<p>Given \(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 15\) (a plane) and \(|\vec{r} - (\hat{j} + 2\hat{k})| = 4\) (a sphere with centre \((0,1,2)\) and radius 4). The centre of the circle formed by their intersection is:</p>
Step-by-Step Solution
Key Concept: The intersection of a plane and sphere is a circle whose center lies on the perpendicular from the sphere's center to the plane. Find this perpendicular foot using the plane equation.
Step 1: Identify the sphere center C = (0, 1, 2) and radius R = 4. The plane equation is: x + 2y + 2z = 15, with normal vector n = (1, 2, 2). Step 2: The center of the intersection circle lies on the line through C perpendicular to the plane. Parametric form: r (t) = (0, 1, 2) + t(1, 2, 2) = (t, 1+2t, 2+2t) Step 3: This point must satisfy the plane equation: t + 2(1+2t) + 2(2+2t) = 15 t + 2 + 4t + 4 + 4t = 15 9t + 6 = 15 t = 1 Step 4: Substitute t = 1: Center = (1, 1+2, 2+2) = (1, 3, 4) Verification: Distance from (0,1,2) to plane = |0 + 2(1) + 2(2) - 15|/√(1+4+4) = |-9|/3 = 3 ✓ ∴ Answer: B (The center of the intersection circle is (1, 3, 4))
Correct Answer: B