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Mathematics · 3D Geometry

JEE Main 2024 — 8 April, Shift 1 — Question 4

Let P(x,y,z)\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z}) be a point in the first octant, whose projection in the xy-plane is the point Q . Let OP=γ\mathrm{OP}=\gamma; the angle between OQ and the positive xx-axis be θ\theta; and the angle between OP and the positive zz-axis be ϕ\phi, where O is the origin. Then the distance of P from the x -axis is :

  1. Option A:

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

    Correct
  2. Option B:

    γ1+cos⁡2θsin⁡2ϕ\gamma \sqrt{1+\cos ^{2} \theta \sin ^{2} \phi}

  3. Option C:

    γ1−sin⁡2θcos⁡2ϕ\gamma \sqrt{1-\sin ^{2} \theta \cos ^{2} \phi}

  4. Option D:

    γ1+cos⁡2ϕsin⁡2θ\gamma \sqrt{1+\cos ^{2} \phi \sin ^{2} \theta}

Answer: A

Step-by-step solution

P(x,y,z),Q(x,y,O);x2+y2+z2=γ2P(x, y, z), Q(x, y, O) ; x^{2}+y^{2}+z^{2}=\gamma^{2}

OQ‾=xi˙^+yj^\overline{\mathrm{OQ}}=x \hat{\dot{i}}+y \hat{j}

cos⁡θ=xx2+y2\cos \theta=\frac{x}{\sqrt{x^{2}+y^{2}}}

cos⁡ϕ=zx2+y2+z2\cos \phi=\frac{\mathrm{z}}{\sqrt{\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}}}

⇒sin⁡2ϕ=x2+y2x2+y2+z2\Rightarrow \sin ^{2} \phi=\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}}

distance of P from x -axis y2+z2\sqrt{\mathrm{y}^{2}+\mathrm{z}^{2}}

⇒γ2−x2⇒γ1−x2γ2\Rightarrow \sqrt{\gamma^{2}-x^{2}} \Rightarrow \gamma \sqrt{1-\frac{x^{2}}{\gamma^{2}}} =γ1−cos⁡2θsin⁡2ϕ=\gamma \sqrt{1-\cos ^{2} \theta \sin ^{2} \phi}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Mathematics
Chapter
3D Geometry
Topic
Direction Cosines and Direction Ratios

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