<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 \(x - y + z = 3\). The co-ordinates of \(Q\) are:</p>
<p>(A) \((1, 2, 0)\)</p>
<p>(B) \((2, 0, 1)\)</p>
<p>(C) \((-1, 0, 4)\)</p>
<p>(D) \((4, 0, -1)\)</p>
Step-by-Step Solution
Key Concept: A point P on the given line projects to Q on the plane x+y+z=3, where Q must also satisfy x-y+z=3. The key is that PQ is perpendicular to the plane (parallel to its normal vector (1,1,1)), so P = Q + t(1,1,1) for some t, and both P and Q satisfy their respective constraints.
Step 1: Find the intersection line of the two planes x+y+z=3 and x-y+z=3. Subtracting: 2y = 0 ⟹ y = 0 So points on intersection have form: x + z = 3, y = 0 ⟹ Q = (a, 0, 3-a) for parameter a. Step 2: Since the perpendicular from P (on the given line) to the plane has direction vector (1,1,1) (normal to x+y+z=3), we have P = Q + t(1,1,1) = (a+t, t, 3-a+t). Step 3: Point P lies on the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z}{1} = s$ (say). So P = (1+2s, -1-s, s). Step 4: Equating both expressions for P: a+t = 1+2s ... (i) t = -1-s ... (ii) 3-a+t = s ... (iii) Step 5: From (ii): t = -1-s. Substitute in (i): a-1-s = 1+2s ⟹ a = 2+3s. Substitute t = -1-s in (iii): 3-a-1-s = s ⟹ 2-a = 2s ⟹ a = 2-2s. Step 6: From 2+3s = 2-2s: 5s = 0 ⟹ s = 0. Thus a = 2. Step 7: Therefore Q = (2, 0, 1). ∴ Answer: B
Correct Answer: B