If $a$ and $b$ are two unit vectors and $\theta$ is the angle between them,then the unit vector along the angular bisector of $a$ and $b$ is given by

The area of the rectangle having vertices $P, Q, R, S$ with position vectors $-\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}, -\hat{i}-\hat{j}+\hat{k}$ respectively is

If the position vectors of the vertices $A, B$,and $C$ of a triangle $ABC$ are $4\hat{i} + 7\hat{j} + 8\hat{k}$,$2\hat{i} + 3\hat{j} + 4\hat{k}$,and $2\hat{i} + 5\hat{j} + 7\hat{k}$ respectively,then the position vector of the point where the bisector of angle $A$ meets $BC$ is:

If $\overrightarrow{F_1} = i - j + k,$ $\overrightarrow{F_2} = -i + 2j - k,$ $\overrightarrow{F_3} = j - k,$ $\vec{A} = 4i - 3j - 2k$ and $\vec{B} = 6i + j - 3k,$ then the scalar product of $(\overrightarrow{F_1} + \overrightarrow{F_2} + \overrightarrow{F_3})$ and $\overrightarrow{AB}$ will be:

If $P=(0,1,2), Q=(4,-2,1)$ and $O=(0,0,0)$,then $\angle POQ=$

If $\overrightarrow{a} \cdot \hat{i}=4$,then $(\overrightarrow{a} \times \hat{j}) \cdot(2 \hat{j}-3 \hat{k})$ is equal to