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 Pre-Calculus and Calculus
 Vectors: Parallelogram Missing a Point
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uberclay
Advanced Member

Canada
159 Posts

Posted - 04/16/2012 :  23:53:43  Show Profile  Reply with Quote
Given the parallelogram ABCD, where A (4, 2), B (-6, 1) and D (-3, -4), find the coordinates of C.

I cannot, for the life of me, wrap my head around this problem. Initially I solved this by observation (B is to A as C is to D) for the solution C (-13, -5), but it's supposed to be solved using Vector addition/subtraction.

The Cartesian Vectors I get are AB (-10, -1) and AD (-7, -6) giving magnitudes |AB| = 101^(1/2) and |AD| = 85^(1/2)

I found the angle between AB and AD by 90 - tan^-1 (-1/-10) + tan^-1 (-7/-6) (this is a pre-dot/cross product lesson)

The cosine law gives the correct magnitude for the diagonal AC;
|AC| = |AB| + |AD| = |AB|^2 + |AD|^2 - 2|AB||AD|Cos, where = 180 - BAD, which is 338.

And here is where I am stuck. How do I get the coordinates out of this magnitude?
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uberclay
Advanced Member

Canada
159 Posts

Posted - 04/17/2012 :  13:36:16  Show Profile  Reply with Quote
Disregard. I have hugely overcomplicated this.

All that needs to be done here is vector addition
I'll work it out and let you know how it goes.

Edited by - uberclay on 04/17/2012 14:14:33
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uberclay
Advanced Member

Canada
159 Posts

Posted - 04/17/2012 :  19:08:13  Show Profile  Reply with Quote
I'm still baffled but I think I got it.

Subtracting the vector OA from OB gives AB = (-10, -1)

And adding the vector OD to AB gives OC

AB + OD = (-10, -1) + (-3, -4) = (-13, -5)

Therefore C = (-13, -5)

I couldn't find an example in the text, which is odd. And my submitted answer (graded as incorrect) was the same logic without the vector notation.

I'm assuming that evaluating for AB establishes the relation between A and B, and when applied to D, gives the location of C.

Edited by - uberclay on 04/17/2012 19:41:33
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Ultraglide
Advanced Member

Canada
299 Posts

Posted - 04/17/2012 :  22:02:40  Show Profile  Reply with Quote
You're making this way too hard. The coordinates of a parallelogram, if they are given as letters, are always cyclic so in this case AB is parallel to DC. Now vector AB is (-6-4, 1-2) or
(-10,-1). Let C have coordinates (x,y) so DC is (x-(-3), y-(-4)) or (x+3, y+4). But AB and DC are equal so x+3=-10 giving x = -13 and y+4=-1 giving y = -5. So (x,y)=(-13, -5). You can check as follows, - 13 - (-3) = -10 and -5 -(-4) = -1.
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uberclay
Advanced Member

Canada
159 Posts

Posted - 04/18/2012 :  16:56:27  Show Profile  Reply with Quote
Well, it makes sense now ; 0 )

Thanks Ultraglide
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