A particle moves from position null to \(\left(11\hat i + 11\hat j + 15\hat k \right)\) due to a uniform force of \(\left(4\hat i + \hat j + 3\hat k\right)\)N. If the displacement is in m, then the work done will be: (Given: \(W=\overrightarrow {F}.\overrightarrow {S}\))
1. \(100~\text{J}\)
2. \(200~\text{J}\)
3. \(300~\text{J}\)
4. \(250~\text{J}\)

The dot product of two mutual perpendicular vector is:

1. \(0\)

2. \(1\)

3. \(\infty\)

4. None of the above

The angle between the two vectors \(\left(- 2 \hat{i} +3 \hat{j} + \hat{k}\right)\) and \(\left(\hat{i} + 2 \hat{j} - 4 \hat{k}\right)\) is:
1. \(0^{\circ}\)
2. \(90^{\circ}\)
3. \(180^{\circ}\)
4. \(45^{\circ}\)

If \(\overrightarrow {A} = 2\hat{i} + \hat{j} - \hat{k},\) \(\overrightarrow {B} = \hat{i} + 2\hat{j} + 3\hat{k},\) and \(\overrightarrow {C} = 6 \hat{i} - 2\hat{j} - 6\hat{k},\) then the angle between \(\left(\overrightarrow {A} + \overrightarrow{B}\right)\) and \(\overrightarrow{C}\) will be:
1. \(30^{\circ}\)
2. \(45^{\circ}\)
3. \(60^{\circ}\)
4. \(90^{\circ}\)

The magnitude of the resultant of two vectors of magnitude \(3\) units and \(4\) units is \(1\) unit. What is the value of their dot product?

1. \(-12\) units

2. \(-7\) units

3. \(-1\) unit

4. \(0\)

$\stackrel{}{\mathrm{}}$\(\overrightarrow A\) 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. \(\frac{(3\hat i -2\hat j)}{2}\)

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. \(t = \frac{\pi}{4\omega}\)

The vector sum of two forces is perpendicular to their vector difference. In that case, the forces:

The angle which the vector \(\overrightarrow{A} = 2 \hat{i} + 3 \hat{j}\) makes with the \(y\text-\)axis, where \(\hat i\) and \(\hat j\) are unit vectors along \(x\text-\) and \(y\text-\)axis, respectively, is:
1. \(\cos^{-1}\left(\frac{3}{5}\right)\)
2. \(\cos^{-1}\left(\frac{2}{3}\right)\)
3. \(\tan^{-1}\left(\frac{2}{3}\right)\)
4. \(\sin^{-1}\left(\frac{2}{3}\right)\)

The unit vector perpendicular to vectors \(\overrightarrow a= \left(3 \hat{i}+\hat{j}\right)  \) and \(\overrightarrow B = \left(2\hat i - \hat j -5\hat k\right)\) is:
1. \(\pm \frac{\left(\right. \hat{i} - 3 \hat{j} + \hat{k} \left.\right)}{\sqrt{11}}\)
2. \(\pm \frac{\left(3 \hat{i} + \hat{j}\right)}{\sqrt{11}}\)
3. \(\pm \frac{\left(\right. 2 \hat{i} - \hat{j} - 5 \hat{k} \left.\right)}{\sqrt{30}}\)
4. None of these