I have two vectors in 3D Cartesian coordinate space such that each vector is described by components .
I would like to find the angle between these two vectors in the xy coordinate plane, ie. viewed from an observer located on the z axis.
Can I simply set the z component of each vector to zero and calculate the angle between them using the definition of the dot product? If not, what do I do to project the 3D vectors onto the 2D xy plane?
Best Answer
Yes, you can.
Let's look at a more general case, where the plane is perpendicular to some unit vector $\hat{n}$, and your vectors are $\vec{p}$ and $\vec{q}$.
Projecting the vectors to the plane yields $$\vec{p}' = \vec{p} - \hat{n}\left(\hat{n} \cdot \vec{p}\right) \tag{1}\label{1}$$ and $$\vec{q}' = \vec{q} - \hat{n}\left(\hat{n} \cdot \vec{q}\right) \tag{2}\label{2}$$
If we consider $\hat{n} = ( 0, 0, 1 )$ and $\vec{p} = ( x_p , y_p , z_p )$, we find that $\vec{p}' = ( x_p , y_p , 0 )$.
So, yes, projecting to the $xy$ plane is the same thing as zeroing out the $z$ component.
(As a reminder to those who have forgotten, the angle $\theta$ between any two three-dimensional vectors fulfills $$\cos(\theta) = \frac{\vec{p}' \cdot \vec{q}'}{\left\lVert\vec{p}'\right\rVert \left\lVert\vec{q}'\right\rVert} \qquad \text{and} \qquad \sin(\theta) = \frac{\left\lVert\vec{p}' \times \vec{q}'\right\rVert}{\left\lVert\vec{p}'\right\rVert \left\lVert\vec{q}'\right\rVert}$$ which are quite often useful.)