Find the distance between the lines $l_{1}$ and $l_{2}$ given by $\vec{r}=\hat{i}+2 \hat{j}-4 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}+3 \hat{j}-5 \hat{k}+\mu(2 \hat{i}+3 \hat{j}+6 \hat{k})$.

Let the vectors $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ be such that $|\overrightarrow{a}|=2, |\overrightarrow{b}|=4$ and $|\overrightarrow{c}|=4$. If the projection of $\overrightarrow{b}$ on $\overrightarrow{a}$ is equal to the projection of $\overrightarrow{c}$ on $\overrightarrow{a}$ and $\overrightarrow{b}$ is perpendicular to $\overrightarrow{c}$,then the value of $|\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}|$ is

The set of all real values of $c$ such that the angle between the vectors $\vec{a} = cx \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $\vec{b} = x \hat{i} + 2 \hat{j} + 2cx \hat{k}$ is an obtuse angle for all real $x$ is:

If $\overrightarrow{a}=-\hat{i}+\hat{j}+2 \hat{k}$, $\overrightarrow{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\overrightarrow{c}=-2 \hat{i}+\hat{j}+3 \hat{k}$, then the angle between $2 \overrightarrow{a}-\overrightarrow{c}$ and $\overrightarrow{a}+\overrightarrow{b}$ 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

If the position vectors of $A$ and $B$ are $6i + j - 3k$ and $4i - 3j - 2k$ respectively,then the work done by the force $\vec{F} = i - 3j + 5k$ in displacing a particle from $A$ to $B$ is ............ $units$.