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Inner products and Circles |
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| Mar4-12, 10:06 AM | #1 |
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Inner products and Circles![]() 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? |
| Mar4-12, 10:25 AM | #2 |
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I think you're missing some terms from the length:
<(x,y),(x,y)>=5x2+2(xy+yx)+y2 |
| Mar4-12, 10:46 AM | #3 |
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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?
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| Mar4-12, 09:58 PM | #4 |
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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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