- #1

fred4321

- 4

- 0

Hi,

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

Q.E.D.

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

to

|u| |v| cos(θ)