If `vecu, vecv, vecw` are non -coplanar vectors and `p,q,` are real numbers then the equality
`[3vecu p vecv p vecw]-[p vecv vecw qvecu]-[2vecw-qvecv qvecu]=0` holds for

Options

A. exactly one value of `(p,q)`
B. exactly two values of `(p,q)`
C. more than two but not all values of `(p,q)`
D. all values of `(p,q)`

Answer

A

Explanation

We have
`[3vecu p vecv p vecw]-[pvecv vecw qvecr]-[2vecw q vecv q vecu]=0`
`implies3p^(2)[vecu vecv vecw]-pq[vecv vecw vecu]-2q^(2)[vecw vecv vecu]=0`
`implies3p^(2)[vecu vecv vecw]-pq[vecu vecv vecw]+2q^(2)[vecu vecv vecw]=0`
`=(3p^(2)-pq+2q^(2))[vecu vecv vecw]=0`
`implies3p^(2)-pq+2q^(2)=0 [ :'[vecu vecv vecw]!=0]`
`impliesp^(2)-1/3p +2/3q^(2)=0`
`=(p-q/6)^(2)+23/36q^(2)=0impliesp-q/6=0, q=0impliesp=0,q=0`
Hence there is exactly one value of `(p,q)`