Let the point Q be the image of the point P(2, 1, -5) in the line L: x-1 2 = y-2 1 = z+1 -2 . If R( , , ) is a point on the line L such that > 0 and the area of the triangle PQR is 18 , then + + is equal to

Step-by-step solution

Let M be the foot of the perpendicular from P(2, 1, -5) to the line L . Any point on L can be written as M(2t+1, t+2, -2t-1) . The direction ratios of PM are (2t-1, t+1, -2t+4) . Since PM is perpendicular to L , we have: 2(2t-1) + 1(t+1) - 2(-2t+4) = 0 9t - 9 = 0 t = 1 Thus, M = (3, 3, -3) . The distance PM is: PM = (3-2)^2 + (3-1)^2 + (-3+5)^2 = 1 + 4 + 4 = 3 Since Q is the image of P in L , M is the midpoint of PQ , and PQ L . The length of the base of PQR is PQ = 2PM = 6 . The area of PQR is given by 1 2 PQ MR ,