Verified Solution • 2025–26 Session moderate

QJEEM24Ma50be: JEE Main 2024 Mathematics - Class 12 JEEM24Ma50be Solution: Let be a point in the first octant, whose projection in the…

Let P(x,y,z)P(x, y, z) be a point in the first octant, whose projection in the xyx y -plane is the point QQ . Let OP=γO P=\gamma ; the angle between OQO Q and the positive xx -axis be θ\theta ; and the angle between OPO P and the positive zz -axis be ϕ\phi , where OO is the origin. Then the distance of PP from the xx -axis is

Multiple Choice Options

  • A γ1−sin⁡2ϕcos⁡2θ\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
  • B γ1+cos⁡2θsin⁡2ϕ\gamma \sqrt{1+\cos ^2 \theta \sin ^2 \phi}
  • C γ1+cos⁡2ϕsin⁡2θ\gamma \sqrt{1+\cos ^2 \phi \sin ^2 \theta}
  • D γ1−sin⁡2θcos⁡2ϕ\gamma \sqrt{1-\sin ^2 \theta \cos ^2 \phi}

Step-by-Step Verified Solution

amp;OP→=xi^+yj^+zk^amp;OQ→=xi^+yj^amp;∣OP∣=γ=x2+y2+z2amp;cos⁡θ=xx2+y2⇒cos⁡2θ=x2γ2−z2=x2γ2−γ2cos⁡2ϕamp;cos⁡ϕ=zx2+y2+z2=zγamp; Distance of P from x-axis =y2+z2amp;d=γ2−x2amp;⇒x2=γ2sin⁡2ϕcos⁡2θamp;⇒d=γ2−γ2sin⁡2ϕcos⁡2θamp;=γ1−sin⁡2ϕcos⁡2θ\begin{aligned}& \overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k} \\& \overrightarrow{O Q}=x \hat{i}+y \hat{j} \\& |O P|=\gamma=\sqrt{x^2+y^2+z^2} \\& \cos \theta=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow \cos ^2 \theta=\frac{x^2}{\gamma^2-z^2}=\frac{x^2}{\gamma^2-\gamma^2 \cos ^2 \phi} \\& \cos \phi=\frac{z}{\sqrt{x^2+y^2+z^2}}=\frac{z}{\gamma} \\& \text { Distance of } P \text { from } x \text {-axis }=\sqrt{y^2+z^2} \\& d=\sqrt{\gamma^2-x^2} \\& \Rightarrow x^2=\gamma^2 \sin ^2 \phi \cos ^2 \theta \\& \Rightarrow d=\sqrt{\gamma^2-\gamma^2 \sin ^2 \phi \cos ^2 \theta} \\& =\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}\end{aligned}

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