , statement B is correct.

lf vectors $\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are given by $\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$ and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$ then which of the following is/are correct?
$a)\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are mutually perpendicular
$\mathrm{b})$ Product of $\vec{\mathrm{A}}\times\vec{\mathrm{B}}$ is same as $\vec{\mathrm{B}}\times\vec{\mathrm{A}}$
$\mathrm{c})$ The magnitude of $\vec{\mathrm{A}}$ and $\vec{\mathrm{B}}$ are equal
$\mathrm{d})$ The magnitude of $\vec{\mathrm{A}}.\vec{\mathrm{B}}$ is zero

  1. a, d are correct

  2. b, c are correct

  3. c, d are correct

  4. b, a are correct


Correct Option: A
Explanation:

$\vec{\mathrm{A}}=5\hat{\mathrm{i}}+6\hat{\mathrm{j}}+3\hat{\mathrm{k}}$  and $\vec{\mathrm{B}}=6\hat{\mathrm{i}}-2\hat{\mathrm{j}}-6\hat{\mathrm{k}}$  is given. 

Now scalar product of this two vector is $\vec{\mathrm{A}} . \vec{\mathrm{B}} = 5\times 6-6\times 2-3\times 6=0$     
So they are mutually perpendicular.

lf $\vec{a}=2\hat{i}+6n\hat{j}+m\hat{k}$ and $\vec{b}=\hat{i}+18\hat{j}+3\hat{k}$ are parallel to each other then the values of $m,n$ are:

  1. 6,6

  2. 6,1

  3. -1,6

  4. -1,-6


Correct Option: A
Explanation:

Since, the vector a is parallel to b, the corresponding coefficients of all the 3 components must bear the same ratio
i.e $ \dfrac{2}{1} = \dfrac{6n}{18} = \dfrac{m}{3} $
Or, $6n = 36, n = 6$
And $m = 6$

$\vec{A}$ and $\vec{B}$ are two vectors in a plane at an angle of $60^{0}$ with each other. $\vec{C}$ is another vector perpendicular to the plane containing vectors $\vec{A}$ and $\vec{B}$. Which of the following relations is possible?

  1. $\vec{A}+\vec{B}=\vec{C}$

  2. $\vec{A}+\vec{C}=\vec{B}$

  3. $\vec{A}\times\vec{B}=\vec{C}$

  4. $\vec{A}\times\vec{C}=\vec{B}$


Correct Option: C
Explanation:

Vector C is perpendicular to both vectors A and B. Hence, it can be equal to their cross product.
Options A and B make vector C in the plane of vectors A  and B which is not possible.
In option D, this is not possible as vector B is not perpendicular to  vector A

If $\vec{A} = 2\hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$, sketch vectors graphically and find the component of $\vec{A}$ along $\vec{B}$ and perpendicular to $\vec{B}$.

  1. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{4}{2}(\hat{i}+\hat{j})$

  2. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$

  3. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{1}{2}(\hat{i}+\hat{j})$

  4. Component of $A$ along $B$; $\dfrac{1}{3}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$


Correct Option: B

Given $\vec{A} = 2\hat{i} + p\hat{j} + q\hat{k}$ and $\vec{B}=5\hat{i}+7\hat{j} + 3\hat{k}$. If $\vec{A}|| \vec{B}$, then the values of $p$ and $q$ are, respectively,

  1. $\dfrac{14}{5}$ and $\dfrac{6}{5}$

  2. $\dfrac{14}{3}$ and $\dfrac{6}{5}$

  3. $\dfrac{6}{5}$ and $\dfrac{1}{3}$

  4. $\dfrac{3}{4}$ and $\dfrac{1}{4}$


Correct Option: A
Explanation:
Given $\vec { A } =2\uparrow +p\hat { j } +q\hat { k } \quad \quad \vec { B } =5\uparrow +7\hat { j } +3\hat { k } $
$A\parallel B\Rightarrow A\times B=0$

Take $det{AB}$ and simplifying,

$\therefore$  $3p-7q=0$ and $6-5q=0$  and  $14-5p=0$
                               $\Rightarrow q=\dfrac { 6 }{ 5 } $                   $p=\dfrac { 14 }{ 5 } $

If the two vectors $\vec{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = \hat{i} + 2 \hat{j} - n \hat{k}$ are perpendicular, then the value of $n$ is:-

  1. $1$

  2. $2$

  3. $3$

  4. $4$


Correct Option: B
Explanation:

Given,

$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \end{array}$
They are perpendicular,
$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \ \therefore \overrightarrow { A } .\overrightarrow { B } =0 \ \Rightarrow \left( { 2\widehat { i } +3\widehat { j } +4\widehat { k }  } \right) .\left( { \widehat { i } +2\widehat { j } -n\widehat { k }  } \right) =0 \ \Rightarrow 2\widehat { i } .\widehat { i } +3\widehat { j } .2\widehat { i } -4\widehat { k } .n\widehat { k } =0\, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { j } =j.\widehat { k } =\widehat { i } .\widehat { k } =0 } \right]  \ \Rightarrow 2+6-4n=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { i } =\widehat { j } .\widehat { j } =\widehat { k } .\widehat { k } =1 } \right]  \ \Rightarrow 4n=8 \ \therefore n=2 \end{array}$
Hence, Option $B$ is correct.

Given $\overline { a } + \overline { b } + \vec { c } + \overline { d } = 0$ , which of the following statements is/are not a correct statement?

  1. $\vec { a } , \vec { b } , \vec { c }$ and $\vec { d }$ must be a null vector.

  2. The magnitude of $( \vec { a } + \vec { c } )$ equals the magnitude of $a( \vec { b } + \vec { d } )$

  3. The magnitude of $\vec { a }$ can never be greater than the sum of the magnitudes of $\vec { b } , \vec { c }$ and $\vec { d }$

  4. $\vec{b}$+$\vec{c}$ must He in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear and in the line of $\vec{a}$ and $\vec{d}$, if they are collinear.


Correct Option: A
Explanation:

In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

Simple example:- a=î,b=2î,c=–3î,d=0

(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c |  …. (#)

Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

This is the explanation of correct solution.

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