The square of the distance of the point of intersection of the lines r= ( i+ j- k )+ (a i- j ), a 0 and r= (4 i- k )+ (2 i+a k ) from the origin is :
- 5
- 10
- 17
- 26
Answer
(3) 17
Solution
Line 1: (1+a , 1- , -1); line 2: (4+2 , 0, -1+a )
Equating y: 1- =0 =1
Equating z: -1=-1+a =0 ( a 0)
Equating x: 1+a=4 a=3
Point of intersection =(4, 0, -1)
Square of its distance from O =42+02+(-1)2=17
Answer: (3)