If the position vectors of $P$ and $Q$ are $\hat{i}+2 \hat{j}-7 \hat{k}$ and $5 \hat{i}-3 \hat{j}+4 \hat{k}$ respectively, then the cosine of the angle between $\overrightarrow{PQ}$ and $z$-axis is

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 $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector,then $(a \cdot b)b + (a \cdot c)c + \frac{a \cdot (b \times c)}{|b \times c|} (b \times c) = $

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 $\angle B=$

In a triangle $PQR$,let $\vec{a}=\vec{QR}, \vec{b}=\vec{RP}$ and $\vec{c}=\vec{PQ}$. If $|\vec{a}|=3, |\vec{b}|=4$ and $\frac{\vec{a} \cdot(\vec{c}-\vec{b})}{\vec{c} \cdot(\vec{a}-\vec{b})}=\frac{|\vec{a}|}{|\vec{a}|+|\vec{b}|}$,then the value of $|\vec{a} \times \vec{b}|^2$ is.

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $a - \sqrt{2}b$ will be a unit vector if $\theta = $