If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\sqrt{3} \vec{c}=\overrightarrow{0}$,then the angle between $\vec{a}$ and $\vec{b}$ is

If the position vectors of the vertices $A, B$ and $C$ of a $\Delta ABC$ are respectively $4\hat{i} + 7\hat{j} + 8\hat{k}$,$2\hat{i} + 3\hat{j} + 4\hat{k}$ and $2\hat{i} + 5\hat{j} + 7\hat{k}$,then the position vector of the point,where the bisector of $\angle A$ meets $BC$ is

The magnitude of the projection of vector $\vec{a} = -\hat{i} + 2\hat{j} - \hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ is . . . . . . .

Let $PQR$ be a triangle. The points $A, B$ and $C$ are on the sides $QR, RP$ and $PQ$ respectively such that $\frac{QA}{AR} = \frac{RB}{BP} = \frac{PC}{CQ} = \frac{1}{2}$. Then $\frac{\operatorname{Area}(\triangle PQR)}{\operatorname{Area}(\triangle ABC)}$ is equal to $........$

If the position vectors of $A$ and $B$ are $6i + j - 3k$ and $4i - 3j - 2k$ respectively,then the work done by the force $\vec{F} = i - 3j + 5k$ in displacing a particle from $A$ to $B$ is ............ $units$.

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero,non-coplanar vectors and $\vec{b_1} = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,$\vec{b_2} = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$,and $\vec{c_1} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b_1}$,$\vec{c_2} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b_1}}{|\vec{b_1}|^2}\vec{b_1}$,$\vec{c_3} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} + \frac{\vec{c} \cdot \vec{b_2}}{|\vec{c}|^2}\vec{b_1}$,$\vec{c_4} = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{c}|^2}\vec{a} - \frac{\vec{b} \cdot \vec{c}}{|\vec{b}|^2}\vec{b_1}$. Then,which of the following is a set of mutually orthogonal vectors?