Find $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$ if $\vec{a}+\vec{b}+\vec{c}=\alpha\vec{d},\vec{b}+\vec{c}+\vec{d}=\beta\vec{a}$

Question

If $\vec{a}+\vec{b}+\vec{c}=\alpha\vec{d}$, $\vec{b}+\vec{c}+\vec{d}=\beta\vec{a}$ and $\vec{a},\vec{b},\vec{c}$ are non-coplanar, then prove that $\vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{0}$

My Attempt $$ \vec{a}+\vec{b}+\vec{c}+\vec{d}=\vec{d}(1+\alpha)=\vec{a}(1+\beta) $$ $$ \vec{d}\times(\vec{a}+\vec{b}+\vec{c})=0\\ \vec{a}\times(\vec{b}+\vec{c}+\vec{d})=0\\ $$ But I have no hint about the sum giving a null vector.

Answer 1

From $\vec{d}(1+\alpha)=\vec{a}(1+\beta)$ you only have two options:

• either $\alpha=-1$ and so the original claims holds;
• or $\vec{d}=\frac{1+\beta}{1+\alpha}\vec{a}$ but then $$\vec{a}+\vec{b}+\vec{c}=\alpha\cdot \frac{1+\beta}{1+\alpha}\vec{a}$$ which contradicts the non-complanarity of $\vec{a},\vec{b}$ and $\vec{c}$