If the angle between the vector $\overrightarrow{A}<mspace\; width="1em"></mspace>and<mspace\; width="1em"></mspace>\overrightarrow{B}$ is  $\theta$, the value of the product $\left(\overrightarrow{B}\times \overrightarrow{A}\right).\overrightarrow{A}$ is equal to:

1. $B{A}^2<mspace\; width="1em"></mspace>\cos \theta$

2. $B{A}^2<mspace\; width="1em"></mspace>\sin \theta$

3. $B{A}^2<mspace\; width="1em"></mspace>\sin \theta \cos \theta$

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  $\overrightarrow{\mathrm{r}}=4\hat{\mathrm{j}}-3\hat{\mathrm{k}},<mspace\; width="1em"></mspace>\mathrm{then\; value\; of}<mspace\; width="1em"></mspace>\left|\overrightarrow{\mathrm{v}}\right|$ will be:

1. $\sqrt{29}<mspace\; width="1em"></mspace>\mathrm{units}$

2. $31<mspace\; width="1em"></mspace>\mathrm{units}$

3. $\sqrt{37}<mspace\; width="1em"></mspace>\mathrm{units}$

4. $\sqrt{41}<mspace\; width="1em"></mspace>\mathrm{units}$

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 $\overrightarrow{\mathrm{F}}=\left(2\hat{\mathrm{i}}-3\hat{\mathrm{j}}+4\hat{\mathrm{k}}<mspace\; width="1em"></mspace>\right)$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. $-17\hat{\mathrm{i}}+6\hat{\mathrm{j}}-13\hat{\mathrm{k}}<mspace\; width="1em"></mspace>$

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)  $150°$

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

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. \(\frac{\hat{i}   +   2 \hat{j}   +   3 \hat{k}}{6}\)
2. \(\frac{6\hat{j} -4 \hat{k}}{\sqrt{52}}\)
3. \(\frac{6\hat{j} +4 \hat{k}}{\sqrt{52}}\)
4. \(\frac{2\hat{i} - \hat{j}}{\sqrt{5}}\)

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

2. $\frac{\mathrm{A}}{\mathrm{B}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{1}{2}$

3. $\overrightarrow{\mathrm{A}}·\overrightarrow{\mathrm{B}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>48$

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

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 $\left|\overrightarrow{\mathrm{a}}<mspace\; width="1em"></mspace>\times <mspace\; width="1em"></mspace>\overrightarrow{\mathrm{b}}\right|=?$