If $\vec{a}, \vec{b}, \vec{c} \in \Delta^{3}$, prove that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}, \vec{a} \times \vec{b} = \vec{a} \times \vec{c}$ and $\vec{a} \neq 0 \implies \vec{b} = \vec{c}$
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The two identities can be written as $$ (b-c)\cdot a=0,\ \ \ \ (b-c)\times a=0. $$ The first equality says that $b-c$ is orthogonal to $a$. When two vectors are orthogonal to each other, the norm of their cross product is the product of their norms. So $$ 0=\|(b-c)\times a\|=\|b-c\|\,\|a\|. $$ If $a\ne0$, then $\|a\|>0$ and so $\|b-c\|=0$.