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If a vector $\left(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}+8\hat{\mathrm{k}}\right)$ is perpendicular to the vector $\left(4\hat{\mathrm{i}}-4\hat{\mathrm{j}}+\mathrm{\alpha }\hat{\mathrm{k}}\right)$. Then the value of $\mathrm{\alpha }$ is:
1.  -1
2.  1/2
3.  -5/2
4.  1

There are two force vectors, one of \(5~\text{N}\) and the other of \(12~\text{N}\). At what angle should the two vectors be added to get the resultant vector of \(17~\text{N}, 7~\text{N},\) and \(13~\text{N}\) respectively:
1. \(0^{\circ}, 180^{\circ}~\text{and}~90^{\circ}\)
2. \(0^{\circ}, 90^{\circ}~\text{and}~180^{\circ}\)
3. \(0^{\circ}, 90^{\circ}~\text{and}~90^{\circ}\)
4. \(180^{\circ}, 0^{\circ}~\text{and}~90^{\circ}\)

If the angle between the two forces increases, the magnitude of their resultant:
1. Decreases
2. Increases
3. Remains unchanged
4. First decreases, then increases

If \(\overrightarrow {P}= \overrightarrow {Q}\), then which of the following is NOT correct?
1. \(\widehat{P}= \widehat{Q}\)
2. \(\left|\overrightarrow {P}\right|= \left|\overrightarrow {Q}\right|\)
3. \(P\widehat{Q}= Q\widehat{P}\)
4. \(\overrightarrow {P}+ \overrightarrow {Q}= \widehat{P}+ \widehat{Q}\)

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

If the vector sum of two vectors $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ is maximum, then the angle $\mathrm{\theta }$ between two vectors will be:
1.  $0°$
2.  $30°$
3.  $45°$
4.  $60°$

For the given figure, which of the following is true?
                          
1. \(\overrightarrow{B}= \overrightarrow{A}+\overrightarrow{C}\)
2. \(\overrightarrow{A}= \overrightarrow{B}+\overrightarrow{C}\)
3. \(\overrightarrow{C}= \overrightarrow{A}+\overrightarrow{B}\)
4. All of these

A force of 5N acts on a particle along a direction making an angle of $60°$ with vertical. Its vertical component will be:
1. 10 N
2. 3 N
3. 4 N
4. 2.5 N

If \(\overrightarrow {A}= 2\hat i + 4\hat j- 5\hat k,\) then the direction cosines of the vector are:
(direction cosines (or directional cosines) of a vector are the cosines of the angles between the vector and the three \(+\)ve coordinate axes.)
1. \(\frac{2}{\sqrt{45}}, \frac{4}{\sqrt{45}}~\text{and}~\frac{-5}{\sqrt{45}}\)
2. \(\frac{1}{\sqrt{45}}, \frac{2}{\sqrt{45}}~\text{and}~\frac{3}{\sqrt{45}}\)
3. \(\frac{4}{\sqrt{45}}, 0~\text{and}~\frac{4}{\sqrt{45}}\)
4. \(\frac{3}{\sqrt{45}}, \frac{2}{\sqrt{45}}~\text{and}~\frac{5}{\sqrt{45}}\)

If $\hat{\mathrm{p}}$ is the unit vector in the direction $\overrightarrow{\mathrm{B}}$, then:
1.  $\hat{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{\left|\overrightarrow{\mathrm{B}}\right|}{\overrightarrow{\mathrm{B}}}$
2.  $\hat{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\left|\overrightarrow{\mathrm{B}}\right|<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}$
3.  $\hat{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{\overrightarrow{\mathrm{B}}}{\left|\overrightarrow{\mathrm{B}}\right|}$
4.  $\hat{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{\overrightarrow{\mathrm{B}}}{\overrightarrow{\mathrm{B}}}$

\(\overrightarrow A\) and \(\overrightarrow {B}\) are two vectors given by \(\overrightarrow {A}= 2\hat i + 3\hat j\) and \(\overrightarrow {B}= \hat i + \hat j\). The component of \(\overrightarrow A\) parallel to \(\overrightarrow B\) is:
1. \(\frac{(2\hat i -\hat j)}{2}\)
2. \(\frac{5}{2}(\hat i - \hat j)\)
3. \(\frac{5}{2}(\hat i + \hat j)\)
4. \(\frac{(3\hat i -2\hat j)}{2}\)

The angle made by the vector $\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\sqrt{3}\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>3\hat{\mathrm{j}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2\hat{\mathrm{k}}$ with y-axis is:
1.  ${\sin }^{-1}\left(\frac{3}{\sqrt{14}}\right)$
2.  ${\sin }^{-1}\left(\frac{\sqrt{7}}{4}\right)$
3.  ${\cos }^{-1}\left(\frac{4}{3}\right)$
4.  ${\cos }^{-1}\left(\frac{3}{5}\right)$

If $\left|\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}\right|<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\sqrt{3}\overrightarrow{\mathrm{A}}.<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}$, then the value of $\left|\overrightarrow{\mathrm{A}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\overrightarrow{\mathrm{B}}\right|$ is:-
1.  ${\left({\mathrm{A}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{B}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{\mathrm{AB}}{\sqrt{3}}\right)}^{1/2}$
2.  A + B
3.  ${\left({\mathrm{A}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{B}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\sqrt{3}\mathrm{AB}\right)}^{1/2}$
4.  ${\left({\mathrm{A}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{B}}^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\mathrm{AB}\right)}^{1/2}$

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 $\overrightarrow{\mathrm{a}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{b}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>3\hat{\mathrm{i}}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2\hat{\mathrm{j}}$, then $\left|\overrightarrow{\mathrm{a}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{b}}\right|=?$ 

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }}<mspace\; width="1em"></mspace>=\overrightarrow{\mathrm{r}<mspace\; width="1em"></mspace>}\times \overrightarrow{\mathrm{F}}$)
1. $6\hat{\mathrm{i}}-6\hat{\mathrm{j}}+12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$
2. $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$
3. $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$
4. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

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. \(4\)

The position of a particle in a rectangular co-ordinate system is \((3, 2, 5)\). Then its position vector will be:
1. \(5\hat i + 6\hat j + 2\hat k\)
2. \(3\hat i + 2\hat j + 5\hat k\)
3. \(5\hat i + 3\hat j + 2\hat k\)
4. None of these

If vectors $\hat{\mathrm{i}}-3\hat{\mathrm{j}}+5\hat{\mathrm{k}}$ and $\hat{\mathrm{i}}-3\hat{\mathrm{j}}-\mathrm{\alpha }\hat{\mathrm{k}}$ are equal vectors, then the value of $\mathrm{\alpha }$ is:
1. -5
2. 2
3. -3
4. 4