<p>The centre of the circle given by \(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 15\) and \(|\vec{r} - (\hat{j} + 2\hat{k})| = 4\) is</p>
Step-by-Step Solution
Key Concept: The first equation represents a plane, and the circle is the intersection of this plane with the sphere defined by the second equation. The center of the circle lies on the plane and is the foot of perpendicular from the sphere's center to the plane.
<p><strong>Step 1:</strong> Identify the sphere and plane. The equation |<strong>r</strong> - (<strong>j</strong> + 2<strong>k</strong>)| = 4 represents a sphere with center C = (0, 1, 2) and radius R = 4. The equation <strong>r</strong> · (<strong>i</strong> + 2<strong>j</strong> + 2<strong>k</strong>) = 15 represents a plane with normal vector <strong>n</strong> = (1, 2, 2).</p><p><strong>Step 2:</strong> Find the distance from sphere's center to the plane. Using the distance formula: d = |0(1) + 1(2) + 2(2) - 15|/√(1² + 2² + 2²) = |0 + 2 + 4 - 15|/√9 = |-9|/3 = 3</p><p><strong>Step 3:</strong> The center of the circle is the foot of perpendicular from C(0, 1, 2) to the plane. The perpendicular direction is along the unit normal: <strong>u</strong> = (1, 2, 2)/3. The center of the circle is: (0, 1, 2) + 3 · (1/3, 2/3, 2/3) = (0, 1, 2) + (1, 2, 2) = (1, 3, 4)</p><p>∴ Answer: C = (1, 3, 4)</p>
Correct Answer: C