3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
If a variable point $P$ moves such that the line passing through $P$ and $Q(0, 0, 2)$ makes an angle $60°$ with $z$-axis, then locus of $P$ is:
x^2 - y^2 - 3(z - 2)^2 = 0
x^2 - y^2 + 3(z - 2)^2 = 0
x^2 + y^2 - 3(z - 2)^2 = 0
x^2 + y^2 + 3(z - 2)^2 = 0
Step-by-Step Solution
Key Concept: The locus of points on lines through a fixed point making a constant angle with an axis forms a cone with that point as vertex.
Point $P(x,y,z)$ lies on a line through $(0,0,2)$ making $60°$ with the $z$-axis. The direction vector satisfies $\frac{(xi+yj+(z-2)k)\cdot k}{\sqrt{x^2+y^2+(z-2)^2}}=\cos 60°$. This simplifies to $\frac{z-2}{\sqrt{x^2+y^2+(z-2)^2}}=\frac{1}{2}$, which gives $x^2+y^2-3(z-2)^2=0$ or $x^2+y^2-3(z-2)^2=0$.
Correct Answer: 3