Question

Three particles $P,Q$ and $R$ are moving along the vectors $\vec{A}=\hat{i}+\hat{j},\vec{B}=\hat{j}+\hat{k}$ and $\vec{C}=-\hat{i}+\hat{j}$ 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 $\vec{A}$ and $\vec{B}$. Similarly particle $Q$ is moving normal to the plane which contains vector $\vec{A}$ and $\vec{C}$. 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_________.

Answer 1

Answer: 3

Direction of $P{\hat{v}}_1=\pm \frac{\vec{A}\times \vec{B}}{|\vec{A}\times \vec{B}|}$
$=\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$
Direction of $Q{\hat{v}}_2=\pm \frac{\vec{A}\times \vec{C}}{|\vec{A}\times \vec{C}|}$
$=\pm \frac{2\hat{k}}{2}=\pm \hat{k}$
Angle between hat $v_1$ and hat $v_2$
$\frac{{\hat{v}}_1\cdot {\hat{v}}_2}{|{\hat{v}}_1||{\hat{v}}_2|}=\frac{\pm 1/\sqrt{3}}{(1)(1)}=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow x=3$