Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}|=1, |\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=2$. If $\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}$,then the value of $\vec{b} \cdot \vec{c}$ is

If $a, b, c$ are coplanar vectors,then

If $\overline{a}, \overline{b}, \overline{c}$ are three vectors such that $|\overline{a}+\overline{b}+\overline{c}|=1$,$\overline{c}=\lambda(\overline{a} \times \overline{b})$ and $|\overline{a}|=\frac{1}{\sqrt{3}}, |\overline{b}|=\frac{1}{\sqrt{2}}, |\overline{c}|=\frac{1}{\sqrt{6}}$,then the angle between $\bar{a}$ and $\bar{b}$ is

If $a$ and $b$ are mutually perpendicular vectors,then $(a + b)^2 = $

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 $\vec{\alpha}$.

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three unit vectors such that $|\vec{a}-\vec{b}|^{2}+|\vec{a}-\vec{c}|^{2}=8$. Then $|\vec{a}+2\vec{b}|^{2}+|\vec{a}+2\vec{c}|^{2}$ is equal to