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Question: Let \(\overset{\rightarrow}{A} = \widehat{i}A\cos\theta + \widehat{j}A\sin\theta\) be any vector. An...

Let A→=i^Acos⁡θ+j^Asin⁡θ\overset{\rightarrow}{A} = \widehat{i}A\cos\theta + \widehat{j}A\sin\theta be any vector. Another vector B→\overset{\rightarrow}{B} which is normal to A is

A

i^Bcos⁡θ+jBsin⁡θ\widehat{i}B\cos\theta + jB\sin\theta

B

i^Bsin⁡θ+jBcos⁡θ\widehat{i}B\sin\theta + jB\cos\theta

C

i^Bsin⁡θ−jBcos⁡θ\widehat{i}B\sin\theta - jB\cos\theta

D

i^Bcos⁡θ−jBsin⁡θ\widehat{i}B\cos\theta - jB\sin\theta

Answer

i^Bsin⁡θ−jBcos⁡θ\widehat{i}B\sin\theta - jB\cos\theta

Explanation

Solution

Dot product of two perpendicular vector will be zero.