If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{c}$,and $|\vec{a} + \vec{b} + \vec{c}| = 1$,then the angle between $\vec{b}$ and $\vec{c}$ is:

If the vectors $a\hat{i} + 2\hat{j} + 3\hat{k}$ and $-\hat{i} + 5\hat{j} + a\hat{k}$ are perpendicular to each other,then $a = $

If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k},$ then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.

The vectors $(a \cdot b) c$ and $(a \cdot c) b$ are:

In a trapezium $ABCD$,$\vec{BC} = \lambda \vec{AD}$ and $\vec{x} = \vec{AC} + \vec{BD}$. If $\vec{x} = p \vec{AD}$,then $p =$

Two adjacent sides of a parallelogram $PQRS$ are given by $\vec{PQ} = \hat{i} + \hat{k}$ and $\vec{PS} = \hat{i} - \hat{j}$. If the side $PS$ is rotated about the point $P$ by an acute angle $\alpha$ in the plane of the parallelogram so that it becomes perpendicular to the side $PQ$, then $\sin^2(\frac{5\alpha}{2}) - \sin^2(\frac{\alpha}{2})$ is equal to: