For any two vectors $\vec{a}$ and $\vec{b}$,$|\vec{a}| |\vec{b}|$ . . . . . . $|\vec{a} \cdot \vec{b}|$.

Let a unit vector $\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\overrightarrow{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$,then $|\hat{u}-\overrightarrow{v}|^2$ is equal to

In $\triangle ABC$,if $S$ is the circumcentre and $O$ is the orthocentre,then $\vec{OA} + \vec{OB} + \vec{OC} = $

The position vectors of the vertices of a $\Delta ABC$ are $4\hat{i}-2\hat{j}$,$\hat{i}+4\hat{j}-3\hat{k}$,and $-\hat{i}+5\hat{j}+\hat{k}$ respectively. Then,$\angle ABC$ is equal to:

If $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = |\vec{a} \cdot \vec{b}|$, then $\theta$ is equal to

If $a$ and $b$ are unit vectors and $\theta$ is the angle between $a$ and $b$,then $\sin \frac{\theta}{2}$ is equal to