Let →A=^iAcosθ+^jAsinθ be any vector. Another vector →B which is normal to→A is
∴y=±Bcosθ
Now for y=Bcosθ, from equation (iii),
⇒x=−Bsinθ
Substituting x and y in equation (i), we get,
⇒→B=−^iBsinθ+^jBcosθ
Now for y=−Bcosθ, from equation (iii),
x=Bsinθ
Substituting x and y in equation (i),
we get,
→B=^iBsinθ−^jBcosθ
Final answer: (c).