$a$ and $b$ are two vectors such that $|a|=\sqrt{3}$ and $|b|=\sqrt{2}$. If $x$ is a unit vector satisfying $x \times a = b$,then $x$ is equal to:

$A$ vector $\vec{a}$ of length $2$ units makes an angle $60^{\circ}$ with each of the $X$-axis and $Y$-axis. If another vector $\vec{b}$ of length $\sqrt{2}$ units makes an angle $45^{\circ}$ with each of the $Y$-axis and $Z$-axis, then $\vec{a} \times \vec{b} = $

Given $\bar{a} = 2\bar{i} + \bar{j} - 2\bar{k}$ and $\bar{b} = \bar{i} + \bar{j}$. 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 $30^{\circ}$,then the value of $|(\bar{a} \times \bar{b}) \times \bar{c}|^2$ is

Let $\overrightarrow{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=((\overrightarrow{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i}$. Then the square of the projection of $\vec{a}$ on $\vec{b}$ is:

If $\overrightarrow{OA} = 3i + 2j - k$ and $\overrightarrow{OB} = i + 3j + k$,then the area of the triangle $OAB$ is

If the diagonals of a parallelogram are represented by the vectors $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} + 3\hat{j} - 4\hat{k}$,then its area in square units is: