If vec{A} = hat{i} A cos theta + hat{j} A sin | vedclass

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(C) Two vectors are perpendicular if their dot product is zero.Let $\vec{B} = B_x \hat{i} + B_y \hat{j}$.For $\vec{A} \cdot \vec{B} = 0$,we have $(A \

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

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(C) Two vectors are perpendicular if their dot product is zero.Let $\vec{B} = B_x \hat{i} + B_y \hat{j}$.For $\vec{A} \cdot \vec{B} = 0$,we have $(A \cos 	heta)(B_x) + (A \sin 	heta)(B_y) = 0$.This implies $B_x \cos 	heta = -B_y \sin 	heta$,or $\frac{B_x}{B_y} = -\frac{\sin 	heta}{\cos 	heta}$.Thus,we can choose $B_x = B \sin 	heta$ and $B_y = -B \cos 	heta$.Therefore,$\vec{B} = \hat{i} B \sin 	heta - \hat{j} B \cos 	heta$.

If $\vec{A} = \hat{i} A \cos \theta + \hat{j} A \sin \theta$ is a vector,then another vector $\vec{B}$ which is perpendicular to $\vec{A}$ is given by:

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