<p>Given the plane containing the line \(\dfrac{x-3}{2} = \dfrac{y+2}{-1} = \dfrac{z-1}{3}\) and also containing its projection on the plane \(2x + 3y - z = 5\). The normal vector of required plane is \((2\hat{i} - \hat{j} + 3\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})\). Which of the following points satisfy the required plane equation?</p>
Step-by-Step Solution
Key Concept: The plane containing a line and its projection onto another plane has a normal vector equal to the cross product of the given line's direction vector and the normal vector of the projection plane. This plane must contain both the original line and all projected points.
Step 1: Identify the direction vector of the given line: d = (2, -1, 3) Step 2: Identify the normal vector of projection plane 2x + 3y - z = 5: n_1 = (2, 3, -1) Step 3: Calculate the normal of the required plane using cross product: N = (2, -1, 3) × (2, 3, -1) = i ((-1)(-1) - (3)(3)) - j ((2)(-1) - (3)(2)) + k ((2)(3) - (-1)(2)) = i (1 - 9) - j (-2 - 6) + k (6 + 2) = -8 i + 8 j + 8 k = 8(-1, 1, 1) Step 4: The plane equation is: -x + y + z = d. Using point (3, -2, 1) from the line: -3 - 2 + 1 = -4 Step 5: Plane equation: -x + y + z = -4 or x - y - z = 4 Verify with answer choices by substituting points to check which satisfies this equation. ∴ Answer: B
Correct Answer: B