JEE Main 2024 · 8 April · morning · Single correct · +4 / −1
Let P(x, y, z) be a point in the first octant, whose projection in the x y-plane is the point Q. Let O P=γ; the angle between O Q and the positive x-axis be θ; and the angle between O P 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)
- (b) γ √(1+cos ² θ sin ² φ)
- (c) γ √(1+cos ² φ sin ² θ)
- (d) γ √(1-sin ² θ cos ² φ)
Answer and solution
Answer (a) γ √(1-sin ² φ cos ² θ)
Solution
Maths sir · 4 steps
Write P in spherical form first; then the distance from the x-axis is just √(OP² - x²).
- Height and shadow. φ is measured from the z-axis, so the height is z = γ cos φ. The shadow on the xy-plane has length OQ = γ sin φ.
- Split the shadow. OQ makes angle θ with the x-axis, so x = γ sin φ cos θ and y = γ sin φ sin θ.
- Distance from x-axis. The foot of the perpendicular from P to the x-axis is (x, 0, 0), so the distance is √(y² + z²). Since x² + y² + z² = γ², this equals √(γ² - x²).
√(y²+z²) = √(γ² - γ²sin² φ cos² θ) = γ√(1-sin² φ cos² θ)
- So. The distance of P from the x-axis is γ√(1-sin² φ cos² θ), which is option (a).
The idea to keep. Distance of a point from the x-axis is √(y²+z²), which is √(OP² - x²). With the angle from the z-axis, sin φ goes into the plane and cos φ goes up.
Answer from the NTA final answer key. Solution by PrepWiser.