Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively,having $\vec{a} \cdot \vec{b} = \sqrt{6}$.

If $\bar{a}$ and $\bar{b}$ are vectors such that $|\bar{a}+\bar{b}|=\sqrt{29}$ and $\bar{a} \times(2 \hat{i}+3 \hat{j}+4 \hat{k})=(2 \hat{i}+3 \hat{j}+4 \hat{k}) \times \bar{b}$,then a possible value of $(\bar{a}+\bar{b}) \cdot(-7 \hat{i}+2 \hat{j}+3 \hat{k})$ is

If the position vectors of the vertices $A, B$ and $C$ of $\triangle ABC$ are $\hat{i}+2\hat{j}-5\hat{k}$,$-2\hat{i}+2\hat{j}+\hat{k}$ and $2\hat{i}+\hat{j}-\hat{k}$ respectively,then $\angle B=$

If $a \cdot b = 0$ and $a + b$ makes an angle $60^{\circ}$ with $a$,then

Let $u$ and $v$ be two non-zero vectors. If $|u+v|=|u-v|$,then:

$\overline{a}, \overline{b}$ and $\overline{c}$ are three vectors such that $\overline{a}+\overline{b}+\overline{c}=\overline{0}$ and $|\overline{a}|=3, |\overline{b}|=5, |\overline{c}|=7$. The angle between $\overline{a}$ and $\overline{b}$ is: