3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade None
Question:
A variable plane makes intercepts on the co-ordinate axes the sum of whose squares is constant and equal to $k^2$. Then the locus of the foot of the perpendicular from the origin to the plane is
x^2+y^2+z^2)(x^2+y^2+z^2) = k^2
k^2(x^2+y^2+z^2)(x^2+y^2+z^2) = 1
(x^2+y^2+z^2)^2 = \frac{1}{k^2}
None of these
Step-by-Step Solution
Key Concept: The locus of feet of perpendiculars satisfying a distance constraint is found by relating the perpendicular distance formula to the given condition.
Let $P(\alpha, \beta, \gamma)$ be the foot of perpendicular from the origin to the plane. The plane equation is $\alpha(x-\alpha) + \beta(y-\beta) + \gamma(z-\gamma) = 0$. Using the intercept form and the condition $OP \cdot OQ = p^2$, we find that $(\alpha^2 + \beta^2 + \gamma^2)(\alpha^2 + \beta^2 + \gamma^2) = k^2$, leading to the locus $(x^2 + y^2 + z^2)(x^2 + y^2 + z^2) = k^2$.
Correct Answer: 1