Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 3$ and $|\vec{b}| = \frac{\sqrt{2}}{3}$. For what angle $\theta$ between $\vec{a}$ and $\vec{b}$ is $\vec{a} \times \vec{b}$ a unit vector?

If $\vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$,$\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$,$\vec{c}=-\hat{i}+\hat{j}-4 \hat{k}$ and $\vec{d}=\hat{i}+\hat{j}+\hat{k}$,then $|(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})|=$

The area (in sq. units) of the parallelogram whose diagonals are along the vectors $8\hat{i} - 6\hat{j}$ and $3\hat{i} + 4\hat{j} - 12\hat{k}$ is:

If $a = 2i + 3j - 5k$,$b = mi + nj + 12k$ and $a \times b = 0$,then $(m, n) = $

Consider the cube in the first octant with sides $OP, OQ$ and $OR$ of length $1$,along the $x$-axis,$y$-axis and $z$-axis,respectively,where $O(0,0,0)$ is the origin. Let $S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right)$ be the centre of the cube and $T$ be the vertex of the cube opposite to the origin $O$ such that $S$ lies on the diagonal $OT$. If $\overrightarrow{p} = \overrightarrow{SP}, \overrightarrow{q} = \overrightarrow{SQ}, \overrightarrow{r} = \overrightarrow{SR}$ and $\overrightarrow{t} = \overrightarrow{ST}$,then the value of $|(\overrightarrow{p} \times \overrightarrow{q}) \times (\overrightarrow{r} \times \overrightarrow{t})|$ is:

$a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}, c=4 \hat{i}+5 \hat{j}-\hat{k}$ are three vectors and a vector $r$ is perpendicular to both the vectors $b$ and $c$. If $r \cdot a=9$,then $r=$