Question
If the straight lines x = 1 + s, y = –3 – λs, z = 1 + λs and x = t/ 2 , y = 1 + t, z = 2 – t with parameters s and t respectively, are co-planar then λ equals
-
–2
-
–1
-
-1/2
-
0
Solution
A.
–2
Given 
are coplanar then plan passing through these lines has normal perpendicular to these lines
⇒ a - bλ + cλ = 0 and a/2 +b -c =0 (where a, b, c are direction ratios of the normal to the plan) On solving, we get λ = -2.
