3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
A variable line passing through the point $P(0, 0, 2)$ always makes angle $60°$ with $z$-axis, intersects the plane $x + y + z = 1$. Then the locus of point of intersection of the line and the plane is:
$x^2 + y^2 + 3(z - 2)^2 = 0$
$x^2 + y^2 = 3(z + 2)^2$
$x^2 + y^2 = 3(z - 2)^2$
$x^2 + y^2 = 2(z-3)^2$
Step-by-Step Solution
Key Concept: Parametric substitution into a plane equation determines the parameter value and hence the intersection point.
For line $\frac{x}{l} = \frac{y}{m} = \frac{z-2}{1/2} = λ$ with direction cosines $l, m, \cos θ$: substituting parametric form into plane equation gives $λ(l + m + 1/2) = -1$. This yields $λ = \frac{-2}{2l+2m+1}$. The intersection point is $x = \frac{-2l}{2l+2m+1}$, $y = \frac{-2m}{2l+2m+1}$, $z = 2 + \frac{-1}{2l+2m+1}$.
Correct Answer: 3