Trigonometric EquationHard
Question
The intersection of the spheres x2 + y2 + z2 + 7x - 2y - z = 13 and x2 + y2 + z2 - 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane
Options
A.x - y - z = 1
B.x - 2y - z = 1
C.x - y - 2z = 1
D.2x - y - z = 1
Solution
Required plane is S1 - S2 = 0
where S1 = x2 + y2 + z2 + 7x - 2y - z - 13 = 0 and
S2 = x2 + y2 + z2 - 3x + 3y + 4z - 8 = 0
⇒ 2x - y - z = 1.
where S1 = x2 + y2 + z2 + 7x - 2y - z - 13 = 0 and
S2 = x2 + y2 + z2 - 3x + 3y + 4z - 8 = 0
⇒ 2x - y - z = 1.
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