If `vecb and vecc` are any two mutually perpendicular unit vectors and `veca` is any vector, then `(veca.vecb)vecb+(veca.vecc)vecc+(veca.(vecbxxvecc))/(|vecbxxvecc|^2)(vecbxxvecc)=` (A) 0 (B) `veca (C) `veca/2` (D) `2veca`

Answer

`veca`

Explanation

Let `vecalpha, vecbeta,vecgamma` be any three mutually perpendicular non-coplanar, unit vectors and `veca` be any vector, then
`veca= (veca.vecalpha)vecalpha+ (veca.vecbeta)+(veca.vecgamma)vecgamma`
Here `vecb, vecc` are two mutually perpendicular vectors,
therefore, `vecb , vecc and (vecb xx vecc)/(|vecb xx vecc|)` are three mutually
Perpendicular non-coplanaar unit vectors. Hence
`veca=(veca .vecb)vecb+(veca.vecc)vecc`
`+(veca.(vecbxxvecc)/(|vecb xx vecc|))(vecb xx vecc)/(|vecb xx vecc|)`
`(veca.vecb)vecb+(veca.vecc)vecc`
`+(veca.(vecbxxvecc))/(|vecb xx vecc|^(2))(vecbxxvecc)`