A particle moves from a point \(\left(\right. - 2 \hat{i} + 5 \hat{j} \left.\right)\) to \(\left(\right. 4 \hat{j} + 3 \hat{k} \left.\right)\) when a force of \(\left(\right. 4 \hat{i} + 3 \hat{j} \left.\right)\) \(\text{N}\) is applied. How much work has been done by the force?

If vectors \(\overrightarrow{{A}}=\cos \omega t \hat{{i}}+\sin \omega t \hat{j}\) and \(\overrightarrow{{B}}=\cos \left(\frac{\omega t}{2}\right)\hat{{i}}+\sin \left(\frac{\omega t}{2}\right) \hat{j}\) are functions of time. Then, at what value of \(t\) are they orthogonal to one another?
1. \(t = \frac{\pi}{4\omega}\)
2. \(t = \frac{\pi}{2\omega}\)
3. \(t = \frac{\pi}{\omega}\)
4. \(t = 0\)

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) 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:

If the angle between the vector $\overrightarrow{\mathrm{A}}\text{\hspace{0.17em}and\hspace{0.17em}}\overrightarrow{\mathrm{B}}$ is \(\theta\), the value of the product $(\overrightarrow{\mathrm{B}}\times \overrightarrow{\mathrm{A}})\cdot \overrightarrow{\mathrm{A}}$ is equal to:

1. zero

2. BA²\(\sin\theta \cos \theta\)

3. BA²\(\cos\theta\)

4. BA²\(\sin\theta\)

$\begin{array}{l}\text{\hspace{0.17em}If\hspace{0.17em}}|\overrightarrow{\mathrm{A}}\times \overrightarrow{\mathrm{B}}|=\sqrt{3}\overrightarrow{\mathrm{A}}\cdot \overrightarrow{\mathrm{B}}\text{, then the value of\hspace{0.17em}}\\ |\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|\text{\hspace{0.17em}is:\hspace{0.17em}}\end{array}$

1. ${({\mathrm{A}}^2+{\mathrm{B}}^2+\frac{\mathrm{AB}}{\sqrt{3}})}^{1/2}$

2. A+B

3. ${({\mathrm{A}}^2+{\mathrm{B}}^2+\sqrt{3}\mathrm{AB})}^{1/2}$

4. ${({\mathrm{A}}^2+{\mathrm{B}}^2+\mathrm{AB})}^{1/2}$