Let $A = \hat{i} + 2 \hat{j}$. If $B$ is a vector in the $XY$ plane such that $(A + B) \cdot B = 15$ and $A \cdot B = 6$,then $|B|$ is

The angle $\theta$ between vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ is . . . . . . .

$ABCD$ is a parallelogram and $P$ is a point on the segment $AD$ dividing it internally in the ratio $3:1$. If the line $BP$ meets the diagonal $AC$ in $Q$, then $AQ:QC$ equals

If $\vec{a} = 4\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, then $(\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) =$

The vector projection of $\overline{AB}$ on $\overline{CD}$,where $A \equiv(2,-3,0), B \equiv(1,-4,-2), C \equiv(4,6,8)$ and $D \equiv(7,0,10)$,is

If $|\vec{a}|=2, |\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$,then $|\vec{a} \cdot \vec{b}|$ is equal to :