Let {\P}( {\x}, {\y}, {\z}) be a point in the first octant, whose projection in the xy-plane is the point Q. Let {\OP}=\gamma; the angle between {\OQ} and the positive x-axis be \theta; and the angle between O P and the positive {\z}-axis be \phi, where {\O} is the origin. Then the distance of {\P} from the {\x}-axis is : [8 Apr 2024 Shift 1]

Correct Answer is : γ√1−sin2φcos2θ

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

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

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

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

JEE Main 8 Apr 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 4 of 90

Marks: +1, -0

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 :

[8 Apr 2024 Shift 1]

γ√1−sin‌2φcos2θ
γ√1+cos2θsin‌2φ
γ√1−sin‌2θcos2φ
γ√1+cos2φsin‌2θ

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