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

Correct option is 2i^B sinθ−j^Bcosθ

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Let $\vec{A} = \hat{i} A \cos θ + \hat{j} A s i n θ$ be any vector. Another vector $\vec{B}$ which is perpendicular to $\vec{A}$ can be expressed as :

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i^B cosθ−j^Bsinθ

i^B sinθ−j^Bcosθ

i^B cosθ+j^Bsinθ

i^B sinθ+j^Bcosθ

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detailed solution

Correct option is B

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

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

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material

Let $\vec{A} = \hat{i} A \cos θ + \hat{j} A s i n θ$ be any vector. Another vector $\vec{B}$ which is perpendicular to $\vec{A}$ can be expressed as :