# crossproducts of linearly independent vectors

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• Feb 4th 2013, 05:59 PM
hollywood
Re: crossproducts of linearly independent vectors
Quote:

Originally Posted by learning
x.a = r3 a.(bxc), x.b = r2 b.(axc), x.c = r1 c.(axb)

No, you can't cancel the dot product.

It's sometimes hard to keep track of which variables are "known" and which are "unknown". Here you are trying to express some vector x (which is therefore "known") as a combination of $a \times b$, $a \times c$, and $b \times c$, where a, b, and c are "known". The other variables - $r_1$, $r_2$, and $r_3$ are the coefficients we're looking for - they are "unknown".

So all you have to do is divide to get your answer:

$r_3 = \frac{x \cdot a}{a \cdot (b \times c)}$
$r_2 = \frac{x \cdot b}{b \cdot (a \times c)}$
$r_1 = \frac{x \cdot c}{c \cdot (a \times b)}$

-Hollywood
• Feb 5th 2013, 08:41 AM
learning
Re: crossproducts of linearly independent vectors
hey, is this a final answer then? is it acceptable to have x = something involving x? i was looking to get x = something with just axb, bxc, axc terms.. i understand everything you wrote is correct but do u think, that, given this wording of the question, that this will count as a solution?
• Feb 5th 2013, 09:14 AM
hollywood
Re: crossproducts of linearly independent vectors
I'm pretty sure that's the answer. Given x, a, b, and c (in any basis), it gives the coefficients to express x in terms of $a \times b$, $a \times c$, and $b \times c$.

- Hollywood
• Feb 5th 2013, 09:20 AM
Hartlw
Re: crossproducts of linearly independent vectors
For some reason I’m having trouble accessing the forum. So I’ll have to paraphrase hollywood’s post #2:

hollywood wrote: “Suppose a1(x1 x x2) + a2(x1 x x3) +a3(x2 x x3) = 0. What happens when you take the dot product with, say, x1?”

This is a very nice answer to the OP. My apologies for missing this. There were so many back and forths I assumed the question was open. If I had one tiny quibble it would be to observe that a remark on why x1.(x2 x x3) unequal to zero (implying a3 = 0) might be in order. One might note:
Geometrically: If vectors are planar (linearly dependent), x1 is perpendicular to (x2 x x3)
Algebraically: x1.(x2 x x3) = det| x1,x2,x3 | unequal to 0 because x1, x2, x3 are linearly independent.
• Feb 5th 2013, 09:30 AM
learning
Re: crossproducts of linearly independent vectors
its six of one half a dozen of the other. thanks for the maths guys. im comfortable with this solution now.
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