$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

Let two non-collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves,so that at any time $t$ the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \cos t + \hat{b} \sin t$. When $P$ is farthest from origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$,then

If $|a \times b| = 4$ and $|a \cdot b| = 2$,then $|a|^2 |b|^2 = $

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

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 $|a \times b|^2 + |a \cdot b|^2 = 36$ and $|a| = 3$,then $|b|$ is equal to