The correct option is C (→a−→c).(→b×→d)=0
Given:
L1:→r=→a+t→b
L2:→r=→c+s→d
We know that →a,→c are the fixed points lying on the line L1,L2 respectively
And →b,→d are the direction ratios lying of the line L1,L2 respectively which are the two vectors coplanarity condition on plane.
For coplanarity we need third vector lying on the plane which can be →m=→a−→c
Hence for Coplanar Lines L1,L2
⇒→m.(→b×→d)=0
⇒(→a−→c).(→b×→d)=0