Application of circularity of hybrid product

[influx]

Homework Statement

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2. Homework Equations
a·(b x c) = b·(c x a)

The Attempt at a Solution

How did they get from the (b) to (c)? In particular, I am referring to the numerator and denominator of the fraction. I mean if you apply a·(b x c) = b·(c x a) then surely the terms inside the brackets (of both numerator and denominator) of part (c) should be the other way round?

I mean for instance r_CD·(k x r_BA) = k·(r_BA x r_CD) and not r_CD·(k x r_BA) = k·(r_CD x r_BA) as they got?

 

[gneill]

Did you happen to notice that the prefixed "-" sign also disappeared from (b) to (c)?...

 

[influx]

Did you happen to notice that the prefixed "-" sign also disappeared from (b) to (c)?...

Ah I never noticed that! So -r_CD·(k x r_BA) = -k·(r_BA x r_CD) but I'm still confused in terms of how they got from -k·(r_BA x r_CD) to k·(r_CD x r_BA)? I know that A x B = -B x A so -k·(r_BA x r_CD) can be written as -k·(-r_CD x r_BA) and I'm guessing the two negatives cancel somehow but don't know how..

 

[gneill]

-r_CD can be written as (-1)r_CD, the scalar -1 being "pulled out" of the vector. That scalar in turn can be moved out of the parentheses and applied to the vector -k.

 

[influx]

-r_CD can be written as (-1)r_CD, the scalar -1 being "pulled out" of the vector. That scalar in turn can be moved out of the parentheses and applied to the vector -k.

So -k·((-1)r_CD x r_BA)=(-1)(-k)·(r_CD x r_BA) = k·(r_CD x r_BA)?

I was thinking it would be something along these lines but I wasn't sure how to pull out the negative when dealing with vectors.

 

[gneill]

So -k·((-1)r_CD x r_BA)=(-1)(-k)·(r_CD x r_BA) = k·(r_CD x r_BA)?

I was thinking it would be something along these lines but I wasn't sure how to pull out the negative when dealing with vectors.

Yes that's good. A "negative" is just a scalar constant, -1. So you can manipulate it as you would any scalar multiplyer of a vector.