Bruce Sherman

2022-10-09

Another proof of $|\stackrel{\to }{a}×\stackrel{\to }{b}{|}^{2}=|\stackrel{\to }{a}{|}^{2}\cdot |\stackrel{\to }{b}{|}^{2}-\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}{\right)}^{2}$
I know how to prove the formula $|\stackrel{\to }{a}×\stackrel{\to }{b}{|}^{2}=|\stackrel{\to }{a}{|}^{2}\cdot |\stackrel{\to }{b}{|}^{2}-\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}{\right)}^{2}$ ,but i need to prove it using the formula $\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}=\left(\stackrel{\to }{a}\cdot \stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{b}\cdot \stackrel{\to }{c}\right)\stackrel{\to }{a}$, how to prove it?

cegukwt

Assume without loss of generality $\stackrel{\to }{a}\ne \stackrel{\to }{0}$. Set $\stackrel{\to }{c}=\stackrel{\to }{a}$ so
$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{a}={a}^{2}\stackrel{\to }{b}-\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)\stackrel{\to }{a}.$
This is a cross product of perpendicular vectors, so its square modulus is
$|\stackrel{\to }{a}×\stackrel{\to }{b}{|}^{2}{a}^{2}={a}^{4}{b}^{2}+{\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)}^{2}{a}^{2}-2{a}^{2}{\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)}^{2}={a}^{2}\left({a}^{2}{b}^{2}-{\left(\stackrel{\to }{a}\cdot \stackrel{\to }{b}\right)}^{2}\right).$
Now cancel the ${a}^{2}$ factors.

Tiana Hill

Slightly differently:
Let $\stackrel{\to }{c}=\stackrel{\to }{a}×\stackrel{\to }{b}$ ,and use $\left(\stackrel{\to }{P}×\stackrel{\to }{Q}\right).\stackrel{\to }{R}=\stackrel{\to }{P}.\left(\stackrel{\to }{Q}×\stackrel{\to }{R}\right)$, and $\stackrel{\to }{P}×\left(\stackrel{\to }{Q}×\stackrel{\to }{R}\right)=\left(\stackrel{\to }{P}.\stackrel{\to }{R}\right)\stackrel{\to }{Q}-\left(\stackrel{\to }{P}.\stackrel{\to }{Q}\right)\stackrel{\to }{R}$, then
$|\stackrel{\to }{a}×\stackrel{\to }{b}{|}^{2}=\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right).\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)=\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right).\stackrel{\to }{c}=\stackrel{\to }{a}.\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=\stackrel{\to }{a}.\left[\stackrel{\to }{b}×\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)\right]$
$=\stackrel{\to }{a}.\left[{b}^{2}\stackrel{\to }{a}-\left(\stackrel{\to }{b}.\stackrel{\to }{a}\right)\stackrel{\to }{b}\right]={a}^{2}{b}^{2}-\left(\stackrel{\to }{a}.\stackrel{\to }{b}{\right)}^{2}.$

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