Question

Find the intersection of the spheres ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}+7x-2y-z=13$ and ${{x}^{2}}+{{y}^{2}}+{{z}^{2}}-3x+3y+4z=8$

• A. $x-y-z=1$
• B. $x-2y-z=1$
• C. $x-y-2z=1$
• D. $2x-y-z=1$

Answer 1

Answer: (D)

$2x-y-z=1$

Answer 2

The correct option is(D): 2 x − y − z =1.

given by x² + y² + z² + 7x - 2y - z - 13 = 0 and x² + y² + z² - 3x + 3y + 4z - 8 = 0. If these spheres intersect, then the equation S - S' = 0 represents the equation of their common intersection plane.

Therefore, by subtracting the second sphere equation from the first, we get:

(x² + y² + z² + 7x - 2y - z - 13) - (x² + y² + z² - 3x + 3y + 4z - 8) = 0

Simplifying further:

x² + y² + z² + 7x - 2y - z - 13 - x² - y² - z² + 3x - 3y - 4z + 8 = 0

This reduces to:

10x - 5y - 5z - 5 = 0

And finally:

2x - y - z = 1.