<p>Two lines \(\dfrac{x-3}{1} = \dfrac{y+1}{3} = \dfrac{z-6}{-1}\) and \(\dfrac{x+5}{7} = \dfrac{y-2}{-6} = \dfrac{z-3}{4}\) intersect at the point R. The reflection of R in the \(xy\)-plane has coordinates:</p>
Step-by-Step Solution
Key Concept: Two lines intersect when a point satisfies both parametric equations simultaneously. The reflection of a point in the xy-plane negates only the z-coordinate while keeping x and y unchanged.
Step 1: Set up parametric equations Line 1: x = 3+s, y = -1+3s, z = 6-s Line 2: x = -5+7t, y = 2-6t, z = 3+4t Step 2: Find intersection point R At intersection: 3+s = -5+7t, -1+3s = 2-6t, 6-s = 3+4t From equation 1: s = -8+7t Substitute in equation 2: -1+3(-8+7t) = 2-6t -1-24+21t = 2-6t 27t = 27 ⟹ t = 1 Therefore s = -8+7(1) = -1 Step 3: Find coordinates of R x = 3+(-1) = 2, y = -1+3(-1) = -4, z = 6-(-1) = 7 So R = (2, -4, 7) Step 4: Reflect R in xy-plane Reflection in xy-plane: (x, y, z) → (x, y, -z) Reflection of R = (2, -4, -7) ∴ Answer: A
Correct Answer: A