If $P=(0,1,2)$, $Q=(4,-2,1)$, and $O=(0,0,0)$, then $\angle POQ$ is equal to

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to:

If in a $\Delta ABC$,$O$ and $O^{\prime}$ are the incentre and orthocentre respectively,then $\vec{O^{\prime}A} + \vec{O^{\prime}B} + \vec{O^{\prime}C}$ is equal to

If $x$ and $y$ are two unit vectors and $\theta$ is the angle between them,then $\frac{1}{2}|x-y|$ is equal to

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:

If the resultant of two forces is of magnitude $P$ and equal to one of them and perpendicular to it,then the other force is