Prove that a (b+c)+b (c+a)+c (a+b)=0

Question

Prove that $\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b})=0$

✓

Answer

Using the distributive law of vector product,
$
\begin{aligned}
& LHS=\vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times(\vec{c}+\vec{a})+\vec{c} \times(\vec{a}+\vec{b}) \\
= & (\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})+(\vec{b} \times \vec{c})+(\vec{b} \times \vec{a})+(\vec{c} \times \vec{a})+(\vec{c} \times \vec{b}) \\
= & (\vec{a} \times \vec{b})-(\vec{c} \times \vec{a})+(\vec{b} \times \vec{c})-(\vec{a} \times \vec{b})+(\vec{c} \times \vec{a})-(\vec{b} \times \vec{c}) \\
= & 0=\operatorname{RHS} \quad \text { Hence proved. }
\end{aligned}
$

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