Tag: physics

Questions Related to physics

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})$
$\Rightarrow \vec{A} \parallel (\vec{B} \times \vec{C})$