Understanding the Cosine Formula for Vector Multiplication in 3D Space

[bezgin]

a.b= a(x) * b(x) + a(y) * b(y) + a(z) * b(z)
and this is equal to a*b*cosm
where a and b is the magnitude of the vectors in space and m is the angle between them.

I really wonder why this equation is true. I couldn't find the proof anywhere and the teacher didn't show it, he only wrote the formula.

I am a really fanatic of proving such theorems; if you can advise me a book that might catch my interest, I'd be glad.

 

[Tide]

$\vec a \cdot \vec b$ is basically the length of the projection of $\vec b$ onto $\vec a$ and simple geometry establishes it.

 

[bezgin]

Well, it might look simple for you but I can't mathematically prove it on my own. Even in 2 dimensions, it's hard. Let us take two vectors u(a,b) and w(c,d). According to the formula cos [arccos(a/sqrt(a^2+b^2)) - arccos(c/sqrt(c^2+d^2] = (a*c + b*d) / sqrt[(a^2+b^2) * (c^2+d^2)]

From this line, it looks like I've managed to do the 2 dimension proof, the 3 dimensions seem to be impossible. More than 3? It's completely another issue.

 

[Tide]

Here's one way to view it.

Suppose you have two vectors. Place the tail of one (say B) at the head of the other (A). Now extend vector A by drawing a dashed line from its head. This line makes an angle (say theta) with vector B.

Now move along the dashed line until you find a point at which the line from that point to the tip of B is at right angles to the dashed line. The segment of the dashed line from the tip of A to this point is the projection of B onto A and it's length is B cos(theta). Multiply that length by the length of A and you have the scalar or dot product.

The problem in 3 dimensions actually reduces to the two dimensional problem since any two vectors in 3D space are coplanar so you can carry out the calculation on this two dimensional plane.

 

[arildno]

For higher, dimensions, it's best to prove the Cauchy-Schwarz inequality:
$$\frac{|\vec{a}\cdot\vec{b}|}{||\vec{a}||||\vec{b}||}\leq1$$

 

[bezgin]

I proved the schwarz inequality during the lesson but it doesn't help me comprehend the formula for the 3-d dimensions. The schwarz inequality only show us that the result of the equation is within the range of cosine function. (-1 and 1) Can someone prove that the angle between the vectors u(a,b,c) and w(x,y,z) is equal to arccos[(a*x+b*y+c*z)/(sqrt((a^2+b^2+c^2) * (x^2+y^2+z^2))]