The position vectors of the vertices of $\triangle ABC$ are $4\hat{i} - 2\hat{j}$,$\hat{i} + 4\hat{j} - 3\hat{k}$,and $-\hat{i} + 5\hat{j} + \hat{k}$ respectively. Then,$m \angle ABC = $

$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane, and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

If $r=b+ta$ and $r=d+sc$ are two skew lines, then the shortest distance between them is

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

The orthogonal projection of vector $a$ on vector $b$ is given by:

The vector projection of $\vec{b}$ on $\vec{a}$ where $\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}$ and $\vec{b}=7 \hat{i}-5 \hat{j}-\hat{k}$ is: