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Question
Let →A=^iAcosθ+^jAsinθ , be any vector. Another vector →B which is normal to →A is :-
Let →A=^iAcosθ+^jAsinθ , be any vector. Another vector →B which is normal to →A is :-
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Solution
The correct option is C ^iBsinθ−^jBcosθ
We know that any vector perpendicular to a vector a^i+b^j can be written as b^i−a^j, such that their dot product becomes zero.
Thus, a vector perpendicular to →A=^iAcosθ+^jAsinθ can be written as →B=^iBsinθ−^jBcosθ since their dot product = ABcosθsinθ−ABsinθcosθ=0
We know that any vector perpendicular to a vector a^i+b^j can be written as b^i−a^j, such that their dot product becomes zero.
Thus, a vector perpendicular to →A=^iAcosθ+^jAsinθ can be written as →B=^iBsinθ−^jBcosθ since their dot product = ABcosθsinθ−ABsinθcosθ=0
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