The distance of the point (2,4,0) from the point of intersection of the lines x+6 3 = y 2 = z+1 1 and x-7 4 = y-9 3 = z-4 2 is

Step-by-step solution

A point on the first line can be expressed in parametric form as x = 3 - 6 , y = 2 , z = - 1 . The second line has parameterization x = 4 + 7 , y = 3 + 9 , z = 2 + 4 . Equating coordinates at the intersection point yields the system: 3 - 4 = 13 2 - 3 = 9 - 2 = 5 From the third equation, = 2 + 5 . Substituting into the second gives 2(2 + 5) - 3 = 9 , which simplifies to = -1 . Then = 2(-1) + 5 = 3 . These values satisfy all equations, giving the intersection point I(3, 6, 2) . The distance from Q(2, 4, 0) to I is (3