# Doubt in polar components of a vector

#### Ozma

##### New member
I'm having troubles in understanding how my physics textbook finds the polar components of the vector $$\displaystyle \vec{p}$$ (I've uploaded two images): by book introduces two orthogonal vectors $$\displaystyle \hat{u}_r$$ and $$\displaystyle \hat{u}_\theta$$ and says that $$\displaystyle \vec{p}=p \cos \theta \hat{u}_r-p \sin \theta \hat{u}_\theta$$.
If I'm not wrong, the components should be the projections of $$\displaystyle \vec{p}$$ onto $$\displaystyle \hat{u}_r$$ and $$\displaystyle \hat{u}_\theta$$, that is $$\displaystyle \vec{p}=p_r \hat{u}_r+p_\theta \hat{u}_\theta$$ (in red and green in the second picture), but while I'm fine with the fact that the projection onto $$\displaystyle \hat{u}_r$$ is $$\displaystyle p_r=p \cos \theta$$ I don't understand the minus sign in the projection onto $$\displaystyle \hat{u}_\theta$$ is $$\displaystyle p_\theta=-p \sin \theta$$.
The main doubt is the fact that since by hypothesis $$\displaystyle \hat{u}_r$$ and $$\displaystyle \hat{u}_\theta$$ are orthogonal, the angle between $$\displaystyle \vec{p}$$ and $$\displaystyle \hat{u}_\theta$$ should be $$\displaystyle \theta+\pi/2$$, so the angle between the extension of $$\displaystyle \vec{p}$$ (the one dashed under $$\displaystyle \vec{p}$$ in my drawing) and $$\displaystyle \hat{u}_\theta$$ should be $$\displaystyle \pi-\pi/2-\theta=\pi/2-\theta$$ and so $$\displaystyle p_\theta=p \cos (\pi/2-\theta)=p \sin \theta$$; so I don't get the minus sign. Where am I doing a mistake?

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• Mel Gibson