Let $\vec{a}=\hat{i}-2\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\vec{c}$ is a vector such that $\vec{b} \times \vec{c}=\vec{b} \times \vec{a}$ and $\vec{c} \cdot \vec{a}=0$,then $\vec{c} \cdot \vec{b}$ is equal to

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=|\vec{b}|=\sqrt{14}$ and $\vec{a} \cdot \vec{b}=-7$,then $\frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|}=$

If $|a| = |b| = 1$ and $|a + b| = \sqrt{3}$,then the value of $(3a - 4b) \cdot (2a + 5b)$ is

If the constant forces $2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle due to which it is displaced from a point $A(4,-3,-2)$ to a point $B(6,1,-3)$, then the work done by the forces is (in $\text{ unit}$)

The position vectors of the vertices $A, B$ and $C$ of a triangle are $2 \hat{i}-3 \hat{j}+3 \hat{k}$,$2 \hat{i}+2 \hat{j}+3 \hat{k}$ and $-\hat{i}+\hat{j}+3 \hat{k}$ respectively. Let $l$ denote the length of the angle bisector $AD$ of $\angle BAC$,where $D$ is on the line segment $BC$. Then $2 l^2$ equals:

If $\overrightarrow{a}, \overrightarrow{b}$ and $\overrightarrow{c}$ are unit vectors such that $\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}$,then the value of $3 \overrightarrow{a} \cdot \overrightarrow{b}+2 \overrightarrow{b} \cdot \overrightarrow{c}+\overrightarrow{c} \cdot \overrightarrow{a}$ is