A vector $\overrightarrow{\mathrm{A}}$, which has magnitude 8.0 is added to a vector $\overrightarrow{\mathrm{B}}$ which lies on the x-axis. The sum of these two vectors lies on the y-axis and has a magnitude twice the magnitude of $\overrightarrow{\mathrm{B}}$. The magnitude of the vector $\overrightarrow{\mathrm{B}}$

1.  8

2.  $\sqrt{2} \times 8$

3.  $\frac{8}{\sqrt{5}}$

4.  $8\sqrt{5}$

If $\overrightarrow{\mathrm{A}} = 2\hat{\mathrm{i}} + 3\hat{\mathrm{j}} + 6\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{B}} = 3\hat{\mathrm{i}} + 2\hat{\mathrm{j}} - \hat{\mathrm{k}}$, then the unit vector in the direction of $\overrightarrow{\mathrm{A}} + \overrightarrow{\mathrm{B}}$ is

1.  $\frac{\left(\hat{\mathrm{i}} - \hat{\mathrm{j}} - \hat{\mathrm{k}}\right)}{\sqrt{3}}$

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

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

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

The forces ${\mathrm{F}}_1 \mathrm{and} {\mathrm{F}}_2$ are acting perpendicular to each other at a point and have resultant R. If force ${\mathrm{F}}_2$ is replaced by $\frac{{\mathrm{R}}^2 - {\mathrm{F}}_1^2}{{\mathrm{F}}_2}$ acting in the direction opposite to that of ${\mathrm{F}}_2$, the magnitude of resultant

(1) Becomes half

(2) Becomes double

(3) Becomes one third

(4) Remains the same

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 \(\overrightarrow {A}\) $\mathrm{and}$ \(\overrightarrow{B}\) are two vectors inclined to each other at an angle \(\theta,\) then the component of \(\overrightarrow {A}\)$\stackrel{}{}$ perpendicular to \(\overrightarrow {B}\) and lying in the plane containing \(\overrightarrow {A}\) and$$ \(\overrightarrow {B}\) will be:
1. \(\frac{\overrightarrow {A} \overrightarrow{.B}}{B^{2}} \overrightarrow{B}\)
2. \(\overrightarrow{A}   -   \frac{\overrightarrow{A} \overrightarrow{.B}}{B^{2}} \overrightarrow{B}\)
3. \(\overrightarrow{A} -\overrightarrow{B}\)
4. \(\overrightarrow{A} + \overrightarrow{B}\)$\stackrel{}{}$

If \(\left|\overrightarrow A\right|\ne \left|\overrightarrow B\right|\) and \(\left|\overrightarrow A \times \overrightarrow B\right|= \left|\overrightarrow A\cdot \overrightarrow B\right|\), then: 

If $\overrightarrow{\mathrm{R}}$ is the resultant of two vectors $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{R}}$' is the difference in them, and $\left|\overrightarrow{\mathrm{R}}\right| = \left|\overrightarrow{\mathrm{R}}\text{'}\right|$, then:

(1) $\overrightarrow{\mathrm{A}} \parallel \overrightarrow{\mathrm{B}}$

(2) $\overrightarrow{\mathrm{A}} \perp \overrightarrow{\mathrm{B}}$

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

(4) $\overrightarrow{\mathrm{A}}$ makes an angle of 120° with $\overrightarrow{\mathrm{B}}$

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, \(\overrightarrow{A} =   \left(\right. 2 \hat{i}   -   5 \hat{j}   +   2 \hat{k} \left.\right)\) and \(\overrightarrow{B} =   \left(4 \hat{i}   -   10 \hat{j}   +   c \hat{k} \right).\) What should be the value of \(c\) so that vector \(\overrightarrow A \) and \(\overrightarrow B\) would becomes parallel to each other?
1. \(1\)
2. \(2\)
3. \(3\)
4. \(4\)

Given below are two statements: