$A$ man goes $10\,m$ towards North,then $20\,m$ towards East. The displacement is ........ $m$.

$\overrightarrow{A} = 4\hat{i} + 3\hat{j}$ and $\overrightarrow{B} = 4\hat{i} + 2\hat{j}$. Find a vector parallel to $\overrightarrow{A}$ but with a magnitude five times that of $\overrightarrow{B}$.

If $|\vec{P}+\vec{Q}|=|\vec{P}|=|\vec{Q}|$,then the angle between $\vec{P}$ and $\vec{Q}$ is: (in $^{\circ}$)

$A$ physical quantity that has a direction is called...

The position vectors of points $A, B, C$ and $D$ are $A = 3\hat{i} + 4\hat{j} + 5\hat{k}$,$B = 4\hat{i} + 5\hat{j} + 6\hat{k}$,$C = 7\hat{i} + 9\hat{j} + 3\hat{k}$ and $D = 4\hat{i} + 6\hat{j}$. Then the displacement vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ are:

The expression $\left( \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} \right)$ is a