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#1 2005-12-18 07:57:41
- krassi_holmz
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vector multiplicated by i
What will happen with the vector if we multiply it by
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#2 2005-12-18 08:01:42
- John E. Franklin
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Re: vector multiplicated by i
Is this a 2-dimensional vector?
What is the coordinate system?
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#3 2005-12-18 08:11:19
- krassi_holmz
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Re: vector multiplicated by i
OK. Let the vector be 2D.
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#4 2005-12-18 08:12:50
- krassi_holmz
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Re: vector multiplicated by i
It isn't so inportant what is the coordinate system.
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#5 2005-12-18 08:15:50
- krassi_holmz
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Re: vector multiplicated by i
But let
v=ai+bj
(here i is vector, not (-1)^(1/2)! (-1)^(1/2) must be italic i)
Last edited by krassi_holmz (2005-12-18 08:18:14)
IPBLE: Increasing Performance By Lowering Expectations.
#6 2005-12-18 11:24:39
- John E. Franklin
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Re: vector multiplicated by i
Refresh my memory. What happens to the function y=x if you multiply
it by :italic(i)
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#7 2005-12-18 18:45:04
- krassi_holmz
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Re: vector multiplicated by i
I use the following for italic
Code:
[i]italic[/i]
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#8 2005-12-18 18:54:02
- krassi_holmz
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Re: vector multiplicated by i
OK. First try to myltiply one point with coordinates {x,y}.
What will happen?
The answer is {ix,iy}. So?
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#9 2005-12-31 10:25:14
- God
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Re: vector multiplicated by i
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).
#10 2005-12-31 10:39:34
- krassi_holmz
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Re: vector multiplicated by i
So the vector multiplicated by i becomes a vector with negative length!
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#11 2005-12-31 11:00:20
- God
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Re: vector multiplicated by i
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 (2005-12-31 11:00:54)
#12 2005-12-31 22:28:06
- krassi_holmz
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Re: vector multiplicated by i
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.