The Cartesian equation of the plane `vecr=(1+lamda-mu)hati+(2-lamda)hatj+(3-2lamda+2mu)hatk` is

Options

A. `2x+y=5`
B. `2x-y=5`
C. `2x+z=5`
D. `2x-z=5`

Answer

c

Explanation

Given plane is
`vecr=(1+lamda-mu)hati+(2-lamda)hatj+(3-2lamda+mu)hatk`
`=(hati+2hatj+3hatk)+lamda(hati-hatj-2hatk)+mu(-hati+2hatk)`
which is a plane passing through `veca=hati+2hatj+3hatk`
and parallel to the vectors `vecb=hati+hatj+3hatkandvecc=-hati+2hatk`
Therefore, it is perpendicular to the vector `vecn=vecbxxvecc=-2hati+hatk`
Hence, equation of plane is `-2(x-1)=(0)(y-2)-(z-3)=0or2x+z=5`