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:

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=1$,$|\bar{b}|=4$,and $\bar{a} \cdot \bar{b}=2$. If $\bar{c}=(2 \bar{a} \times \bar{b})-3 \bar{b}$,then the angle between $\bar{b}$ and $\bar{c}$ is

$A$ unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ is . . . . . . where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The unit vectors orthogonal to $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and coplanar with $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ are

If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two perpendicular lines,then the direction cosines of the line which is perpendicular to both the lines will be:

If $a \times b = b \times c \ne 0,$ where $a, b$ and $c$ are coplanar vectors,then for some scalar $k$