# Prove that if $p_1 \times p_2$ is positive, then $p_1$ is clockwise from $p_2$?

In Introduction to Algorithms (CLRS), Exercise 33-1-1, we are asked to prove that if $$p_1 \times p_2$$ is positive then $$p_1$$ is clockwise from $$p_2$$ and if it's negative, then $$p_1$$ is counter-clockwise from $$p_2$$.

This is a picture from CLRS

I found the following proof: https://sites.math.rutgers.edu/~ajl213/CLRS/Ch33.pdf

The proof above says to consider the angles both vectors make with the x-axis. We know the angle is $$\arctan y/x$$. We also know the cross product $$p_1 \times p_2 = x_1y_2 - y_1x_2$$. If this product is greater than zero, then this means that $$y_2/x_2 > y_1/x_1$$. Since $$\arctan$$ is monotone then the angle $$p_2$$ makes with the x-axis is larger. which means you need to move clockwise direction to get from $$p_2$$ to $$p_1$$

BUT the question is asking to prove that $$p_1$$ is clockwise relative $$p_2$$. What am I missing? is the above proof wrong?

I drew the following based on the proof above:

• You have been correct all the time. Except in the very last moment. Instead of "therefore, $p_2$ is clockwise relative to $p_1$", it should have been, "therefore, $p_2$ is counterclockwise relative to $p_1$." (I assume that $x_1>0$ and $x_2>0$ in the illustration.) Apr 14, 2020 at 1:42

In two-dimensional space, $$p_1 \times p_2$$ is equal to $$|p_1||p_2|\sin\theta$$, where $$\theta$$ is the angle from $$p_1$$ to $$p_2$$. Depending which textbook you're using, this is probably the definition of the two-dimensional cross product—I'm not sure what definition Introduction to Algorithms uses.
$$|p_1|$$ and $$|p_2|$$ both have to be non-negative (by the definition of vector magnitude), so if neither of them is zero, the sign of the whole expression will be the sign of $$\sin\theta$$.
Now, how do we define "clockwise"? The easiest definition I know is is, $$p_1$$ is clockwise from $$p_2$$ if the angle(*) from $$p_1$$ to $$p_2$$ is less than 180 degrees. Try a few examples and you'll see how this definition works—remember that angles are traditionally measured with counter-clockwise being positive.
Now, we know that if $$p_1 \times p_2$$ is positive, then $$\sin\theta$$ must be positive. And if $$\sin\theta$$ is positive, then $$\theta$$ must be less than 180 degrees(*). So by the definition of clockwise, $$p_1$$ is clockwise from $$p_2$$.