The vector equation of the plane passing through the points `hati+hatj-2hatk,2hati-hatj+hatk` and `hati+hatj+hatk`, is

Options

A. `vecr.(9hati+3hatj-hatk)=-14`
B. `vecr.(9hati+3hatj-hatk)=14`
C. `vecr.(3hati+9hatj-hatk)=14`
D. none of these

Answer

B

Explanation

Let A,B,C be the points with position vectors
`hati+hatj-2hatk,2hati-hatj+hatk` and `hati+2hatj+hatk` respectively. Then,
`vec(AB)=hati-2hatj+3hatk` and `vec(AB)=-hati+3hatj+0hatk`
The vector normal to the plane containing points A,B and C is
`vecn=vec(AB)xxvec(AC)=|(hati,hatj,hatk),(1,-2,3),(-1,3,0)|=-9hati-3hatj+hatk`
The required plane passes through the point having position vector `veca=hati+hatj-2hatk` and is normal to the vector `vecn=-9hati-3hatj+hatk`.
So its vector equation is
`impliesvecr.vecn=veca.vecnimpliesveci.(9hati+3hatj-hatk)=14`