Question

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 :

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

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

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

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

Answer 1

Correct option is (1) \(\gamma \sqrt{1-\sin ^{2} \phi \cos ^{2} \theta}\)

OQ = OP cos (90 – ϕ)

\( \mathrm{OR}=\mathrm{OQ} \cos \theta\)

\(\mathrm{OR}=\mathrm{OP} \sin \phi \cos \theta\)

\(\mathrm{OR}=\gamma \sin \phi \cos \theta\)

\(\mathrm{PR}=\sqrt{\gamma^2-\mathrm{OR}^2}\)

\(\mathrm{PR}=\gamma \sqrt{1-\sin ^2 \phi \cos ^2 \theta}\)