Let $a, b$ and $c$ be unit vectors such that $a \cdot b = 0 = a \cdot c$ and the acute angle between $b$ and $c$ is $\frac{\pi}{3}$,then $|a \times b - a \times c|$ is equal to

$A$ unit vector perpendicular to both the vectors $\hat{i}+\hat{j}$ and $\hat{j}+\hat{k}$ is

If $\vec{x}$ is a unit vector such that $\vec{x} \times (\hat{i} - 2\hat{j} + \hat{k}) = -\hat{i} + \hat{k}$,then $\vec{x}$ is:

For any vector $\vec{a} \in \mathbb{R}^3$,$|\vec{a} \times \hat{i}|^2 + |\vec{a} \times \hat{j}|^2 + |\vec{a} \times \hat{k}|^2 = $ . . . . . . .

Let $\bar{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\bar{b} = \hat{i} - \hat{j} + \hat{k}$. If $\bar{c}$ is a vector such that $\bar{a} \cdot \bar{c} = |\bar{c}|$,$|\bar{c} - \bar{a}| = 2\sqrt{2}$ and the angle between $\bar{a} \times \bar{b}$ and $\bar{c}$ is $60^{\circ}$,then $|(\bar{a} \times \bar{b}) \times \bar{c}|$ is equal to

The adjacent sides of a parallelogram are $\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. Then,the area of the parallelogram is . . . . . . sq. units.