$A$ particle is acted upon by constant forces $4\hat{i} + \hat{j} - 3\hat{k}$ and $3\hat{i} + \hat{j} - \hat{k}$. The displacement of the particle from the point $\hat{i} + 2\hat{j} + 3\hat{k}$ to the point $5\hat{i} + 4\hat{j} + \hat{k}$ is given. Find the total work done by the forces in units.

The value of $c$ such that for all real $x$, the vectors $\vec{a} = cxi - 6j + 3k$ and $\vec{b} = xi + 2j + 2cxk$ make an obtuse angle is:

If $G(\bar{g})$,$H(\bar{h})$,and $P(\bar{p})$ are respectively the centroid,orthocenter,and circumcentre of a triangle and $x \bar{p} + y \bar{h} + z \bar{g} = \overline{0}$,then $x, y, z$ are respectively:

Let $\overrightarrow{OA}$ and $\overrightarrow{OB}$ be two sides of a triangle. The median $\overrightarrow{AM}$ is perpendicular to the angle bisector $\overrightarrow{OL}$ and $|\overrightarrow{AM}|:|\overrightarrow{OL}|=1:2$. The angle between $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is

If $\vec{a}=\hat{i}+(\tan \theta) \hat{j}+\left(\frac{3}{\sqrt{\sin \frac{\theta}{2}}}\right) \hat{k}$ and $\vec{b}=\tan \theta(\hat{j}-\hat{i})-\left(2 \sqrt{\sin \frac{\theta}{2}}\right) \hat{k}$ are orthogonal vectors and $\vec{c}=(\sin 2 \theta) \hat{i}-2 \hat{j}+2 \hat{k}$ makes an obtuse angle with $X$-axis, then $\theta=$

Let $(p \wedge q)$ denote the angle between vectors $p$ and $q$. If $a + b + c = 0$, $|a| = 7$, $|b| = 5$, and $|c| = 3$, find the correct relation.