Hello, schteve!

I have 3 vectors $\displaystyle \vec a, \vec b, \vec c$ such that: .$\displaystyle a+b+c\:=\:0$

This implies they are all the same length . . . . not true

and they finish where they start.

The question then asks me to prove that: .$\displaystyle \vec a\times\vec b \:=\: \vec b\times\vec c \:=\: \vec c\times\vec a$

Earboth showed you why the three cross products have the same *direction.*

Now we must prove they have the same *magnitude* (length).

The three vectors form a triangle. Code:

C
*
* *
|b| * * |a|
* *
* *
* *
A * * * * * * * B
|c|

Law of Sines: .$\displaystyle \frac{|a|}{\sin A} \:=\:\frac{|b|}{\sin B} \quad\Rightarrow\quad|a|\sin B \:=\:|b|\sin A$

Multiply both sides by $\displaystyle |c|\!:\;\;|a||c|\sin B \:=\:|b||c|\sin A$ .[1]

Recall that the magnitude of a cross product is the area

. . of the parallelogram determined by the two vectors. Code:

* - - - - - - *
/:::::::::::::/
v /:::::::::::::/
/:θ:::::::::::/
* - - - - - - *
u

That is: .$\displaystyle \text{Area} \:=\:|\vec u \times \vec v| \:=\:|u||v|\sin\theta$

Hence: .$\displaystyle \begin{array}{ccc}|\vec a \times \vec c| &=& |a||c|\sin B \\ \\[-4mm] |\vec b \times \vec c| &=& |b||c|sin A \end{array}$

According to [1], these two magnitudes are equal: .$\displaystyle |\vec a \times \vec c| \:=\:|\vec b \times \vec c|$

In a similar fashion, we can prove that all three magnitudes are equal.