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November 08, 2025, 04:09:51 am

Author Topic: Vector Proofs of Geometry  (Read 2204 times)  Share 

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rhcpfox1

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Vector Proofs of Geometry
« on: November 13, 2007, 04:03:24 pm »
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Does anybody have any examples they can upload on Vector proofs?
I struggle with the proofs and any examples you may have would be greatly appreciated.
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Galelleo

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Vector Proofs of Geometry
« Reply #1 on: November 13, 2007, 04:27:53 pm »
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sry mate, i havent got a scanner... but ill try to draw you one.

if A is a point (1,1) and B is a point (1,3), and C is point (2,2), if the midpoint of AB is M... prove that AB is perpendicular to CM.

Ok... so the idea is to prove that somethings perpendicular... and we know that the easiest way to do that is to equate the scalar dot product to 0. (because we know that a.b=|a||b|cos(theta)... and cos(theta) = 0 when perpendicular, and the magnitudes wont be 0.

So... we can establish that its asking us... CM.AB = 0.

So first we find AB... which is AO + OB (easier way to do this is OB-OA, because we know that AO = -OA)

So OB-OA = 1i + 3j - (1i + 1j) = 2j = AB

Now we need to find CM. CM is a little harder because we need to use addition of vectors. we know that AM is .5AB, because its the midpoint... and we know that CA + AM will equal CM. So.. we find CA... (OA-OC), then we find AM (.5AB) and add em together... this will give us

(1i+1j) - (2i + 2j)= -1i - 1j = CA
.5( 2j) = 1j = AM

CA+AM = -1i = CM.

okay... now CM.AB is going to be (-1i).(2j) ... and because theyre in differetn directions they equal zero, as a general rule for scalar products (ie... i.j = 0, while i.i and j.j would equal 1)

therefore, as CM.AB = -i.2j = 0, CM and AB are perpendicular.

Hope that helps. If any bit needs clarifying just ask... im not very good at explaining, generally.
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AppleXY

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Vector Proofs of Geometry
« Reply #2 on: November 13, 2007, 05:54:52 pm »
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*gulp* i'm so bad at vector proofs :( *gulp*


need some srs practice.

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rhcpfox1

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Vector Proofs of Geometry
« Reply #3 on: November 13, 2007, 06:04:27 pm »
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Quote from: "AppleXY"
*gulp* i'm so bad at vector proofs :( *gulp*


need some srs practice.


Lol at least it's not needed for Actuarial right Apple?  :P

Thanks Galelleo, it did make sense,  I think i just need to sink my teeth into a few to learn how to go about proving certain things. I guess I can always wish for a little luck that a nasty one doesn't appear on the exam.
Here's hoping
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Galelleo

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Vector Proofs of Geometry
« Reply #4 on: November 13, 2007, 06:10:53 pm »
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I think that our chances are that the vector proof question iwll be scalar product = 0, perpendicular proof...

especially since i dont think there was any scalar product on the exam 1.

you just need to be familiar with addition of vectors, basic geometric shapes, and dot product of vectors.

Also, it helps if you draw out the poitns they give you,and remember that each point is given from the origin. so if point a is (x,y) then OA is xi + yi.
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Rosetta_Stoned

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Vector Proofs of Geometry
« Reply #5 on: November 16, 2007, 10:25:08 pm »
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I love vector proofs!

Shame none of them came up in a full question :(
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AppleXY

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Vector Proofs of Geometry
« Reply #6 on: November 18, 2007, 10:50:29 am »
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Quote from: "rhcpfox1"
Quote from: "AppleXY"
*gulp* i'm so bad at vector proofs :( *gulp*


need some srs practice.


Lol at least it's not needed for Actuarial right Apple?  :P



Indeed. :p No way, buddy, its all calculus, fin math, stats and prob down here :D

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