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 and $OM$ is the projection of $\vec{a}$ on $\vec{b}$ then $l(AM)$ is equal to...

Let $\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k},\ \vec{b} = 3\hat{i} + 2\hat{j} + 2\hat{k},\ \vec{c} = 4\hat{i} - 3\hat{j} + \hat{k}$, then the vectors $\vec{a},\ \vec{b},\ \vec{c}$ are

Let $\bar{a} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}),\ \bar{b} = (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}),\ \bar{c} = (c_1\hat{i} + c_2\hat{j} + c_3\hat{k})$ be three non-zero vectors such that $\bar{a}$ is a unit vector perpendicular to both $\bar{b}$ and $\bar{c}$. If the angle between $\bar{b}$ and $\bar{c}$ is $\dfrac{\pi}{3}$ then $\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}^2 = $

Two adjacent sides of a parallelogram ABCD are given by $\overline{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overline{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. The side AD is rotated by an acute angle $\alpha$ in the plane of the parallelogram so that AD becomes AD'. If AD' makes a right angle with the side AB, then the cosine of the angle $\alpha$ is given by