Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

If $3 \hat{j}$,$4 \hat{k}$ and $3 \hat{j}+4 \hat{k}$ are the position vectors of the vertices $A$,$B$,and $C$ respectively of $\triangle ABC$,then the position vector of the point in which the bisector of $\angle A$ meets $BC$ is

Find the angle between the diagonals of parallelogram $PQRS$,if $\vec{PQ} = 3\hat{i} - 2\hat{j} + 2\hat{k}$ and $\vec{PS} = \hat{i} - 2\hat{k}$.

The components of a vector $a$ along and perpendicular to the non-zero vector $b$ are respectively:

Let $a$ and $b$ be two unit vectors and $\theta$ is the angle between them. Then,$a+b$ is a unit vector,if

If $|a|=2$ and $|b|=3$ and the angle between $a$ and $b$ is $120^{\circ}$,then the length of the vector $\left|\frac{a}{2}-\frac{b}{3}\right|$ is