Inner products and Circles


by TranscendArcu
Tags: circles, products
TranscendArcu
TranscendArcu is offline
#1
Mar4-12, 10:06 AM
P: 288

Mostly I'd like to look at the third part of the problem. I'm not sure if this is the correct way to derive the equation:

So, finding the length of a given vector given this inner product:
[itex]<(x,y),(x,y)> = 5x^2 + y^2[/itex].

Taking the length, we have

[itex]|(x,y)| = \sqrt{5x^2 + y^2}[/itex], which we define as equaling 1. Squaring both sides we find,

[itex]5x^2 + y^2 = 1[/itex]. I think this is the equation of the circle, but I'm not sure. If it is, then my picture has y-intercepts at 1,-1 and x-intercepts at -sqrt(1/5),sqrt(1/5).

Is this correct?
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Bacle2
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#2
Mar4-12, 10:25 AM
Sci Advisor
P: 1,168
I think you're missing some terms from the length:

<(x,y),(x,y)>=5x2+2(xy+yx)+y2
TranscendArcu
TranscendArcu is offline
#3
Mar4-12, 10:46 AM
P: 288
Whoops. You're right. My real equation is [itex]5x^2 -2(xy+xy) +y^2 =1[/itex]. This changes shape of the circle (it's more elongated in quadrants I and III now), but the intercepts remain the same I think. No?

Bacle2
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#4
Mar4-12, 09:58 PM
Sci Advisor
P: 1,168

Inner products and Circles


I think if you do a rotation of the plain, you may be able to get rid of the mixed xy-terms.


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