If $a = 3i - j + 2k$ and $b = 2i + j - k$,then evaluate $a \times (a \cdot b)$.

If with reference to the right-handed system of mutually perpendicular unit vectors $\hat{i}, \hat{j}$ and $\hat{k}$,$\vec{\alpha} = 3\hat{i} - \hat{j}$ and $\vec{\beta} = 2\hat{i} + \hat{j} - 3\hat{k}$,then express $\vec{\beta}$ in the form $\vec{\beta} = \vec{\beta}_{1} + \vec{\beta}_{2}$,where $\vec{\beta}_{1}$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_{2}$ is perpendicular to $\vec{\alpha}$.

$a = 4 \hat{i} + 3 \hat{j}$ and $b$ are two vectors in the $XOY$ plane,and $a$ is perpendicular to $b$. $A$ vector $c$ lying in the same plane and having projections $1$ and $2$ on $a$ and $b$ respectively is:

Let $a = i + 2j + k$,$b = i - j + k$,and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is:

If $x$ and $y$ are two unit vectors and the angle between them is $\phi$,then $\frac{1}{2} |x - y| = $

If $\vec{a}$ and $\vec{b}$ are two vectors such that $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} < 0$ and $|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|$,then the angle between the vectors $\vec{a}$ and $\vec{b}$ is