Let L1: r=( i- j+2 k)+ ( i- j+2 k), R, L2: r=( j- k)+ (3 i+ j+p k), R, and L3: r= ( i+m j+n k), R be three lines such that L1 is perpendicular to L2 and L3 is perpendicular to both L1 and L2. Then, the point which lies on L3 is
- (1,7,-4)
- (1,-7,4)
- (-1,7,4)
- (-1,-7,4)
Answer
(3) (-1,7,4)
Solution
L1 L2: (1)(3)+(-1)(1)+(2)(p)=0 p=-1
L3 both L3 d1 d2= vmatrix i& j& k\\1&-1&2\\3&1&-1 vmatrix=- i+7 j+4 k
∵ L3 passes through the origin, its points are (-1,7,4)
=1 (-1,7,4) lies on L3
Answer: (3)