A vector $\overrightarrow{A}$ points vertically downward, and another vector $\overrightarrow{B}$ points towards the east. Then, the vector product $\overrightarrow{A}\times \overrightarrow{B}$ is

Three coplanar vectors $\overrightarrow{A}$, $\overrightarrow{B}$ and $\overrightarrow{\mathrm{C}}$ have magnitudes $4$, $3$ and $2$, respectively. If the angle between any two vectors is $120°$, then which of the following vectors may be equal to $\frac{3\overrightarrow{A}}{4}+\frac{\overrightarrow{B}}{3}+\frac{\overrightarrow{C}}{2}$?

Consider three vectors,

$\overrightarrow{A}=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-2\hat{\mathrm{k}}$,

$\overrightarrow{B}=5\hat{\mathrm{i}}+n\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and 

$\overrightarrow{C}=-\hat{\mathrm{i}}+2\hat{\mathrm{j}}+3\hat{\mathrm{k}}$

If these three vectors are coplanar, then the value of $n$ will be

For the situation shown below, what will be the magnitude of minimum force (in Newton) that can be applied in $y$ direction so that, the resultant force is along the east direction?

The column$‐\mathrm{I}$ shows vector diagrams relating three vectors $\overrightarrow{a}$, $\overrightarrow{b}$ and $\overrightarrow{c}$. Match the vector equations in the column$‐\mathrm{II}$, with vector diagrams in the column$‐\mathrm{I}$.

	Column$‐\mathrm{I}$		Column$‐\mathrm{II}$
$\left(A\right)$		$\left(p\right)$	$\overrightarrow{a}-(\overrightarrow{b}+\overrightarrow{c})=0$
$\left(B\right)$		$\left(q\right)$	$\overrightarrow{b}-\overrightarrow{c}=\overrightarrow{a}$
$\left(C\right)$		$\left(r\right)$	$\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}$
$\left(D\right)$		$\left(s\right)$	$\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$

If $\left|\overrightarrow{a}\right|=3\sqrt{2}$ and $\left|\overrightarrow{b}\right|=6\sqrt{2}$, then $x-$component of $\left(\overrightarrow{a}+\overrightarrow{b}\right)$ is