I would like to know how I can extend the angle between two vectors in 3D space to the x-y plane. So, there are two vectors in 3D space, and the angle between them is found using the definition of the dot product. I would like to somehow get the value of the angle that would result from extending the angle between the two vectors to the x-y plane.
I hope that what I'm asking can be followed.
Thank you.
Edit:
Sorry, I find it hard to word what I am after. I'll try and explain:
The two 3D vectors have some plane in common on which they both lie. I would like the angles (lieing on that plane), that the vectors each make with the x-y plane.
If that makes no sense, here's another attempt:
Imagine an arc between the two vectors in 3D space. That arc can represent the angle between the vectors (which can be found using the dot product). Now extend that arc until it intercepts the x-y plane. Now the arc represents an angle I'm after.
Edit 2:
Image representing what I'm after (I'm after the angle represented by the smallest arc, and the angle represented by the largest arc (denoted by theta)). i44.tinypic.com/2a80wgw.png
Best Answer
If I read your diagram correctly, the plane defined by vectors $v_1$ and $v_2$ intersects the $x$-$y$ plane at a vector that I'll call $v_3$, the magnitude of which is irrelevant. You want the angle from $v_3$ to $v_2$ and also the angle from $v_3$ to $v_1$. So you can define $v_3 = v_1 + \alpha v_2$ where $\alpha$ is such that the $z$-component of $v_3$ is zero. This value $\alpha$ is well defined given that the $z$-component of $v_2$ is not zero. Once you find $v_3$, you know that you will use the dot product to find the angles.
(At first I thought, if you wanted the angle between the two vectors' projections onto the XY-plane, you could just set each z-component to zero and take the dot product. Of course there is no such angle if either vector has x-component and y-component equal to zero.)