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 $\displaystyle a \times b$, $\displaystyle a \times c$, and $\displaystyle b \times c$, where a, b, and c are "known". The other variables - $\displaystyle r_1$, $\displaystyle r_2$, and $\displaystyle r_3$ are the coefficients we're looking for - they are "unknown".

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

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

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

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

-Hollywood

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?

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 $\displaystyle a \times b$, $\displaystyle a \times c$, and $\displaystyle b \times c$.

- Hollywood

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.

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.