Let $\vec{a} \times \vec{b} = 7 \hat{i} - 5 \hat{j} - 4 \hat{k}$ and $\vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k}$. If the length of the projection of $\vec{b}$ on $\vec{a}$ is $\frac{8}{\sqrt{14}}$,then $|\vec{b}| = $

If $\overline{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\overline{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\overline{c}=3 \hat{i}+\hat{j}$ such that $\overline{b}+\lambda \overline{a}$ is perpendicular to $\overline{c}$,then $\lambda$ is

$A$ particle acted on by constant forces $4i + j - 3k$ and $3i + j - k$ is displaced from the point $i + 2j + 3k$ to the point $5i + 4j + k$. The total work done by the force is ............... $unit$.

If $\vec{a}, \vec{b}$ and $\vec{c}$ are unit vectors such that $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$,then $|\vec{a} \times \vec{c}|$ is equal to

The position vectors of the vertices of a triangle $ABC$ are $4i - 2j$,$i + 4j - 3k$,and $-i + 5j + k$ respectively. Then $\angle ABC = $

If $4 \hat{i}+7 \hat{j}+8 \hat{k}$,$2 \hat{i}+3 \hat{j}+4 \hat{k}$ and $2 \hat{i}+5 \hat{j}+7 \hat{k}$ 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 $\angle B$ meets $CA$ is: