Question

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

Direction of $P$,

${\overrightarrow{v}}_1=\frac{\overrightarrow{A}\times \overrightarrow{B}}{|\overrightarrow{A}\times \overrightarrow{B}|}=\frac{(\hat{i}+\hat{j})\times (\hat{j}+\hat{k})}{|(\hat{i}+\hat{j})\times (\hat{j}+\hat{k})|}=\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$

Direction of $Q$,

${\overrightarrow{v}}_2=\frac{\overrightarrow{A}\times \overrightarrow{C}}{|\overrightarrow{A}\times \overrightarrow{C}|}=\frac{(\hat{i}+\hat{j})\times (-\hat{i}+\hat{j})}{|(\hat{i}+\hat{j})\times (-\hat{i}+\hat{j})|}=\hat{k}$

Angle between ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$,

$\cos \theta =\frac{{\overrightarrow{v}}_1.{\overrightarrow{v}}_2}{|{\overrightarrow{v}}_1||{\overrightarrow{v}}_2|}=\frac{\left(\frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\right).\left(\hat{k}\right)}{\left(1\right)\times \left(1\right)}=\frac{1/\sqrt{3}}{\left(1\right)\left(1\right)}=\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Given, $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{x}}\right)$

$\therefore x=3$