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Prove that if $$a=b \times c$$

$$b=c \times a$$

$$c=a \times b$$ then $a \perp b$, $a \perp c$, $b \perp c$, and $|a|=|b|=|c|=1$

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What have you tried? Please don't just give us an order to do something for you. –  Stefan Dec 4 '12 at 18:47
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Also, $a=b=c=0$ contradicts the last statement. –  copper.hat Dec 4 '12 at 18:47
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up vote 2 down vote accepted

$a=b \times c \perp b$ and the analogous statements follow simply from the fact that a vector product is orthogonal to its factors. Note that $|x \times y|=|x||y|sin(\angle (x,y))$ so $a \perp b \rightarrow |c|=|a||b|sin(90°)=|a||b|$. Analogously, $|b|=|a||c|$ and $|a|=|b||c|$. Multiplying these three equalities, you obtain $|a||b||c|=(|a||b||c|)^2$, so $|a||b||c|\in \{0,1\}$.

If $|a||b||c|=0$ then WLOG $|a|=0$, i. e. $a=0$, implying b=0 x c=0 and c=0 x b=0.

If $|a||b||c|=1$ then $|a|^2=|a|*(|b||c|)=|a||b||c|=1$, i. e. $|a|=1$ and analogously, $|b|=|c|=1$.

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