If a, b, c are non-coplanar vectors and is a real number, then [( a+ b) ^2 b c]=[ a b+ c b…

Question

If $\vec a,\vec b,\vec c$ are non-coplanar vectors and $\lambda$ is a real number, then $\big[\lambda(\vec a+\vec b)\ \ \lambda^2\vec b\ \ \lambda\vec c\big]=[\vec a\ \ \vec b+\vec c\ \ \vec b]$ holds for:

Accepted Answer

LHS $=\lambda^4[\vec a+\vec b\ \ \vec b\ \ \vec c]=\lambda^4[\vec a\ \vec b\ \vec c]$ (since adding $\vec b$ to $\vec a$ doesn't change the triple product).

RHS $=[\vec a\ \ \vec b+\vec c\ \ \vec b]=[\vec a\ \vec b\ \vec b]+[\vec a\ \vec c\ \vec b]=0-[\vec a\ \vec b\ \vec c]=-[\vec a\ \vec b\ \vec c]$

So $\lambda^4[\vec a\ \vec b\ \vec c]=-[\vec a\ \vec b\ \vec c]$. Since $[\vec a\ \vec b\ \vec c]
e0$, $\lambda^4=-1$, which has no real solution.

Solutions · 1