prove the scalar triple product a,b,c are vectors $(a-b)\cdot ((b-c) \times (c-a))=0$

Question

prove the scalar triple product a,b,c are vectors

$$(a-b)\cdot ((b-c) \times (c-a))=0$$

and

$$(a+b)\cdot ((b+c) \times (c+a))=(a-b)\cdot ((b-c)\times (c-a))$$

I attempted expanding many ways i could not crack it there is too many unknowns. I think it maybe solved if squared but i was unable to solve it. i reached but i am unsure if its right.

|a-b||a||b|sin(θ)cos(θ)

|a-b||a||b|*|a-b||a||b|=0 it looks easy but its a wolf in sheep's clothes. i also did

a-b=k and b-c=v and c-a=w

k1(v2w3-v3w2)-k2(v1w3-v3w1)+k3(v1w2-v2w1)=0?

any help is appreciated

Answer 1

(i) $a-b$, $b-c$, $c-a$ are coplanar so the triple scalar product is $0$.

(ii) Your second equation is false: take as an example $a=(-1,1,1)$, $b=(1,-1,1)$, $c=(1,1,-1)$.

Answer 2

For the first one, use the facts that $a-b=-(c-a)-(b-c)$ and $v\cdot(v\times w) = w\cdot(v\times w) = 0$.

The second one is false.

Answer 3

If you know the determinant formulation of the scalar triple product, notice that the sum of the three input vectors (i.e. the the differences) is zero, hence they are linearly dependant, and the determinant of a matrix with linearly dependent rows or columns is zero.