Let `vec(alpha)=ahati+bhatj+chatk, vec(beta)=bhati+chatj+ahatk` and `vec(gamma)=chati+ahatj+bhatk` be three coplnar vectors with `a!=b`, and `vecv=hati+hatj+hatk`. Then `vecv` is perpendicular to

Options

A. `vecalpha`
B. `vecbeta`
C. `vecgamma`
D. none of these

Answer

a,b,c

Explanation

It is given that `vecalpha, vecb and vecgamma` are coplanar vectors, therefore
` [vecalpha vecbeta vecgamma] =0`
`Rightarrow |{:(a,b,c),(b,c,a),(c,a,b):}|=0`
` or 3abc -a^(3) -b^(3) -c^(3) =0`
` or a^(3) +b^(3) +c^(3) -3abc =0`
` or (a + b +C0 (a^(2) +b^(2) + c^(2) -ab -bc-ca)=0`
or a+b+c =0
` [ a^(2) +b^(2) +c^(2)-ab -bc-ca ne 0)`
` Rightarrow vecv. vecalpha = vecv. vecbeta = vecv. vecgamma = 0`
Hence, `vecv` is perpendicular to `vecalpha ,vecbeta and vecgamma`