Let P be the foot of the perpendicular from the point Q(10,−3,−1) on the line 7x−3=−1y−2=−2z+1. Then the area of the right angled triangle PQR , where R is the point (3,−2,1), is
(A)915
(B)30
(C)815
(D)330✓ Correct
Answer: (D)
Solution
The line is 7x−3=−1y−2=−2z+1=λ, so a general point on it is (7λ+3,−λ+2,−2λ−1).
For this point to be the foot of the perpendicular from Q(10,−3,−1), the vector from this point to Q must be perpendicular to the line's direction (7,−1,−2):