Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $|a|=|b|=|c|=1$

share|cite|improve this question
What have you tried? Please don't just give us an order to do something for you. – Stefan Dec 4 '12 at 18:47
Also, $a=b=c=0$ contradicts the last statement. – copper.hat Dec 4 '12 at 18:47
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, $|b|=|c|=1$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.