# Polar coordinates unit vector conversion confusion

My question stems from the following graphic:

It aims to explain the following relationship:

$$\hat r = cos\ \theta \ \hat i + sin\ \theta \ \hat j$$ $$\hat \theta = -sin\ \theta \ \hat i + cos\ \theta \ \hat j$$

Now, judging from the graphic, $\hat r$ is derived from considering trigonometric properties of the triangle formed by hypotenuse $\hat r$ and its sine and cosine components as its legs. Since $\theta$ is at the tail of $\hat r$, its $x$ component is $cos \ \theta$ and its y component $sin \ \theta$. However, going by the logic of the derivation of $\hat r$, I'd expect the $x$ component of $\hat \theta$ to be $cos\ \theta_2$ and $y$ to be $sin\ \theta_2$. Note that I also added the subscript $2$ to differentiate it between the $theta$ used in finding the components of $\hat r$, but it highlights my two main confusions.

• Why is theta taken to be at the head of $\hat \theta$ so that its $y$ and $x$ components are the way they are, and if that's not why $x$ is $sin\ \theta$ and $y$ is $cos\ \theta$, what's the reason?
• Sine and cosine $\theta$ were written twice in each triangle in question in this graphic. Are they the same magnitude? And if not, why haven't they been distinguished by something like $\theta$ and $\theta_2$?

I feel like its highlighting a key trigonometric property I've forgotten, but since I don't know exactly what I'm not understanding I can't look up such a detail.

$\hat \theta$ is a unit vector which is at right angles to the unit vector $\hat r$.
If this is so then $\hat \theta \cdot \hat r =0$

So you need to find a vector $\hat \theta = a \hat x + b \hat j$ such that the conditions $a^2+b^2 =1$ and $\hat \theta \cdot \hat r =0$ are satisfied.
You will find that there are two unit vectors $\hat \theta$ which satisfy those conditions and by convention the vector which is pointing in the anti clockwise is the one which is used.

I have annotated your diagram to try and show you what is going on.

You have a unit vector $\hat r$ in green with components $(+\cos \theta, +\sin \theta)$ shown in violet.

The unit vector $\hat r$ can be rotated with the locus of the head of the vector shown as a dashed black line which is an arc of a circle of radius one.

As you want $\hat \theta$ shown in red to be at right angles to $\hat r$ it has components $(-\sin \theta, +\cos \theta)$ shown in blue.

Simple geometry can be used to show that all angles $\theta$ are the same because $\hat \theta$ is perpendicular to $\hat r$.

• Thanks so much, this was really helpful. If I understand correctly, would it be correct to say that the $\theta$ used to derive the components of $\hat \theta$ is the angle next to the head of $\hat \theta$, since it's the only candidate that could be the same $\theta$ used in $\hat r$? – sangstar Sep 25 '17 at 17:37
• @sangstar I have labelled that angle $\theta$. – Farcher Sep 25 '17 at 17:45
• Right, and that's because the other possible angle candidate has to be $90 - \theta$ since its complimentary to the angle of $\hat r$? – sangstar Sep 25 '17 at 17:47