How Do Squared Vector Components Relate to the Dot Product?

[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} = \sumr_{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...

 

[Astronuc]

in LaTeX, use \sqrt for square root.

yes, if one does \vec{r}\cdot\vec{r}, one is doing r₁² + r₂² + r₃²

http://hyperphysics.phy-astr.gsu.edu/hbase/vsca.html#vsc1

http://mathworld.wolfram.com/DotProduct.html

http://mathworld.wolfram.com/InnerProduct.html

http://en.wikipedia.org/wiki/Dot_product

 

[Oblio]

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

 

[Astronuc]

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

 

[FedEx]

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

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}