Find the projection of the vector $\hat{i}-\hat{j}$ on the vector $\hat{i}+\hat{j}$.

If $a, b$ and $c$ are unit vectors such that $a+b+c=0$,then $a \cdot b+b \cdot c+c \cdot a$ is equal to

If $\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$,then the angle between the vectors $(2 \bar{a}+\bar{b})$ and $(\bar{a}+2 \bar{b})$ is

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u} \times \hat{v}| = \frac{\sqrt{3}}{2}$. If $\vec{A} = \lambda \hat{u} + \hat{v} + (\hat{u} \times \hat{v})$,then $\lambda$ is equal to: