3D Geometry
Intersection of spheres
Grade 12
Question:
<p>Two spheres \(x^2 + y^2 + z^2 + 7x - 2y - z - 13 = 0\) and \(x^2 + y^2 + z^2 - 3x + 3y + 4z - 8 = 0\) intersect. The plane of intersection is:</p>
<p>\(10x - 5y - 5z - 5 = 0\)</p>
<p>\(2x - y - z = 1\)</p>
<p>\(10x - 5y - 5z = 0\)</p>
<p>Both (1) and (2)</p>
Step-by-Step Solution
Key Concept: When two spheres intersect, their plane of intersection is obtained by subtracting one sphere equation from the other, which eliminates the quadratic terms and yields a linear equation representing the radical plane.
Step 1: Write the two sphere equations: S_1: x^2 + y^2 + z^2 + 7x - 2y - z - 13 = 0 S_2: x^2 + y^2 + z^2 - 3x + 3y + 4z - 8 = 0 Step 2: Subtract S_2 from S_1 to eliminate quadratic terms: (x^2 + y^2 + z^2 + 7x - 2y - z - 13) - (x^2 + y^2 + z^2 - 3x + 3y + 4z - 8) = 0 Step 3: Simplify: 7x - (-3x) - 2y - 3y - z - 4z - 13 - (-8) = 0 7x + 3x - 2y - 3y - z - 4z - 13 + 8 = 0 10x - 5y - 5z - 5 = 0 Step 4: Divide by 5: 2x - y - z - 1 = 0 ∴ Answer: D (The plane of intersection is 2x - y - z - 1 = 0)
Correct Answer: D