# Solve this following

Question:

Three particles $\mathrm{P}, \mathrm{Q}$ and $\mathrm{R}$ are moving along the vectors $\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}, \overrightarrow{\mathrm{B}}=\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{C}}=-\hat{\mathrm{i}}+\hat{\mathrm{j}}$ respectively. They strike on a point and start to move in different directions. Now particle $P$ is moving normal to the plane which contains vector $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$. Similarly particle $\mathrm{Q}$ is moving normal to the plane which contains vector $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{C}}$. The angle between the direction of motion of $P$ and $Q$

is $\cos ^{-1}\left(\frac{1}{\sqrt{x}}\right)$. Then the value of $x$ is

Solution:

Direction of $P \hat{v}_{1}=\pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\pm \frac{\hat{\mathrm{i}}-\hat{j}+\hat{k}}{\sqrt{3}}$

Direction of $\mathrm{Q} \hat{\mathrm{v}}_{2}=\pm \frac{\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{C}}}{|\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{C}}|}=\pm \frac{2 \hat{\mathrm{k}}}{2}=\pm \hat{\mathrm{k}}$

Angle between $\hat{v}_{1}$ and $\hat{v}_{2}$

$\frac{\hat{\mathrm{v}}_{1} \cdot \hat{\mathrm{v}}_{2}}{\left|\hat{\mathrm{v}}_{1}\right|\left|\hat{\mathrm{v}}_{2}\right|}=\frac{\pm 1 / \sqrt{3}}{(1)(1)}=\pm \frac{1}{\sqrt{3}}$

$\Rightarrow x=3$