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 $\cos^{-1}\left(\dfrac{1}{\sqrt x}\right)$. Then the value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 3
$\vec A\times\vec B = (\hat i + \hat j)\times(\hat j + \hat k) = \hat i - \hat j + \hat k$
$\vec A\times\vec C = (\hat i + \hat j)\times(-\hat i + \hat j) = 2\hat k$
$$\cos\theta = \frac{(\hat i - \hat j + \hat k)\cdot2\hat k}{\sqrt3\times2} = \frac{1}{\sqrt3} \Rightarrow x = 3$$
Solution by Sreeraj P, M.Sc Physics