3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
$P$ is a point on the plane $lx + my + nz = p$. A point $Q$ is taken on the line $OP$ such that $OP.OQ = p^2$. Then the locus of $Q$ is:
$p(lx + my + nz) = x^2 + y^2 + z^2
$(lx + my + nz)(x^2 + y^2 + z^2) = p^2
$lx + my + nz = (x^2 + y^2 + z^2)p
None of these
Step-by-Step Solution
Key Concept: Section formula combined with the plane equation and distance condition determines the locus algebraically.
Let $OP : PQ = \lambda : 1$, so $P = \left(\frac{\lambda a}{\lambda - 1}, \frac{\lambda b}{\lambda - 1}, \frac{\lambda c}{\lambda - 1}\right)$. Since $P$ lies on plane $lx + my + nz = p$, we get $\frac{\lambda}{\lambda - 1} = \frac{p}{la + mb + nc}$. Using $OP \cdot OQ = p^2$ yields the locus $x^2 + y^2 + z^2 = (lx + my + nz)p$.
Correct Answer: 1