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

If $\bar{a} = 4\hat{i} + \hat{j} + \hat{k}$, $\bar{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\bar{c} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, then $(\bar{a} + \bar{b}) \cdot (\bar{b} + \bar{c}) = $

If $\bar{a} = 2\hat{i} + \hat{j} - \hat{k}$, $\bar{b} = \hat{i} + 3\hat{k}$ and $\bar{c}$ is a unit vector, then the maximum value of the scalar triple product $[\bar{a}\ \bar{b}\ \bar{c}]$ is

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three coplanar unit vectors. A unit vector $\bar{d}$ is perpendicular to them. If $(\bar{a} \times \bar{b}) \times (\bar{c} \times \bar{d}) = \dfrac{3}{26}\hat{i} - \dfrac{2}{13}\hat{j} + \dfrac{6}{13}\hat{k}$ and the angle between $\bar{a}$ and $\bar{b}$ is $30^\circ$, then $\bar{c}$ is equal to...