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

(A) $\gamma \sqrt{1-\sin ^{2} \phi \cos ^{2} \theta}$

(B) $\gamma \sqrt{1+\cos ^{2} \theta \sin ^{2} \phi}$

(C) $\gamma \sqrt{1-\sin ^{2} \theta \cos ^{2} \phi}$

(D) $\gamma \sqrt{1+\cos ^{2} \phi \sin ^{2} \theta}$

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Three Dimensional Geometry

Let

P (x, y, z)

be a point in the first octant, whose projection in the xy -plane is the point Q . Let

OP = γ

; the angle between OQ and the positive x -axis be

θ

; and the angle between OP and the positive z -axis be

ϕ

, where O is the origin. Then the distance of P from the x -axis is :

(A) $γ 1 - sin^{2} ϕ cos^{2} θ$ ✓ Correct
(B) $γ 1 + cos^{2} θ sin^{2} ϕ$
(C) $γ 1 - sin^{2} θ cos^{2} ϕ$
(D) $γ 1 + cos^{2} ϕ sin^{2} θ$

Answer: (A)

Solution

With

O P = γ

cos θ = \frac{x}{x ^{2} + y ^{2}}

and

cos ϕ = \frac{z}{γ}

, so

sin^{2} ϕ = \frac{x ^{2} + y ^{2}}{γ ^{2}}

. The distance of

P

from the x-axis is

y^{2} + z^{2} = γ 1 - \frac{x ^{2}}{γ ^{2}} = γ 1 - sin^{2} ϕ cos^{2} θ

. Option (1).

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Single correct option

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