The vector equation of the plane `r=(2hati+hatk)+lamda(hati)+mu(hati+2hatj-3hatk)` in scalar product formm is `r*(3hati+2hatk)=alpha`, then `alpha`= . . .

Options

A. 2
B. 3
C. 1
D. 0

Answer

A

Explanation

Given, equation of plane
`r=(2hati+hatk)+lamdahati+mu(hati+2hatj-3hatk)` . . . (i)
Here, plane (i) passing through a (let)=`2hati+hatk`
and parallel to vector b(let)=`hati` and
`c=hati+2hatj-3hatk`
We know that equation of plane passing through a point a and parallel to non-parallel vectors b and c is
`r*(bxxc)=a*(bxxc)=["a b c"]`
Now, `["a b c"]=|(2,0,1),(1,0,0),(1,2,-3)|=2(0)-0+1(2-0)=2`
and `bxxc=|(hati,hatj,hatk),(1,0,0),(1,2,-3)|=3hati+2hatk`
`thereforer*(3hati+2hatk)=2`
therefore, `alpha=2`.