The angle between the two vectors $\left(-2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + \hat{\mathrm{k}}\right)$ and $\left(\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - 4\hat{\mathrm{k}}\right)$ is:

1. 0°

2. 90°

3. 180°

4. 45°

If a unit vector $\hat{j}$ is rotated through an angle of 45° anticlockwise, then the new vector will be:

1.  $\sqrt{2}\hat{\mathrm{i}} + \sqrt{2}\hat{\mathrm{j}}$

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

3.  $\frac{1}{\sqrt{2}}\hat{\mathrm{i}} + \frac{1}{\sqrt{2}}\hat{\mathrm{j}}$

4.  $-\frac{1}{\sqrt{2}}\hat{\mathrm{i}} + \frac{1}{\sqrt{2}}\hat{\mathrm{j}}$

If $\overrightarrow{\mathrm{a}} = 2\hat{\mathrm{i}} + \hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{b}} = 3\hat{\mathrm{i}} + 2\hat{\mathrm{j}}$, then $\left|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}\right|=?$ 

1.  1

2.  $\sqrt{65}$

3.  8

4.  4 

$\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ are two vectors given by $\overrightarrow{\mathrm{A} }= 2\hat{\mathrm{i}} + 3\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{B}} = \hat{\mathrm{i}} + \hat{\mathrm{j}}$. The component of $\stackrel{\to \; }{\mathrm{A}}$parallel to $\overrightarrow{\mathrm{B}}$ is:

1.  $\frac{1}{2}\left(2\hat{\mathrm{i}} - \hat{\mathrm{j}}\right)$

2.  $\frac{5}{2}\left(\hat{\mathrm{i}} - \hat{\mathrm{j}}\right)$

3.  $\frac{5}{2}\left(\hat{\mathrm{i}} + \hat{\mathrm{j}}\right)$

4.  $\frac{\left(3\hat{\mathrm{i}} - 2\hat{\mathrm{j}}\right)}{2}$

If a vector is inclined at angles, $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{and} \mathrm{\gamma }$ with x, y, and z-axis respectively, then the value of ${\sin }^2 \mathrm{\alpha } + {\sin }^2 \mathrm{\beta }+ {\sin }^2 \mathrm{\gamma }$ is equal to

1. 0

2. 1

3. 2

4. $\frac{1}{2}$

A force of \(20\) N acts on a particle along a direction, making an angle of \(60^\circ\) with the vertical. The component of the force along the vertical direction will be:

If $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ are two vectors inclined to each other at an angle $\theta$, then the component of $\overrightarrow{\mathrm{A}}$ perpendicular to $\overrightarrow{\mathrm{B}}$ and lying in the plane containing $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ will be:

1.  $\frac{\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{.B}}}{{\mathrm{B}}^2}\overrightarrow{\mathrm{B}}$

2.  $\overrightarrow{\mathrm{A}} - \frac{\overrightarrow{\mathrm{A}}\overrightarrow{\mathrm{.B}}}{{\mathrm{B}}^2}\overrightarrow{\mathrm{B}}$

3.  $\overrightarrow{\mathrm{A}} - \overrightarrow{\mathrm{B}}$

4.  $\overrightarrow{\mathrm{A}} + \overrightarrow{\mathrm{B}}$

If $\left|\overrightarrow{\mathrm{A}}\right| \ne \left|\overrightarrow{\mathrm{B}}\right|$ and $\left|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}\right|$ = $\left|\overrightarrow{\mathrm{A}} · \overrightarrow{\mathrm{B}}\right|$, then 

1.  $\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}$

2.  $\overrightarrow{\mathrm{A}} \parallel \overrightarrow{\mathrm{B}}$

3.  $\overrightarrow{\mathrm{A}}$ is antiparallel to $\overrightarrow{\mathrm{B}}$

4.  $\overrightarrow{\mathrm{A}}$ is inclined to $\overrightarrow{\mathrm{B}}$ at an angle of 45$°$

Two forces of the same magnitude are acting on a body in the East and North directions, respectively. If the body remains in equilibrium, then the third force should be applied in the direction of:

1. North-East

2. North-West

3. South-West

4. South-East

Given are two vectors,  $\overrightarrow{\mathrm{A}}= (2\hat{\mathrm{i}} - 5\hat{\mathrm{j}} + 2\hat{\mathrm{k}})$ and $\overrightarrow{\mathrm{B}} = (4\hat{\mathrm{i}} - 10\hat{\mathrm{j}} + \mathrm{c}\hat{\mathrm{k}})$. What should be the value of c so that vector $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ would become parallel to each other?

1.  1

2.  2

3.  3

4.  4