Mathematics in Physics

JEE Main level, numerical answer, medium

26 of 121

Question 26 of 121

Three particles P,QP,Q and RR are moving along the vectors A⃗=i^+j^,B⃗=j^+k^\vec{A}=\hat{i}+\hat{j},\vec{B}=\hat{j}+\hat{k} and C⃗=−i^+j^\vec{C}=-\hat{i}+\hat{j}, respectively. They strike on a point and start to move in different directions. Now particle PP is moving normal to the plane which contains vector A⃗\vec{A} and B⃗.\vec{B}. Similarly particle QQ is moving normal to the plane which contains vector A⃗\vec{A} and C⃗.\vec{C}. The angle between the direction of motion of PP and QQ is cos⁡−1(1x).{\cos }^{-1}\left(\dfrac{1}{\sqrt{x}}\right). Then the value of xx is ..

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Answer: 3

Direction of P,v^1=±A⃗×B⃗∣A⃗×B⃗∣=±i^−j^+k^3P,{\hat{v}}_{1}=\pm \dfrac{\vec{A}\times \vec{B}}{|\vec{A}\times \vec{B}|}=\pm \dfrac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}
Direction of Q,v^2=±A⃗×C⃗∣A⃗×C⃗∣=±2k^2=±k^Q,{\hat{v}}_{2}=\pm \dfrac{\vec{A}\times \vec{C}}{|\vec{A}\times \vec{C}|}=\pm \dfrac{2\hat{k}}{2}=\pm \hat{k}
Angle between v^1{\hat{v}}_{1} and v^2{\hat{v}}_{2},
v^1⋅v^2∣v^1∣∣v^2∣=±13(1)(1)=±13\dfrac{{\hat{v}}_{1}\cdot {\hat{v}}_{2}}{\left|{\hat{v}}_{1}\right|\left|{\hat{v}}_{2}\right|}=\dfrac{\pm \dfrac{1}{\sqrt{3}}}{(1)(1)}=\pm \dfrac{1}{\sqrt{3}}
⇒x=3\Rightarrow x=3

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