The distance of the line x-2 y-6 z-3 y-2 z+3 2 = 3 = 4 from the point (1, 4, 0) along the line x 1 = 2 = 3 is :
Step-by-Step Solution
Key Concept: Apply the core result for lines and planes in three dimensions and simplify using the given constraints.
Line passing through (1, 4, 0) and parallel to y-2 y-4 is L : x z+3 x-1 z = = = = 1 2 3 1 2 3 (3) Any point on L : (\lambda + 1, 2\lambda + 4, 3\lambda) x-2 y-6 z-3 Any point on 2 = 3 = 4 is (2\mu + 2, 3\mu+ 6, 4\mu + 3) ⎫ ⎪ ⎪ ⎪ ⎪ \lambda + 1 = 2\mu + 2 ⎬ \lambda = 1\mu = 0 2\lambda + 4 = 3\mu + 6 ⎪ ⎪ ⎪ ⎭ ⎪ 3\lambda = 4\mu + 3 Point: (2, 6, 3) 2 2 2 Distance = \sqrt(2 - 1) + (6 - 4) + (3 - 0) = \sqrt1 + 4 + 9 = \sqrt14
Correct Answer: 3