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Linear Algebra
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vectors: solve x for AB=CD (vectors) A(x,1),B(4,x+3),C(x,x+2) and D(2x,x+6) where should I start?
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Is this is saying that A=<x,1>?
A(x,1) is a point. I guess is AB is (4-x,x+3-1), but it didn't work for me.
\[\vec{AB}=\vec{CD}\]
the says the that x=2.
in other words: B-A = D-C
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of course i see no reason why A-B = C-D would not work equally as well
yes \[\frac{ \rightarrow }{ AB}=\frac{ \rightarrow }{ CD}\]
A( x, 1 ) C( x , x+2) -B(-4,-x-3) -D(-2x,-x-6) --------- ----------- (x-4,-2-x) = ( -x , 0x-4) so equate like components x-4 = -x -2-x = -4
Ok. I did right, but in the end instead put x-4=-x I put x-4 -x =0 <--- wrong. Again thank you amistre64.
youre welcome
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