If the angle between the vector $\overrightarrow{A} and \overrightarrow{B}$ is  $\theta$, the value of the product $\left(\overrightarrow{B}\times \overrightarrow{A}\right).\overrightarrow{A}$ is equal to:

1. $B{A}^2 \cos \theta$

2. $B{A}^2 \sin \theta$

3. $B{A}^2 \sin \theta \cos \theta$

4. zero

A vector A points, vertically upward and, B points towards north. The vector product $\mathrm{A}\times \mathrm{B}$ is -

1.  along west

2.  along east

3.  zero

4.  vertically downward

The linear velocity of a rotating body is given by $\overrightarrow{\mathrm{v}}=\overrightarrow{\mathrm{\omega }}\times \overrightarrow{\mathrm{r}}$, where $\mathrm{\omega }$ is the angular velocity and r is the radius vector. The angular velocity of a body, $\overrightarrow{\mathrm{\omega }}=\hat{\mathrm{i}}-2\hat{\mathrm{j}}+2\hat{\mathrm{k}}$ and their radius vector is  $\overrightarrow{\mathrm{r}}=4\hat{\mathrm{j}}-3\hat{\mathrm{k}}, \mathrm{then\; value\; of} \left|\overrightarrow{\mathrm{v}}\right|$ will be:

1.  $\sqrt{29} \mathrm{units}$

2.  $31 \mathrm{units}$

3.  $\sqrt{37} \mathrm{units}$

4.  $\sqrt{41} \mathrm{units}$

If \(\theta\) is the angle between two vectors $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{b}}$, and $|\overrightarrow{\mathrm{a}}\times \overrightarrow{\mathrm{b}}|=\overrightarrow{\mathrm{a}}.\overrightarrow{\mathrm{b}}$, then \(\theta\) is equal to:
1. \(0^\circ\)
2. \(180^\circ\)
3. \(135^\circ\)
4. \(45^\circ\)

What is the torque of a force $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}} \right)$newton acting at a point $\overrightarrow{\mathrm{r}}=\left(3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}} \right)$ metre about the origin? (Given: $\overrightarrow{\mathrm{\tau }} =\overrightarrow{\mathrm{r} }\times \overrightarrow{\mathrm{F}}$)

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

2.  $17\hat{\mathrm{i}}-6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

3.  $-6\hat{\mathrm{i}}+6\hat{\mathrm{j}}-12\hat{\mathrm{k}}$

4.  $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}$

The scalar product of two vectors is 8 and the magnitude of vector product is $8\sqrt{3}$. The angle between them is:

(1)  $30°$

(2)  $60°$

(3)  $120°$

(4)  $150°$

$\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$ are two vectors and θ is the angle between them. If $\left|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}\right|=\sqrt{3}\left(\overrightarrow{A}.\overrightarrow{B}\right)$, then the value of θ will be:

1. 60^o

2. 45^o

3. 30^o

4. 90^o

The value of the unit vector, which is perpendicular to both A = \(\hat{i} + 2\hat{j} + 3 \hat{k}\) and B = \(\hat{i} - 2\hat{j} - 3 \hat{k}\) $\mathrm{is\; equal\; to}$:

1.  $\frac{\hat{\mathrm{i}} + 2\hat{\mathrm{j}} + 3\hat{\mathrm{k}}}{6}$

2.  $\frac{6\hat{\mathrm{j}} - 4\hat{\mathrm{k}}}{\sqrt{52}}$

3.  $\frac{6\hat{\mathrm{j}} + 4\hat{\mathrm{k}}}{\sqrt{52}}$

4.  $\frac{2\hat{\mathrm{i}} - \hat{\mathrm{j}}}{\sqrt{5}}$

Which of the following option is not true, if $\overrightarrow{\mathrm{A}} = 3\hat{\mathrm{i}} + 4\hat{\mathrm{j}}$ and $\overrightarrow{\mathrm{B}} = 6\hat{\mathrm{i}} + 8\hat{\mathrm{j}}$, where \(\mathrm{A}\) and \(\mathrm{B}\) are the magnitudes of $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$?
1. $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}} = \overrightarrow{0}$

2. $\frac{\mathrm{A}}{\mathrm{B}} = \frac{1}{2}$

3. $\overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{B}} = 48$

4. \(\mathrm{A}=5\)

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