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3-1.Vectors
hard
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 ...... .
A$11$
B$47$
C$5$
D$3$
(JEE MAIN-2021)
Solution
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 $\hat{v}_{2}$
$\cos \theta = \frac{\hat{v}_{1}, \hat{v}_{2}}{\left|\hat{v}_{1}\right|\left|\hat{v}_{2}\right|}=\frac{\pm 1 / \sqrt{3}}{(1)(1)}=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow x=3$
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 $\hat{v}_{2}$
$\cos \theta = \frac{\hat{v}_{1}, \hat{v}_{2}}{\left|\hat{v}_{1}\right|\left|\hat{v}_{2}\right|}=\frac{\pm 1 / \sqrt{3}}{(1)(1)}=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow x=3$
Standard 11
Physics
