If $\vec{a}$ and $\vec{b}$ are two unit vectors with $(\vec{a}, \vec{b}) = \theta$ and $|\vec{a} - \vec{b}| = 1$,then $2|\vec{a} + \vec{b}| \cos \frac{\theta}{2} =$

If the vectors $ai + bj + ck$ and $\pi + qj + rk$ are perpendicular, then

If $a, b, c$ are the $p^{th}, q^{th}, r^{th}$ terms of an $H.P.$ and $\vec{u} = (q-r)\hat{i} + (r-p)\hat{j} + (p-q)\hat{k}$ and $\vec{v} = \frac{\hat{i}}{a} + \frac{\hat{j}}{b} + \frac{\hat{k}}{c}$,then:

In a parallelogram $OACB$,$\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,and the foot of the perpendicular drawn from point $B$ to $AC$ is $M$. If $\vec{a} \cdot \vec{b} = 1$ and $|\vec{a}| = |\vec{b}| = 2$,then $|\overrightarrow{BM}|$ is:

Let $\vec{p}$ and $\vec{q}$ be the position vectors of points $P$ and $Q$ respectively,with respect to the origin $O$,and let $|\vec{p}|=p, |\vec{q}|=q$. The points $R$ and $S$ divide the line segment $PQ$ internally and externally in the ratio $2:3$ respectively. If $\vec{OR}$ and $\vec{OS}$ are perpendicular,then:

If the moduli of $a$ and $b$ are equal and the angle between them is $120^\circ$ and $a \cdot b = -8$,then $|a|$ is equal to