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Dot product geometric proof

  1. Nov 5, 2011 #1
    Dot product proof question?

    I'm having trouble understanding the proof of the dot product in three dimensions (not using the cosine rule approach).

    Here's what I have for the 2D proof:
    u = u1 i + u2 j
    v = v1 i + v2 j
    u.v = u1v1 + u2v2
    u.v = |u| |v| cos(θ)
    => u1v1 + u2v2 = |u| |v| cos(θ)
    x = XOV - XOU
    => u1v1 + u2v2 = |u| |v| cos(XOV - XOU)
    u1v1 + u2v2 = |u| |v| (cos(XOV)cos(XOU) + sin(XOV)sin(XOU))
    u1v1 + u2v2 = |u| |v| ( u2/|v| * u1/|u| + v2/|v| * v1/|u|)
    u1v1 + u2v2 = |u| |v| ( u1u2/(|v||u|) + v1v2/(|v||u|) )
    u1v1 + u2v2 = u1v1 + u2v2

    Now when I go to do it in 3D:
    u = u1 i + u2 j + u3 z
    v = v1 i + v2 j + u3 z
    u.v = u1v1 + u2v2 + u3v3
    u.v = |u| |v| cos(θ)
    u1v1 + u2v2 = |u| |v| ( sqrt(u1^2+v1^2)/|u| * sqrt(u2^2+v2^2)/|v| + u3/|u| * v3/|v|)
    u1v1 + u2v2 = sqrt(u1^2+v1^2) * sqrt(u2^2+v2^2) + u3v3

    and it doesn't seem to work out. The sqrt(...) parts are because I tried to find the angle between the vector and the x-y plane. I think I found the angle between the two vectors in three dimension incorrectly. Also, I reckon that I should be able to use my 2D proof for the 3D proof.

    P.S. I am aware of the cosine rule approache to proving it, but I don't really like that method. There should be a way to go straight from:
    u1v1 + u2v2 + u3v3
    |u| |v| cos(θ)
  2. jcsd
  3. Nov 5, 2011 #2
    I think of it this way. Say you want to take the dot product of some vector v with i. It's clear what it's doing. It just gives you the first component.

    So, you take the projection of v1 onto i, which is what you want to prove. Then, it's clear what happens when you take the dot product of v1 with a multiple of i. Again, you get what you want. Same for v dot j and v dot k.

    Now, if you want to take v1 dot some vector v2, well, v2 is just a combination of i, j, and k. And the dot product is distributive.

    So, what you need is that if you project a vector onto a sum of two (or three, in this case) other vectors, it's the same thing as projecting onto each of the vectors and then adding the result.

    And what does the thing you are trying to prove say?

    It says that the dot product is obtained by projecting v1 onto v2 and then multiplying the lengths of the projection and v2.

    So, that's basically it.

    Maybe that's a little hard to follow. I leave it as an exercise to think about it until it makes sense.

    Intuition and understanding is much more valuable than pushing symbols around. Once you understand, then you will know how to push the symbols around to prove what you want.
  4. Nov 6, 2011 #3
    I feel like I 'get' the dot product; my issue is, I can't seem to show it mathematically.
  5. Nov 6, 2011 #4
    But I think I did show it mathematically, though, if you pursue the argument all the way. Just focus on one step at a time. First, prove it when u is some vector and v = i. Just focus on that. One step at a time.
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