For a unit vector $\vec{a}$,if $(\vec{x}-\vec{a}) \cdot (\vec{x}+\vec{a}) = 15$,then find $|\vec{x}|$.

If $\overline{a}$ and $\overline{b}$ are two unit vectors such that $\overline{a}+2\overline{b}$ and $5\overline{a}-4\overline{b}$ are perpendicular to each other,then the angle between $\overline{a}$ and $\overline{b}$ is

The vector $\overline{a}=\alpha \hat{i}+2 \hat{j}+\beta \hat{k}$ lies in the plane of the vectors $\bar{b}=\hat{i}+\hat{j}$ and $\bar{c}=\hat{j}+\hat{k}$ and bisects the angle between $\bar{b}$ and $\bar{c}$. Then which one of the following gives possible values of $\alpha$ and $\beta$?

Three vectors $\vec a, \vec b, \vec c$ are inclined at an acute angle with each other such that $|\vec a| = 2, |\vec b| = 3, |\vec c| = 9$ and the lengths of the projections of $\vec a$ on $\vec b$,$\vec b$ on $\vec c$,and $\vec c$ on $\vec a$ respectively are in geometric progression. If the angle between $\vec a$ and $\vec b$ is $\frac{5\pi}{12}$ and the angle between $\vec c$ and $\vec a$ is $\frac{\pi}{12}$,then the angle between $\vec b$ and $\vec c$ is:

If $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{k}$, then the point of intersection of the lines $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is

If the position vectors of $A, B, C, D$ are $\bar{i}+2\bar{j}+2\bar{k}, 2\bar{i}-\bar{j}, \bar{i}+\bar{j}+3\bar{k}$ and $4\bar{j}+5\bar{k}$ respectively,then the quadrilateral $ABCD$ is a