JEE Main · 2004 · 24 Apr 2004 · Offline · Three Dimensional Geometry

The intersection of the spheres

x^{2} + y^{2} + z^{2} + 7 x - 2 y - z = 13

and

x^{2} + y^{2} + z^{2} - 3 x + 3 y + 4 z = 8

is the same as the intersection of one of the sphere and the plane

(A) $x - y - z = 1$
(B) $x - 2 y - z = 1$
(C) $x - y - 2 z = 1$
(D) $2 x - y - z = 1$ ✓ Correct

Answer: (D)

Solution

The intersection of two spheres lies on the plane found by subtracting their equations, since both share the same

x^{2} + y^{2} + z^{2}

part. Subtracting the second from the first gives

(7 x - 2 y - z) - (- 3 x + 3 y + 4 z) = 13 - 8

, that is

10 x - 5 y - 5 z = 5

. Dividing by

5

, the plane is

2 x - y - z = 1

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option

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