The vector equation of the plane containing he line `vecr=(-2hati-3hatj+4hatk)+lamda(3hati-2hatj-hatk)` and the point `hati+2hatj+3hatk` is

Options

A. `vecr.(hati+3hatk)=10`
B. `vecr.(hati-3hatk)=10`
C. ` vecr.(3hati+hatk)=10`
D. none of these

Answer

A

Explanation

The required plane passes through the points `P(-2hati-3hatj+4hatk)` and `Q(hati+2hatj+3hatk)` and is parallel to the vector `vecb=3hati-2hatj-hatk`.
So it is normal to the vector
`vecn=vec(PQ)xxvecb=|(veci,vecj,veck),(-3,-5,1),(3,-2,-1)|=7hati+21hatk`
So, equation of the plane is
`{vecr-(-2hati-3hatj4hatk)}.(7hati+21hatk)=0impliesvecr.(hati+3hatk)=10`