Let $A=(3,4,0), B=(4,4,4), C=(-6,2,3)$ and $D=(1,1,2)$. If $\theta$ is the acute angle between the lines $AB$ and $CD$,then $\cos \theta=$

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$,$\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$,$\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then $x^2+y^2=$

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

Let $\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}$,$\vec{b}=4 \hat{i}+\hat{j}+7 \hat{k}$,and $\vec{c}=\hat{i}-3 \hat{j}+4 \hat{k}$ be three vectors. If a vector $\vec{p}$ satisfies $\vec{p} \times \vec{b}=\vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a}=0$,then $\vec{p} \cdot(\hat{i}-\hat{j}-\hat{k})$ is equal to

Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$,$\vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. $A$ vector $\vec{V}$ in the plane of $\vec{a}$ and $\vec{b}$,whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$,is given by:

If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}, 2\bar{i}-\bar{j}, \bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively,then the quadrilateral $ABCD$ is a