Let $\overrightarrow{OA}=2 \overrightarrow{a}$,$\overrightarrow{OB}=6 \overrightarrow{a}+5 \overrightarrow{b}$ and $\overrightarrow{OC}=3 \overrightarrow{b}$,where $O$ is the origin. If the area of the parallelogram with adjacent sides $\overrightarrow{OA}$ and $\overrightarrow{OC}$ is $15$ sq. units,then the area (in sq. units) of the quadrilateral $OABC$ is equal to :

$a, b, c, d$ are coplanar vectors,then $(a \times b) \times (c \times d)$ is equal to

Let $\bar{a}, \bar{b}, \bar{c}$ be vectors such that $\bar{a} \neq \bar{o}, \bar{b} \neq \bar{o}, \bar{a} \times \bar{c} = \bar{b}$ and $\bar{b} \times \bar{c} = \bar{a}$. Then:

Let $A(2, 3, 5)$ and $C(-3, 4, -2)$ be opposite vertices of a parallelogram $ABCD$. If the diagonal $\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$,then the area of the parallelogram is equal to:

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,

Let $\overrightarrow{a}$ be a non-zero vector parallel to the line of intersection of the two planes passing through the origin and containing the vectors $(\hat{i}+\hat{j}, \hat{i}+\hat{k})$ and $(\hat{i}-\hat{j}, \hat{j}-\hat{k})$ respectively. If $\theta$ is the angle between the vector $\vec{a}$ and the vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$ and $\vec{a} \cdot \vec{b}=6$,then the ordered pair $(\theta, |\vec{a} \times \vec{b}|)$ is equal to