Let overrightarrow{A} = hat{i}Acostheta + hat | vedclass

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(C) Two vectors are normal (perpendicular) to each other if their dot product is zero.Let $\overrightarrow{B} = \hat{i}B_x + \hat{j}B_y$.For $\overrig

Vedclass · Accepted Answer

$\hat{i}B\sin	heta - \hat{j}B\cos	heta$

Vedclass · Answer

(C) Two vectors are normal (perpendicular) to each other if their dot product is zero.Let $\overrightarrow{B} = \hat{i}B_x + \hat{j}B_y$.For $\overrightarrow{A} \cdot \overrightarrow{B} = 0$:$(A\cos	heta)(B_x) + (A\sin	heta)(B_y) = 0$.If we choose $\overrightarrow{B} = B(\hat{i}\sin	heta - \hat{j}\cos	heta)$,then:$\overrightarrow{A} \cdot \overrightarrow{B} = (A\cos	heta)(B\sin	heta) + (A\sin	heta)(-B\cos	heta) = AB\sin	heta\cos	heta - AB\sin	heta\cos	heta = 0$.Thus,the vector $\hat{i}B\sin	heta - \hat{j}B\cos	heta$ is normal to $\overrightarrow{A}$.

Let $\overrightarrow{A} = \hat{i}A\cos\theta + \hat{j}A\sin\theta$ be any vector. Another vector $\overrightarrow{B}$ which is normal to $\overrightarrow{A}$ is

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