The position vectors of the vertices of $\triangle ABC$ 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,$m \angle ABC = $

Let $a$ and $b$ be two unit vectors inclined at an angle $\theta$,then $\sin(\theta/2)$ is equal to

$\bar{a}, \bar{b}, \bar{c}$ are unit vectors. If $\bar{a}, \bar{b}$ are perpendicular vectors,$(\bar{a}-\bar{c}) \cdot(\bar{b}+\bar{c})=0$ and $\bar{c}=l \bar{a}+m \bar{b}+n(\bar{a} \times \bar{b})$ ($l, m, n$ are scalars),then $n^2=$

If $|a| = 2$,$|b| = 5$ and $|a \times b| = 8$,then $a \cdot b$ is equal to

Let $O$ be the origin and $PQR$ be an arbitrary triangle. If a point $S$ satisfies the condition $\overrightarrow{OP} \cdot \overrightarrow{OQ} + \overrightarrow{OR} \cdot \overrightarrow{OS} = \overrightarrow{OR} \cdot \overrightarrow{OP} + \overrightarrow{OQ} \cdot \overrightarrow{OS} = \overrightarrow{OQ} \cdot \overrightarrow{OR} + \overrightarrow{OP} \cdot \overrightarrow{OS}$,then the point $S$ is the:

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.