Proof of Circle Theorem by VectorsDate: 05/03/2001 at 06:09:29 From: Marko Stojovic Subject: Proof of circle theorem by vectors A question requires me to prove, using the vector scalar product, that the angle in a semicircle is always 90 degrees (the hypotenuse being the diameter, and the sides meeting on the perimeter). I have called the sides leading away from the angle vectors a and b (both directed away from the angle), and have tried to prove that a.b = 0. I have tried this by means of drawing a radius from the centre to the angle (calling this vector d, leading away from the angle), and have expressed the diameter of the circle as two opposite vectors, equal in magnitude, leading away from the centre (e and f). I then tried to express a.b as (d+e).(d+f) = d.(e+f) = d.(e-e) = d.0 As I understand it, both vectors have to be positive for the scalar product to work, so I can't proceed beyond this point. Please help... Date: 05/04/2001 at 14:35:32 From: Doctor Floor Subject: Re: Proof of circle theorem by vectors Dear Marko, Thanks for your interesting question. Let X be a point on the circle, let M be its center, and let TU be a diameter. Let vector XM be given by (t,u) and MT by (v,w). Then we know that t^2+u^2 = v^2+w^2 = r^2 where r is the radius of the circle. Now we have: vector XT = (t,u)+(v,w) = (t+v,u+w) and vector XU = (t,u)-(v,w) = (t-v,u-w). Their internal product is equal to (t+v)(t-v) + (u+w)(u-w) = t^2 - v^2 + u^2 - w^2 = t^2 + u^2 - v^2 - w^2 = r^2 - r^2 = 0 and we see that the two vectors must be perpendicular, as desired. If you need more help, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
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