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

Hint:

Directions normal to the planes are along \(\vec{A}\times\vec{B}\) and \(\vec{A}\times\vec{C}\); take the angle between them.

Answer 1

Correct Answer: 3

P moves along \(\vec{n}_1 = \vec{A}\times\vec{B}\) and Q along \(\vec{n}_2 = \vec{A}\times\vec{C}\).
\(\vec{A}\times\vec{B} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1&1&0\\ 0&1&1\end{vmatrix} = \hat i(1-0) - \hat j(1-0) + \hat k(1-0) = \hat i - \hat j + \hat k\).
\(\vec{A}\times\vec{C} = \begin{vmatrix}\hat i & \hat j & \hat k\\ 1&1&0\\ -1&1&0\end{vmatrix} = \hat k(1\cdot1 - 1\cdot(-1)) = 2\hat k\).
\(\cos\theta = \dfrac{\vec{n}_1\cdot\vec{n}_2}{|\vec{n}_1||\vec{n}_2|} = \dfrac{2}{\sqrt{3}\cdot 2} = \dfrac{1}{\sqrt{3}}\).
Comparing with \(\cos^{-1}(1/\sqrt{x})\): \(x = 3\).
Answer: 3