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**krassi_holmz****Real Member**- Registered: 2005-12-02
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What will happen with the vector if we multiply it by

?IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
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Is this a 2-dimensional vector?

What is the coordinate system?

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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OK. Let the vector be 2D.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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It isn't so inportant what is the coordinate system.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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But let

**v**=a**i**+b**j**

(here **i** is vector, not (-1)^(1/2)! (-1)^(1/2) must be italic *i*)

*Last edited by krassi_holmz (2005-12-17 09:18:14)*

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**John E. Franklin****Member**- Registered: 2005-08-29
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Refresh my memory. What happens to the function y=x if you multiply

it by :italic(i)

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I use the following for *italic*

`[i]italic[/i]`

If I tell you it won't be interesting

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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OK. First try to myltiply one point with coordinates {x,y}.

What will happen?

The answer is {*i*x,*i*y}. So?

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**God****Member**- Registered: 2005-08-25
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So it exists in a complex 4-D coordinate system but only exists at the origin on a 2-D real system where the origin of the vector is (0,0).

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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So the vector multiplicated by *i* becomes a vector with negative length!

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**God****Member**- Registered: 2005-08-25
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The length (aka. magnitude) of the vector is still it's absolute value.

So for example, if you had a vector <1+i, i-1>, the absolute value of your x component is sqrt(2), the absolute value of your y component is sqrt(2), and so the length of the vector is 2.

*Last edited by God (2005-12-30 12:00:54)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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I think you are right. But in Minklovski's 2D space the length between two points (x1,y1) and (x2,y2) is

L=sqrt(|x1-x2|²+(i|y1-y2|)²)=sqrt(|x1-x2|²-|y1-y2|²)

IPBLE: Increasing Performance By Lowering Expectations.

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