<p>If the plane \(2ax - 3ay + 4az + 6 = 0\) passes through the midpoint of the line joining the centres of the spheres \(x^2 + y^2 + z^2 + 6x - 8y - 2z = 13\) and \(x^2 + y^2 + z^2 - 10x + 4y - 2z = 8\), then <em>a</em> equals</p>
Step-by-Step Solution
Key Concept: Rewrite each sphere equation in standard form to find their centers, then calculate the midpoint of the line joining these centers. Substitute this midpoint into the plane equation to solve for 'a'.
Step 1: Find the center of the first sphere Rewrite x^2 + y^2 + z^2 + 6x - 8y - 2z = 13 in standard form by completing the square: (x^2 + 6x + 9) + (y^2 - 8y + 16) + (z^2 - 2z + 1) = 13 + 9 + 16 + 1 (x + 3)^2 + (y - 4)^2 + (z - 1)^2 = 39 Center C_1 = (-3, 4, 1) Step 2: Find the center of the second sphere Rewrite x^2 + y^2 + z^2 - 10x + 4y - 2z = 8 in standard form: (x^2 - 10x + 25) + (y^2 + 4y + 4) + (z^2 - 2z + 1) = 8 + 25 + 4 + 1 (x - 5)^2 + (y + 2)^2 + (z - 1)^2 = 38 Center C_2 = (5, -2, 1) Step 3: Find the midpoint of C_1 and C_2 Midpoint M = [(-3 + 5)/2, (4 - 2)/2, (1 + 1)/2] = (1, 1, 1) Step 4: Substitute M(1, 1, 1) into the plane equation 2a(1) - 3a(1) + 4a(1) + 6 = 0 2a - 3a + 4a + 6 = 0 3a + 6 = 0 a = -2 ∴ Answer: C (a = -2)
Correct Answer: C