Let a line L1 pass through the origin and be perpendicular to the lines L2 : r=(3+t) i+(2t-1) j+(2t+4) k and L3 : r=(3+2s) i+(3+2s) j+(2+s) k, t, s R. If (a, b, c), a Z, is the point on L3 at a distance of 17 from the point of intersection of L1 and L2, then (a+b+c)2 is equal to .
Answer
4
Solution
Directions: L2: (1, 2, 2), L3: (2, 2, 1)
L1 is perpendicular to both: vmatrix i & j & k\ 1 & 2 & 2\ 2 & 2 & 1 vmatrix=-2 i+3 j-2 k
1: (x, y, z)= (-2, 3, -2)
L1 2: 3+t=-2 , 2t-1=3 , 2t+4=-2
First and third 3+t=2t+4 t=-1, =-1 (second: -3=-3, true)
P(2, -3, 2)
Point of L3: Q(3+2s, 3+2s, 2+s); PQ2=17
(1+2s)2+(6+2s)2+s2=17 9s2+28s+20=0 (s+2)(9s+10)=0
s=- 109 gives a= 79 (rejected); s=-2 gives Q(-1, -1, 0)
a+b+c=-2 (a+b+c)2=4
Answer: 4