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`

If `vecb and vecc` are any two mutually perpendicular unit vectors and

1.	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» Correct Answer - `veca` 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)`

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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`

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