The area of the parallelogram having diagonals $\vec{d_1} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{d_2} = \hat{i} - 3\hat{j} + 4\hat{k}$ is:

If $\overrightarrow {A} = 2\hat i + 3\hat j - \hat k$ and $\overrightarrow {B} = - \hat i + 3\hat j + 4\hat k$,a unit vector perpendicular to both $\overrightarrow {A}$ and $\overrightarrow {B}$ will be

The diagonals of a parallelogram are represented by vectors $\vec{A} = 5\hat{i} - 4\hat{j} + 3\hat{k}$ and $\vec{B} = 3\hat{i} - 2\hat{j} - \hat{k}$. What is the area of the parallelogram (in $\sqrt{3}$)?

If $|\hat{a} \cdot \hat{b}| = \frac{1}{2}$,then $|\hat{a} - \hat{b}|$ may be $:-$

The diagonals of a parallelogram are $2\,\hat{i}$ and $2\,\hat{j}$. What is the area of the parallelogram in square units?

If $\overrightarrow{A} \times \overrightarrow{B} = \overrightarrow{B} \times \overrightarrow{A}$,then the angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is