The distance of the point (2,3,4) from the line 1-x= y 2 = 1 3 (1+z) is

Step-by-step solution

The given line equation is 1-x = y 2 = 1+z 3 . Rewriting this in standard symmetric form x-x₁ a = y-y₁ b = z-z₁ c , we get x-1 -1 = y-0 2 = z-(-1) 3 . Let the point on the line be P(1- , 2 , 3 -1) . The direction vector of the line is v = - i + 2 j + 3 k . Let A = (2, 3, 4) . The vector AP = (1- -2) i + (2 -3) j + (3 -1-4) k = (-1- ) i + (2 -3) j + (3 -5) k . Since AP is perpendicular to the line, AP v = 0 . (-1)(-1- ) + 2(2 -3) + 3(3 -5) = 0 1+ + 4 -6 + 9 -15 = 0 14 - 20 = 0 = 20 14 = 10 7 . Substituting = 10 7 in