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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

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What do you mean by "extending" the angle? How do you extend an angle?! –  user2468 Feb 28 '12 at 4:20
    
There is no explanation of what you mean by "extending" and of how the 3D relates to the XY-plane. If you don't edit your question I don't think there's much we can do. If you didn't have 1 rep I'd be downvoting at the moment, but instead I'll just way welcome to MSE and try to be clear next time if you want a clear answer =) –  Patrick Da Silva Feb 28 '12 at 4:22
    
I've edited the question. –  user968243 Feb 28 '12 at 4:42
    
I rolled back to your original question; I hope that's what you wanted. –  Rahul Mar 14 '12 at 0:43
    
Yep, thank you! –  user968243 Mar 14 '12 at 0:50
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2 Answers 2

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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.)

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Thank you for the reply. Unfortunately, I don't have enough 'reputation' to add in image to my explanation. Here is a link to a diagram of what I'm after: I'm after theta (the third arc) and I'm also after the angle represented by the first arc. i44.tinypic.com/2a80wgw.png Thanks. –  user968243 Feb 28 '12 at 5:12
    
If I read it 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. –  minopret Feb 28 '12 at 5:39
    
Thank you. That's exactly what I'm after! I'd vote you up but I can't. Sorry. –  user968243 Feb 28 '12 at 5:41
    
I can add that comment as a separate answer and if you're happy with it, I think you can mark it as "accepted". –  minopret Feb 28 '12 at 5:43
    
Don't worry about it, I've already marked your original answer as accepted! –  user968243 Feb 28 '12 at 6:09
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You question is reasonable to think about, I myself did realized this point of view until a professor tells me one thing:

"Let's evaluate the angle between two vector in a Hilbert space (not necessary finite dimension or Hilbert at all), since you have two vectors you have the plane generated by them this plane has an inner product this plane can be considered unless an isometry the well known $\mathbb{R}^2$"

And sincerely I don't have a more elegant answer than that, which tell us that the important is the abstract aspect of $\mathbb{R}^2$.

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Hm, I'm still puzzled by the meaning of "extending" an angle! –  user2468 Feb 28 '12 at 4:43
    
It is not extending any angle the definition is the same in $\mathbb{R}^2$ since it is an abstract entity that is isometric to any plane in a Hilbert space, it should be considered the space of the vectors that you want to evaluate the angle. –  checkmath Feb 28 '12 at 4:50
    
Did my edit help understand the angle I'm after @J.D.? I'm not sure if @chessmath and I are on the same page... –  user968243 Feb 28 '12 at 4:53
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