If $\overrightarrow{OA} = 3i + 2j - k$ and $\overrightarrow{OB} = i + 3j + k$,then the area of the triangle $OAB$ is

The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

If $\overline{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overline{OB} = 4\hat{i} + \hat{k}$ are the position vectors of the points $A$ and $B$, then the position vector of a point on the line passing through $B$ and parallel to the vector $\overline{OA} \times \overline{OB}$ which is at a distance of $\sqrt{189}$ units from $B$ is

Let $\vec{a}=3 \hat{i}+\hat{j}-\hat{k}$ and $\vec{c}=2 \hat{i}-3 \hat{j}+3 \hat{k}$. If $\vec{b}$ is a vector such that $\vec{a}=\vec{b} \times \vec{c}$ and $|\vec{b}|^2=50$,then $|72-| \vec{b}+\vec{c}|^2 |$ is equal to $..........$.

For two given vectors $\bar{a}$ and $\bar{b}$,if the vectors $\overline{A}$ and $\overline{B}$ are such that $\overline{A}+\overline{B}=\bar{a}$,$\overline{A} \times \overline{B}=\bar{b}$,and $\overline{A} \cdot \bar{a}=1$,then $\overline{A}=$

If $\bar{a}, \bar{b}, \bar{c}$ are three coplanar vectors such that $|\bar{a}|=1, |\bar{b}|=2$,$\bar{b} \cdot \bar{c}=8$,and the angle between $\bar{b}$ and $\bar{c}$ is $45^{\circ}$,then $|\bar{a} \times (\bar{b} \times \bar{c})|=$