# Clarification on a dot product

1. Sep 20, 2007

### Oblio

This technically a homework question, but needed for homework and understanding for homework to come. Just hope to get it cleared up, thanks again!

1. This formula is given:

r$$_{1}$$s$$_{1}$$ + r$$_{2}$$s$$_{s}$$+ r$$_{e}$$s$$^{3}$$ = $$\sum$$r$$_{n}$$s$$_{n}$$ (with the limits etc. not too important).

Then, in respect to scalar products, the magnitude of any vector is denoted by l r l or by Pythoagora's theorem: square root of[r(1)$$^{2}$$ + r(2)$$^{2}$$ + r(3)$$^{2}$$]
(couldn't find square root in latex)

and that THAT is the same as square root of [r . r]
This last step I do not follow...

2. Sep 20, 2007

### Astronuc

Staff Emeritus
Last edited: Sep 20, 2007
3. Sep 20, 2007

### Oblio

I know that it's true, (in reading my text) but how does 3 squared terms boil down to r^2?

4. Sep 20, 2007

### Astronuc

Staff Emeritus
This is a vector operation.

One squares the corresponding vector components represented in each of three dimensions, which are orthogonal in the Cartesian system.

The magnitude r is given as the sqrt of the sum of the squares, i.e. r = sqrt (r12 + r22 + r32), so

r2 = r12 + r22 + r32

Last edited: Sep 21, 2007
5. Sep 20, 2007

### FedEx

See $$\vec{r}\cdot\vec{r}$$ is nothing but $$\sca{r}\sca{r}\cos\theta$$. But $$\theta$$ = 0 and so $$\cos\theta$$ = 1.

Hence $$\vec{r}\cdot\vec{r}$$ = $$^{}r^2$$

Now a projection of a vector on an axis is known as the component of the vector on that axis. So take a cartesian system and draw any arbritary vector and drop perpendiculars on the x y and z axis. Now try using pythagoras theorem for this and you will see that how

$$\sca{r^2}$$ = $$\sqrt{r^2_x + r^2_y + r^2_z}$$