# There is a formula such as

1. Oct 7, 2004

### 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.

2. Oct 7, 2004

### Tide

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

3. Oct 7, 2004

### 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.

Last edited: Oct 7, 2004
4. Oct 7, 2004

### 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.

5. Oct 7, 2004

### arildno

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

6. Oct 9, 2004

### 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))]