Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the vectors $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular to each other,then the angle between $\vec{a}$ and $\vec{b}$ is:

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by:

If $\overrightarrow{a} \cdot \hat{i} = \overrightarrow{a} \cdot (2 \hat{i} + \hat{j}) = \overrightarrow{a} \cdot (\hat{i} + \hat{j} + 3 \hat{k}) = 1$,then $\overrightarrow{a}$ is equal to :

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then the angle between $a$ and $b$ is

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

$M$ and $N$ are the midpoints of the sides $BC$ and $CD$ of a parallelogram $ABCD$ respectively,then $\overline{AM} + \overline{AN} =$