MHT CET202015 Oct 2020Morning ShiftMathematicsThree Dimensional GeometryActual
The equation of a plane containing the lines r =( +2 -4 k )+ (2 +3 +6 k ) and r =( +3 +4 k )+ ( + - k ) is
Options
- A9 x+8 y+z+11=0
- B9 x-8 y-z-11=0
- C9 x-8 y-z+11=0
- D9 x-8 y+z+11=0
Correct answer
D. 9 x-8 y+z+11=0
Step-by-step solution
Normal vector of a plane would be perpendicular to both the given lines and parallel to their cross product. Now ₁ ₂= | array ccc i & j & k 2 & 3 & 6 1 & 1 & -1 array |=-9 i +8 j - k i.e. 9,-8,1 are d.r. of normal to a plane Let a = i +3 j +4 k and n =9 i -8 j + k r (9 i -8 j + k ) =9(1)+(-8)(3)+4 1=9-24+4 r (9 i -8 j + k ) =-11 9 x -8 y + z +11=0 is the equation of plane.