<p>A perpendicular is drawn from a point on the line \(\dfrac{x-1}{2} = \dfrac{y+1}{-1} = \dfrac{z}{1}\) to the plane \(x + y + z = 3\) such that the foot of the perpendicular \(Q\) also lies on the plane \(x - y + z = 3\). Then the co-ordinates of \(Q\) are:</p>
Step-by-Step Solution
Key Concept: A point on the given line can be parameterized; the foot of perpendicular Q must satisfy both plane equations AND lie on the normal direction from some point on the line to the plane x + y + z = 3.
Step 1: Parameterize the line: Point P on the line is (1 + 2t, -1 - t, t) Step 2: Q is the foot of perpendicular from P to plane x + y + z = 3. The normal to this plane is n = (1, 1, 1). Step 3: Q lies on the line through P in direction of normal: Q = (1 + 2t, -1 - t, t) + λ(1, 1, 1) = (1 + 2t + λ, -1 - t + λ, t + λ) Step 4: Q must satisfy x + y + z = 3: (1 + 2t + λ) + (-1 - t + λ) + (t + λ) = 3 → 2t + 3λ = 3 ... (i) Step 5: Q must satisfy x - y + z = 3: (1 + 2t + λ) - (-1 - t + λ) + (t + λ) = 3 → 2 + 2t + λ = 3 → 2t + λ = 1 ... (ii) Step 6: From (ii): λ = 1 - 2t. Substitute in (i): 2t + 3(1 - 2t) = 3 → 2t + 3 - 6t = 3 → -4t = 0 → t = 0 Step 7: When t = 0, λ = 1. Therefore Q = (1 + 0 + 1, -1 - 0 + 1, 0 + 1) = (2, 0, 1) Verification: 2 + 0 + 1 = 3 ✓ and 2 - 0 + 1 = 3 ✓ ∴ Answer: Q = (2, 0, 1)
Correct Answer: B