In $\triangle OAB$, $O(0, 0, 0)$, $A(6, 2, -3)$ and $B(4, 0, 3)$ are the vertices. Let $\vec{a}$ and $\vec{b}$ be position vectors of points $A$ and $B$ respectively. If $OM$ is the projection of $\vec{a}$ on $\vec{b}$, then the length $l(AM)$ is equal to...

If $|\bar{a}|=2, |\bar{b}|=3$ and $\bar{a}, \bar{b}$ are mutually perpendicular vectors,then the area of the triangle whose vertices are $0, \bar{a}+2\bar{b}, \bar{a}-2\bar{b}$ is

$\bar{a}, \bar{b}, \bar{c}$ are nonzero vectors such that $\bar{a}$ is perpendicular to $\bar{b}$ and $\bar{c}$,$|\bar{a}|=1, |\bar{b}|=2, |\bar{c}|=1$ and $\bar{b} \cdot \bar{c}=1$. There is a nonzero vector $\bar{d}$ coplanar with $\bar{a}+\bar{b}$ and $2\bar{b}-\bar{c}$. If $\bar{d} \cdot \bar{a}=1$,then $|\bar{d}|^2=$ (Note that $x$ and $y$ are parameters involved when we write $\bar{d}=x(\bar{a}+\bar{b})+y(2\bar{b}-\bar{c})$)

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

$\vec{c}$ is a vector along the bisector of the internal angle between the vectors $\vec{a}=4 \hat{i}+7 \hat{j}-4 \hat{k}$ and $\vec{b}=12 \hat{i}-3 \hat{j}+4 \hat{k}$. If the magnitude of $\vec{c}$ is $3 \sqrt{13}$,then $\vec{c}=$

If $a, b, c$ are three mutually perpendicular vectors such that the magnitudes of $b$ and $c$ are $1/2$ times and $\sqrt{3}/2$ times that of $a$, respectively, then the angle between the vectors $a+b+c$ and $b$ is