If $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$,then find $\vec{a} \cdot \vec{b}$.

$(\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$ if and only if . . . . . . (where $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$).

The projection of $\overrightarrow{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\overrightarrow{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

Three vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy the condition $\vec{a}+\vec{b}+\vec{c}=\vec{0}.$ Evaluate the quantity $\mu=\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a},$ if $|\vec{a}|=1, |\vec{b}|=4$ and $|\vec{c}|=2.$

Given $\vec{a}=3 \hat{i}-\hat{j}$,$\vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}$ and $\vec{b}=\overrightarrow{b_1}+\overrightarrow{b_2}$,where $\overrightarrow{b_1}$ is parallel to $\vec{a}$ and $\overrightarrow{b_2}$ is perpendicular to $\vec{a}$,then $\overrightarrow{b_2}$ is equal to

If $a$ and $b$ are unit vectors and $\alpha$ is the angle between them, then $a+b$ is a unit vector when $\cos \alpha=$