The vectors $(a \cdot b) c$ and $(a \cdot c) b$ are:

Let $a, b, c$ be the position vectors of the vertices of a triangle $ABC$. The vector area of triangle $ABC$ is

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

Given that $\vec{a}, \vec{b}, \vec{p}$,and $\vec{q}$ are four vectors such that $\vec{a} + \vec{b} = \mu \vec{p}$,$\vec{b} \cdot \vec{q} = 0$,and $|\vec{b}|^2 = 1$,then $|(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$ is equal to:

If $a, b$ and $c$ are mutually perpendicular vectors of the same magnitude,then the cosine of the angle between $a$ and $a+b+c$ is

The points $O, A, B, C, D$ are such that $\overrightarrow{OA} = \vec{a}$,$\overrightarrow{OB} = \vec{b}$,$\overrightarrow{OC} = 2\vec{a} + 3\vec{b}$,and $\overrightarrow{OD} = \vec{a} - 2\vec{b}$. If $|\vec{a}| = 3|\vec{b}|$,then the angle between $\overrightarrow{BD}$ and $\overrightarrow{AC}$ is: