<p>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 \(2x + 3y - z = 5\), contains which one of the following points?</p>
Step-by-Step Solution
Key Concept: The plane containing a line and its projection on another plane is perpendicular to the given plane and contains the original line. Find this plane by using the normal vectors of the given plane and the direction of the line.
Step 1: The line passes through point A(3, -2, 1) with direction vector d = (2, -1, 3). Step 2: The plane containing the line and its projection must be perpendicular to the given plane 2x + 3y - z = 5. The normal to the given plane is n_1 = (2, 3, -1). Step 3: The normal to our required plane is perpendicular to both d (direction of original line) and n_1 (normal of given plane). Step 4: Normal vector: n = d × n_1 = (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, 8, 8) or simplified (-1, 1, 1) Step 5: Plane equation using point A(3, -2, 1) and normal (-1, 1, 1): -1(x - 3) + 1(y + 2) + 1(z - 1) = 0 -x + 3 + y + 2 + z - 1 = 0 -x + y + z + 4 = 0 or x - y - z = 4 Step 6: Check which option satisfies this plane equation. ∴ Answer: B
Correct Answer: B