Let A be the point of intersection of the lines L1: x-71= y-50= z-3-1 and L2: x-13= y+34= z+75. Let B and C be the points on the lines L1 and L2 respectively such that AB=AC= 15. Then the square of the area of the triangle ABC is :
- 63
- 57
- 60
- 54
Answer
(4) 54
Solution
Any point of L1 is (7+t, 5, 3-t) and any point of L2 is (1+3s, -3+4s, -7+5s).
At the point of intersection, 5=-3+4s, so s=2; then 7+t=1+6 gives t=0, and z checks: 3-0=-7+10=3.
A=(7, 5, 3)
The directions are b1= i- k and b2=3 i+4 j+5 k, so
= |3+0-5| 2 50= 210= 15
= 1- 125= 245
B lies on L1 and C on L2, BAC= or - ; in both cases BAC= 245. Hence
Area = 12 AB AC BAC= 12 15 245= 3 242
(Area)2= 9 244=54
Answer: (4)