Let P be the image of the point Q(7,-2,5) in the line L: x-12= y+13= z4 and R(5,p,q) be a point on L. Then the square of the area of PQR is .
Answer
957
Solution
L: (2 +1, 3 -1, 4 ); R at x=5 =2 R=(5,5,8).
Foot T of the perpendicular from Q: QT (2,3,4)=0 29 -29=0 =1, T=(3,2,4)
[figure]
T is the mid-point of QP and RT PQ Area = 12 2 QT RT=QT RT
QT2=16+16+1=33, RT2=4+9+16=29
(Area)2=33 29=957
Answer: 957