Three particles \({P}, {Q}\) and \({R}\) are moving along the vectors \({A}=\hat{{i}}+\hat{{j}}, {B}=\hat{{j}}+\hat{{k}}\) and \({C}=-\hat{{i}}+\hat{{j}}\) 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 ...... .

Pick an option to check your answer.

Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

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 \(\hat{v}_{2}\)

\(\cos \theta = \frac{\hat{v}_{1}, \hat{v}_{2}}{\left|\hat{v}_{1}\right|\left|\hat{v}_{2}\right|}=\frac{\pm 1 / \sqrt{3}}{(1)(1)}=\pm \frac{1}{\sqrt{3}}\)

\(\Rightarrow x=3\)

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app