If \(\left|\overrightarrow {v_1}+\overrightarrow {v_2}\right|= \left|\overrightarrow {v_1}-\overrightarrow {v_2}\right|\) and \(\overrightarrow {v_1}\) and \(\overrightarrow {v_2}\) are non-zero vectors, then:
1. \(\overrightarrow {v_1}\) is parallel to \(\overrightarrow {v_2}\)
2. \(\overrightarrow {v_1} = \overrightarrow {v_2}\)
3.  \(\overrightarrow {v_1}\) and \(\overrightarrow {v_2}\) are mutually perpendicular 
4. \(\left|\overrightarrow {v_1}\right|= \left|\overrightarrow {v_2}\right|\)

The component of vector \(\overrightarrow{A} = 3 \hat{i} + \hat{j} + \hat{k}\) along the direction of \(\hat{i} - \hat{j}\) is:
1. \(\sqrt{2}\)
2. \(2\)
3. \(\sqrt{3}\)
4. \(3\)

A force is \(60^{\circ}\) inclined to the horizontal. If its rectangular component in the horizontal direction is \(50\) N, then the magnitude of the force in the vertical direction is:

If vector \(\overrightarrow{A}   =   \cos \omega t \hat{i}   +   \sin \omega t \hat{j}\) and \(\overrightarrow{B} =\cos \frac{\omega t}{2} \hat{i} + \sin \frac{\omega t}{2} \hat{j}\) are functions of time, then the value of \(t\) at which they are orthogonal to each other will be:
1. \(t = \frac{\pi}{2\omega}\)
2. \(t = \frac{\pi}{\omega}\)
3. \(t=0\)
4. \(t = \frac{\pi}{4\omega}\)

At what angle must the two forces \((x+y)\) and \((x-y)\) act so that the resultant comes out to be \(\sqrt{x^2+y^2}\)?
1. \(\cos^{-1}\left(-\frac{x^2+y^2}{2(x^2-y^2)}\right )\)
2. \(\cos^{-1}\left(-\frac{2(x^2-y^2)}{(x^2+y^2)}\right )\)
3. \(\cos^{-1}\left(-\frac{x^2+y^2}{x^2-y^2}\right )\)
4. \(\cos^{-1}\left(-\frac{x^2-y^2}{x^2+y^2}\right )\)

The acceleration of a particle is given by \(a=3t\) at \(t=0\), \(v=0\), \(x=0\). The velocity and displacement at \(t = 2~\text{sec}\) will be:
\(\left(\text{Here,} ~a=\frac{dv}{dt}~ \text{and}~v=\frac{dx}{dt}\right)\)
1. \(6~\text{m/s}, 4~\text{m}\)
2. \(4~\text{m/s}, 6~\text{m}\)
3. \(3~\text{m/s}, 2~\text{m}\)
4. \(2~\text{m/s}, 3~\text{m}\)

$\stackrel{}{\mathrm{}}$\(\overrightarrow{A}\) 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 \(\overrightarrow{a}\) is a vector and \(x\) is a non-zero scalar, then which of the following is correct?

A body is moving according to the equation \(x = at +bt^2-ct^3\) where \(x\) represents displacement and \(a, b~\text{and}~c\) are constants. The acceleration of the body is: (\(\text{Given:}~ a=\frac{d^2x}{dt^2}\))
1. \(a+ 2bt\)
2. \(2b+ 6ct\)
3. \(2b- 6ct\)
4. \(3b- 6ct^2\)