I'm reviewing some basics in the textbook Electromagnetics with Applications and the worked problem 1-6-1 has some ambiguity i hope you can help me with.
The section is simply describing 3 dimensional vectors in Cartesian coordinates. The text shows a figure that is very similar to the figure i show below (copied from here).
The problem adds 2 vectors with a result of $\hat{x}10$ – $\hat{y}4$ + $\hat{z}0$. It asks for the angle this vector makes with respect to the x-axis, $\alpha$.
I calculated it as $\alpha = cos^-{^1}\frac{10}{\sqrt{116}} = 21.8⁰$, but the textbook gives the answer as $\alpha=-21.8⁰$.
I think what they did was to actually calculate the angle to the x-axis that the projection of the vector makes with the x-y plane, which is clearly -21.8⁰.
FINALLY, my question: What constitutes a negative angle $\alpha$ in the figure above?
Best Answer
Consider you've only $x$-axis and $\vec v$ and nothing else (consider them in one plane). Now, you just have to measure the angle ($\alpha$) between these two. If it is obtuse then its $\cos$ will be negative.