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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Three Dimensional Geometry

Let P(x, y, z) be a point in the first octant, whose projection in...

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

Options

  1. Aγ1−sin⁡2ϕcos⁡2θ\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}
  2. Bγ1−sin⁡2θcos⁡2ϕ\gamma \sqrt{1-\sin ^2 \theta \cos ^2 \phi}
  3. Cγ1+cos⁡2ϕsin⁡2θ\gamma \sqrt{1+\cos ^2 \phi \sin ^2 \theta}
  4. Dγ1+cos⁡2θsin⁡2ϕ\gamma \sqrt{1+\cos ^2 \theta \sin ^2 \phi}

Correct answer

A. γ1−sin⁡2ϕcos⁡2θ\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}

Step-by-step solution

P(x,y,z),Q(x,y,O);x2+y2+z2=γ2OQ‾=xi^+yj^cos⁡θ=xx2+y2cos⁡ϕ=zx2+y2+z2⇒sin⁡2ϕ=x2+y2x2+y2+z2\begin{aligned} & \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z}), \mathrm{Q}(\mathrm{x}, \mathrm{y}, \mathrm{O}) ; \mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2=\gamma^2 \\ & \overline{\mathrm{OQ}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}} \\ & \cos \theta=\frac{\mathrm{x}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2}} \\ & \cos \phi=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}} \\ & \Rightarrow \sin ^2 \phi=\frac{\mathrm{x}^2+\mathrm{y}^2}{\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2}\end{aligned} distance of P\mathrm{P} from x\mathrm{x}-axis y2+z2\sqrt{\mathrm{y}^2+\mathrm{z}^2}⇒γ2−x2⇒γ1−x2γ2=γ1−cos⁡2θsin⁡2ϕ\begin{aligned} & \Rightarrow \sqrt{\gamma^2-x^2} \Rightarrow \gamma \sqrt{1-\frac{x^2}{\gamma^2}} \\ & =\gamma \sqrt{1-\cos ^2 \theta \sin ^2 \phi}\end{aligned}$$
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