In triangle $ABC$,the point $P$ divides $BC$ internally in the ratio $3:4$ and $Q$ divides $CA$ internally in the ratio $5:3$. If $AP$ and $BQ$ intersect in a point $G$,then $G$ divides $AP$ internally in the ratio

If $ 2 \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| $,then the angle between $ \vec{a} $ and $ \vec{b} $ is: (in $^{\circ}$)

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:

Let $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$, $\vec{b}=2\hat{i}+\hat{j}-\hat{k}$, $\vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a}\times\vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9p^{2}$ is equal to:

The position vectors of the points $A$ and $B$ with respect to $O$ are $2 \hat{i}+2 \hat{j}+\hat{k}$ and $2 \hat{i}+4 \hat{j}+4 \hat{k}$. The length of the internal bisector of $\angle BOA$ of $\triangle AOB$ is:

Let two non-collinear unit vectors $\hat{a}$ and $\hat{b}$ form an acute angle. $A$ point $P$ moves,so that at any time $t$ the position vector $\overline{OP}$,where $O$ is the origin,is given by $\hat{a} \cos t + \hat{b} \sin t$. When $P$ is farthest from origin $O$,let $M$ be the length of $\overline{OP}$ and $\hat{u}$ be the unit vector along $\overline{OP}$,then