KCET2015MathematicsDeterminants
Equation of line passing through the point ( (2,3,1) ) and parallel to the line of intersection of the plane ( x-2 y-z+5=0 ) and ( x+y+3 z=6 ) is
Options
- A( x-2 5 = y-3 -4 = z-1 3 )
- B( x-2 -5 = y-3 -4 = z-1 3 )
- C( x-2 5 = y-3 4 = z-1 3 )
- D( x-2 4 = y-3 3 = z-1 2 )
Correct answer
B. ( x-2 -5 = y-3 -4 = z-1 3 )
Step-by-step solution
Given equation of planes, [ array l P₁: x-2 y-z+5=0 (1) P₂: x+y+3 z=6 (2) array ] and point ( P(2,3,1) ). Normal vector of Eq. (1) is given by, [ N ₁= i -2 j + k (3) ] Similarly, normal vector of Eq. (2) is given by [ N ₂= i + j +3 k (4) ] Vector perpendicular to the normal to the plans are: [ aligned b &= N ₁ N ₂= | array ccc i & j & k 1 & -2 & 1 1 & 1 & 3 array | &= i (-6+1)- j (3+1)+ k (1+2)=-5 i -4 j +3 k aligned ] So, required equation of line passing through the point ( (2,3,1) ) is, [ x-2 -5 = y-3 -4 = z-1 3