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I have a problem in vector algebra. In this Wolfram-page the last two formulas (9) and (10) are: $$ \frac{ | (x_2 - x_1) \times (x_1 - x_0) | }{|x_2 - x_1 |} = \frac{ | (x_0 - x_1) \times (x_0 - x_2) | }{|x_2 - x_1 |} $$ I've tried to apply properties founded in this other Wolfram-crossproduct-page but I still don't understand it. How is possible? And why the do this vectors manipulation?

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I've edited the question a bit as it's not related to vector calculus. Hope you don't mind. –  Rahul Jan 13 '11 at 3:05
    
@Rahul thanks : ) –  nkint Jan 13 '11 at 10:57
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1 Answer

up vote 2 down vote accepted

$\begin{align}|(x_2-x_1)\times(x_1-x_0)|&=|(x_1-x_2)\times(x_0-x_1)|\\&=|(x_0-x_1)\times(x_1-x_2)|\\&=|(x_0-x_1)\times((x_0-x_1)+(x_1-x_2))|\\&=|(x_0-x_1)\times(x_0-x_2)|\end{align}$

where the second is because you are in magnitude signs and the third is because the cross product of a vector with itself (or any parallel vector) is zero.

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absolute value?? i thought the $|v \times w|$ was the magnitude of the cross product between v and w is magnitude not absolute value.. –  nkint Jan 12 '11 at 23:09
    
You are right it is magnitude, but the magnitude is always positive. I should have said it that way and will fix. I just wanted to justify that changing the order of the factors did not change the result, whereas for the cross product itself it changes the sign. –  Ross Millikan Jan 12 '11 at 23:28
    
ok now it's more clear. but the remaining question is.. why they do that manipulation? –  nkint Jan 13 '11 at 10:59
    
@nkint: Just to give you more options for using the result, I think. –  Ross Millikan Jan 13 '11 at 13:38
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