The shortest distance between the lines $r = 3i + 5j + 7k + \lambda(i + 2j + k)$ and $r = -i - j - k + \mu(7i - 6j + k)$ is

The vectors $\vec{AB} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{BC} = \hat{i} - 2\hat{k}$ are the adjacent sides of a parallelogram. The angle between its diagonals is

If $3\vec{a} - 5\vec{b}$ and $2\vec{a} + \vec{b}$ are perpendicular to each other,and $\vec{a} + 4\vec{b}$ and $-\vec{a} + \vec{b}$ are also perpendicular to each other,and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$,then find $\cos \theta$.

Let $ABC$ be a triangle. Let $u = \overrightarrow{AB}$ and $v = \overrightarrow{AC}$. If $D$ is the midpoint of $BC$,then the length of the median of $\triangle ABD$ through the vertex $B$ is:

If $\overline{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}$,$\overline{b}=\hat{i}-2 \hat{j}+\hat{k}$,and $\overline{c}=\hat{i}+\hat{j}-\hat{k}$ are three vectors,and there exists a vector $\overline{r}$ such that $\overline{r} \times \overline{a}=\overline{b}$ and $\overline{r} \cdot \overline{c}=3$,then the value of $|\overline{r}|$ is:

Two adjacent sides of a parallelogram $ABCD$ are given by $\overrightarrow{AB} = 2\hat{i} + 10\hat{j} + 11\hat{k}$ and $\overrightarrow{AD} = -\hat{i} + 2\hat{j} + 2\hat{k}$. 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