3D Geometry
Sphere
Grade 12
Question:
<p>The intersection of the spheres \(x^2 + y^2 + z^2 + 7x - 2y - z = 13\) and \(x^2 + y^2 + z^2 - 3x + 3y + 4z = 8\) is the same as the intersection of one of the sphere and the plane</p>
<p>\(x - y - z = 1\)</p>
<p>\(x - 2y - z = 1\)</p>
<p>\(x - y = 2z = 1\)</p>
<p>\(2x - y - z = 1\)</p>
Step-by-Step Solution
Key Concept: The intersection of two spheres lies on a plane obtained by subtracting their equations. This plane passes through the circle of intersection and can replace either sphere in determining the intersection locus.
Step 1: Write the two sphere equations:
S_1: x^2 + y^2 + z^2 + 7x - 2y - z = 13
S_2: x^2 + y^2 + z^2 - 3x + 3y + 4z = 8 Step 2: Subtract S_2 from S_1 to eliminate quadratic terms:
(x^2 + y^2 + z^2 + 7x - 2y - z) - (x^2 + y^2 + z^2 - 3x + 3y + 4z) = 13 - 8
7x - 2y - z + 3x - 3y - 4z = 5
10x - 5y - 5z = 5 Step 3: Simplify the plane equation:
2x - y - z = 1 Step 4: Verify the concept: The intersection of spheres S_1 and S_2 (which is a circle) equals the intersection of the plane 2x - y - z = 1 with either sphere S_1 or S_2. This plane is the radical plane of the two spheres. ∴ The required plane is 2x - y - z = 1 (Answer: A)
Correct Answer: A