$\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) + \hat{j} \cdot (\hat{j} \times \hat{k}) = $ . . . . . . .

If $A, B, C, D$ are the points with position vectors $\hat{i}-\hat{j}+\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}-3 \hat{k}$ and $3 \hat{i}-2 \hat{j}+\hat{k}$ respectively,find the projection of $\overrightarrow{AB}$ on $\overrightarrow{CD}$.

$A$ vector $a$ makes equal acute angles with the coordinate axes. Then the projection of vector $b = 5\hat{i} + 7\hat{j} - \hat{k}$ on $a$ is

Let $a = i + 2j + k$,$b = i - j + k$,$c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has projection $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is

Two adjacent sides of a parallelogram $ABCD$ are given by $\overline{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overline{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side $AD$ is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that $AD$ becomes $AD'$. If $AD'$ makes a right angle with the side $AB$,then $\cos \alpha = $

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: