<p>The image of the line \(\dfrac{x-1}{3} = \dfrac{y-3}{1} = \dfrac{z-4}{-5}\) in the plane \(2x - y + z + 3 = 0\) is the line</p>
<p>\(\dfrac{x-3}{3} = \dfrac{y+5}{1} = \dfrac{z-2}{-5}\)</p>
<p>\(\dfrac{x-3}{-3} = \dfrac{y+5}{-1} = \dfrac{z-2}{5}\)</p>
<p>\(\dfrac{x+3}{3} = \dfrac{y-5}{1} = \dfrac{z-2}{-5}\)</p>
<p>\(\dfrac{x+3}{-3} = \dfrac{y-5}{-1} = \dfrac{z+2}{5}\)</p>
Step-by-Step Solution
Key Concept: To find the image of a line in a plane, reflect two distinct points on the line across the plane, then find the line through their images. The reflected line will be parallel to the original if the original is parallel to the plane.
Step 1: Identify a point on the given line. Let P_1 = (1, 3, 4) (at parameter t=0). Step 2: Find another point on the line. At t=1: P_2 = (1+3, 3+1, 4-5) = (4, 4, -1). Step 3: Reflect P_1 = (1, 3, 4) in plane 2x - y + z + 3 = 0. The normal vector is n = (2, -1, 1). The perpendicular from P_1 is: (x,y,z) = (1,3,4) + λ(2,-1,1). Substituting in plane equation: 2(1+2λ) - (3-λ) + (4+λ) + 3 = 0 → 2+4λ-3+λ+4+λ+3 = 0 → 6λ+6=0 → λ=-1. Image P_1' = (1-2, 3+1, 4-1) = (-1, 4, 3). Step 4: Reflect P_2 = (4, 4, -1) in the plane. (x,y,z) = (4,4,-1) + μ(2,-1,1). Substituting: 2(4+2μ) - (4-μ) + (-1+μ) + 3 = 0 → 8+4μ-4+μ-1+μ+3 = 0 → 6μ+6=0 → μ=-1. Image P_2' = (4-2, 4+1, -1-1) = (2, 5, -2). Step 5: Find the line through P_1' = (-1, 4, 3) and P_2' = (2, 5, -2). Direction vector: P_2' - P_1' = (3, 1, -5). The image line is: $\dfrac{x+1}{3} = \dfrac{y-4}{1} = \dfrac{z-3}{-5}$ ∴ Answer: B
Correct Answer: B