<p>The plane containing the line \(\dfrac{x-1}{1} = \dfrac{y-2}{2} = \dfrac{z-3}{3}\) and parallel to the line \(\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}\) passes through the point</p>
Step-by-Step Solution
Key Concept: A plane containing a given line must satisfy the line's point and direction vector. For the plane to be parallel to another line, that line's direction vector must be parallel to the plane (perpendicular to the plane's normal). Use the cross product of two direction vectors to find the normal vector.
Step 1: Identify given information. The plane contains line L_1: (x-1)/1 = (y-2)/2 = (z-3)/3 with direction d_1 = (1,2,3) and point P_1 = (1,2,3). The plane is parallel to line L_2: (x)/1 = (y)/1 = (z)/1 with direction d_2 = (1,1,1). Step 2: The plane contains L_1, so d_1 lies in the plane. The plane is parallel to L_2, so d_2 is parallel to the plane. Therefore, the normal vector n is perpendicular to both: n = d_1 × d_2 = (1,2,3) × (1,1,1) = (2·1 - 3·1, 3·1 - 1·1, 1·1 - 2·1) = (-1, 2, -1) Step 3: The plane equation using point (1,2,3) and normal (-1,2,-1):
-1(x-1) + 2(y-2) - 1(z-3) = 0
-x + 1 + 2y - 4 - z + 3 = 0
-x + 2y - z = 0 or x - 2y + z = 0 Step 4: Check which given point satisfies x - 2y + z = 0. The answer is the point that satisfies this equation. ∴ Answer: D
Correct Answer: D