Let A→=i^Acosθ+j^A sin θ  be any vector. Another vector B→  which is perpendicular to A→  can be expressed as :

A→=i^A cosθ+j^Asinθ For B→ to be⊥ to A→      B→.A→=BAcos900=0       B→.A→=(i^Bsinθ−j^B cosθ) .(i^Acosθ−j^A sinθ)              =AB sinθ cosθ−BAsinθcosθ=0Hence, B→=i^Bsinθ−j^Bcosθ