Simplification of an expression in scalar triple product terms

Question

Apparently, for three non-coplanar vectors $\vec a, \vec b ,$and $\vec c$, and any arbitrary vector $\vec r$, the following holds true; $$\vec c((\vec a×\vec b) .\vec r)+\vec a((\vec b×\vec c) .\vec r)+\vec b((\vec c×\vec a) .\vec r)=[\vec a \vec b \vec c]\vec r$$ where [.] signifies the scalar triple product.

I've tried to arrive at this but failed. Obviously the BAC-CAB rule doesn't work, and I wasted time with that. Factoring out the $\vec r$ doesn't seem possible. I haven't seen any general equations for scalar triple products that might help, either.

Can someone help me with the proof here?,

Answer 1

Hint: Since $\vec a, \vec b, \vec c$ are non coplaner. It follows that they are linearly independent. Any vector in $\mathbb R^3$ can be written as their linear combination.
So let $\vec r=x\vec a+ y\vec b+z\vec c$, where $x,y,z$ are scalers.

So left hand side becomes: $x\vec a[\vec a \;\vec b\; \vec c]+(?)+(?)$

Can you take it from here?

Answer 2

First , we know that a.(bxc) can be expressed as [a b c] , where a,b,c are vectors

Now ,a,b,c are non-coplanar vectors ,therefore any arbitrary vector 'r' can be expressed as a linear combination of a,b and c .

Therefore , let r= la + mb + nc ,..(1)

where l,m,n are constants and a,b,c are vectors .

Now doing dot product with (axb) with both sides of equation (1) , we get

r.(axb)= n{c.(axb)}. OR [r a b] = n[c a b] , therefore

n= [r a b]/[c a b]..(2)

Similarly ,doing dot product with (bxc) and () with (1) , we get respectively

l=[r b c]/[a b c] ...(3)

m=[r a c]/[b a c]....(4)

Now putting these values in (1)

We get , r= [r b c]/[a b c]a + [r a c]/[b a c]b + [r a b]/[c a b]c ..(5)

Now , as [a b c]=[b c a]=[c a b]

(5) can be simplified as

r[a b c] =[r b c]a + [r a c]b + [r a b]c

Hence , proved .