Let P be the foot of the perpendicular from the point Q(10,-3,-1) on the line x-37= y-2-1= z+1-2. Then the area of the right angled triangle PQR, where R is the point (3,-2,1), is
- 30
- 9 15
- 3 30
- 8 15
Answer
(3) 3 30
Solution
Foot of perpendicular: P=(7 +3, 2- , -1-2 ) with QP (7,-1,-2)
7(7 -7)-(5- )+4 =0 =1, P=(10,1,-3)
[figure]
QP=(0,4,-2), QR=(-7,1,2) QP QR=0, so the right angle is at Q
QP=2 5, QR=3 6 Area = 12(2 5)(3 6)=3 30
Answer: (3)