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

Options

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

Answer

C

Explanation

We have `vecr = (1+lambda-mu)hati+(2-lambda)hatj+(3-2lambda+2mu)hatk`
`rArr vecr=(hati+2hatj+3hatk)+lambda(hati-hatj-2hatk)+mu(-hati+2hatk)`
This is a plane passing through `veca = hati+2hatj+3hatk` and parallel to the vectors `vecb=hati-hatj-2hatk` and `vecc=-hati+2hatk`
Therefore, it is perpendicular to the vector `vecn=vecb xx vecc = -2hati-hatk`
Hence, its vector equation is
`(vecr-veca)=veca.vecn`
`rArr vecr. (-2hati-hatk)=-2-3`
`rArr vecr.(2hati+hatk)=5`
So, the Cartesian equation is
`(xhati+yhatj+zhatk).(2hati+hatk)=5`
or `2x+z=5`