# $\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}$$\ne \stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$

$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{b}·\stackrel{\to }{c}\right)\stackrel{\to }{a}$

$\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)\stackrel{\to }{c}$

よって

$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}\ne \stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$

となり，結合法則は成り立たない

## ■説明

$\stackrel{\to }{a}=\left({a}_{1},{a}_{2},{a}_{3}\right)$$\stackrel{\to }{b}=\left({b}_{1},{b}_{2},{b}_{3}\right)$$\stackrel{\to }{c}=\left({c}_{1},{c}_{2},{c}_{3}\right)$

とおく．

### ●$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}$

$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}=\left\{\left({a}_{1},{a}_{2},{a}_{3}\right)×\left({b}_{1},{b}_{2},{b}_{3}\right)\right\}×\left({c}_{1},{c}_{2},{c}_{3}\right)$

$=\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2},{a}_{3}{b}_{1}-{a}_{1}{b}_{3},{a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right)×\left({c}_{1},{c}_{2},{c}_{3}\right)$

$=\left(\left({a}_{3}{b}_{1}-{a}_{1}{b}_{3}\right){c}_{3}-\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right){c}_{2},\left({a}_{1}{b}_{2}-{a}_{2}{b}_{1}\right){c}_{1}-\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right){c}_{3},\left({a}_{2}{b}_{3}-{a}_{3}{b}_{2}\right){c}_{2}-\left({a}_{3}{b}_{1}-{a}_{1}{b}_{3}\right){c}_{1}\right)$

$=\left({a}_{3}{b}_{1}{c}_{3}-{a}_{1}{b}_{3}{c}_{3}-{a}_{1}{b}_{2}{c}_{2}+{a}_{2}{b}_{1}{c}_{2},{a}_{1}{b}_{2}{c}_{1}-{a}_{2}{b}_{1}{c}_{1}-{a}_{2}{b}_{3}{c}_{3}+{a}_{3}{b}_{2}{c}_{3},{a}_{2}{b}_{3}{c}_{2}-{a}_{3}{b}_{2}{c}_{2}-{a}_{3}{b}_{1}{c}_{1}+{a}_{1}{b}_{3}{c}_{1}\right)$

$=\left(\left({a}_{3}{b}_{1}{c}_{3}+{a}_{2}{b}_{1}{c}_{2}\right)-\left({a}_{1}{b}_{3}{c}_{3}+{a}_{1}{b}_{2}{c}_{2}\right),\left({a}_{1}{b}_{2}{c}_{1}+{a}_{3}{b}_{2}{c}_{3}\right)-\left({a}_{2}{b}_{1}{c}_{1}+{a}_{2}{b}_{3}{c}_{3}\right),\left({a}_{2}{b}_{3}{c}_{2}+{a}_{1}{b}_{3}{c}_{1}\right)-\left({a}_{3}{b}_{2}{c}_{2}+{a}_{3}{b}_{1}{c}_{1}\right)\right)$

$=\left(\left({a}_{3}{c}_{3}+{a}_{2}{c}_{2}\right){b}_{1}-\left({b}_{3}{c}_{3}+{b}_{2}{c}_{2}\right){a}_{1},\left({a}_{1}{c}_{1}+{a}_{3}{c}_{3}\right){b}_{2}-\left({b}_{1}{c}_{1}+{b}_{3}{c}_{3}\right){a}_{2},\left({a}_{2}{c}_{2}+{a}_{1}{c}_{1}\right){b}_{3}-\left({b}_{2}{c}_{2}+{b}_{1}{c}_{1}\right){a}_{3}\right)$

$=\left(\left({a}_{3}{c}_{3}+{a}_{2}{c}_{2}\right){b}_{1}-\left({b}_{3}{c}_{3}+{b}_{2}{c}_{2}\right){a}_{1}+{a}_{1}{b}_{1}{c}_{1}-{a}_{1}{b}_{1}{c}_{1},\left({a}_{1}{c}_{1}+{a}_{3}{c}_{3}\right){b}_{2}-\left({b}_{1}{c}_{1}+{b}_{3}{c}_{3}\right){a}_{2}+{a}_{2}{b}_{2}{c}_{2}-{a}_{2}{b}_{2}{c}_{2},\left({a}_{2}{c}_{2}+{a}_{1}{c}_{1}\right){b}_{3}-\left({b}_{2}{c}_{2}+{b}_{1}{c}_{1}\right){a}_{3}+{a}_{3}{b}_{3}{c}_{3}-{a}_{3}{b}_{3}{c}_{3}\right)$

$=\left(\left({a}_{3}{c}_{3}+{a}_{2}{c}_{2}+{a}_{1}{c}_{1}\right){b}_{1}-\left({b}_{3}{c}_{3}+{b}_{2}{c}_{2}+{b}_{1}{c}_{1}\right){a}_{1},\left({a}_{1}{c}_{1}+{a}_{3}{c}_{3}+{a}_{2}{c}_{2}\right){b}_{2}-\left({b}_{1}{c}_{1}+{b}_{3}{c}_{3}+{b}_{2}{c}_{2}\right){a}_{2},\left({a}_{2}{c}_{2}+{a}_{1}{c}_{1}+{a}_{3}{c}_{3}\right){b}_{3}-\left({b}_{2}{c}_{2}+{b}_{1}{c}_{1}+{b}_{3}{c}_{3}\right){a}_{3}\right)$

$=\left(\left({a}_{3}{c}_{3}+{a}_{2}{c}_{2}+{a}_{1}{c}_{1}\right){b}_{1},\left({a}_{1}{c}_{1}+{a}_{3}{c}_{3}+{a}_{2}{c}_{2}\right){b}_{2},\left({a}_{2}{c}_{2}+{a}_{1}{c}_{1}+{a}_{3}{c}_{3}\right){b}_{3}\right)-\left(\left({b}_{3}{c}_{3}+{b}_{2}{c}_{2}+{b}_{1}{c}_{1}\right){a}_{1},\left({b}_{1}{c}_{1}+{b}_{3}{c}_{3}+{b}_{2}{c}_{2}\right){a}_{2},\left({b}_{2}{c}_{2}+{b}_{1}{c}_{1}+{b}_{3}{c}_{3}\right){a}_{3}\right)$

$=\left({a}_{3}{c}_{3}+{a}_{2}{c}_{2}+{a}_{1}{c}_{1}\right)\left({b}_{1},{b}_{2},{b}_{3}\right)-\left({b}_{3}{c}_{3}+{b}_{2}{c}_{2}+{b}_{1}{c}_{1}\right)\left({a}_{1},{a}_{2},{a}_{3}\right)$

$=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{b}·\stackrel{\to }{c}\right)\stackrel{\to }{a}$

すなわち

$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}$$=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{b}·\stackrel{\to }{c}\right)\stackrel{\to }{a}$　･･････(1)

### ●$\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$

$\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)\stackrel{\to }{c}$

$\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)=\left({a}_{1},{a}_{2},{a}_{3}\right)×\left\{\left({b}_{1},{b}_{2},{b}_{3}\right)×\left({c}_{1},{c}_{2},{c}_{3}\right)\right\}$

$=\left({a}_{1},{a}_{2},{a}_{3}\right)×\left({b}_{2}{c}_{3}-{b}_{3}{c}_{2},{b}_{3}{c}_{1}-{b}_{1}{c}_{3},{b}_{1}{c}_{2}-{b}_{2}{c}_{1}\right)$

$=\left({a}_{2}\left({b}_{1}{c}_{2}-{b}_{2}{c}_{1}\right)-{a}_{3}\left({b}_{3}{c}_{1}-{b}_{1}{c}_{3}\right),{a}_{3}\left({b}_{2}{c}_{3}-{b}_{3}{c}_{2}\right)-{a}_{1}\left({b}_{1}{c}_{2}-{b}_{2}{c}_{1}\right),{a}_{1}\left({b}_{3}{c}_{1}-{b}_{1}{c}_{3}\right)-{a}_{2}\left({b}_{2}{c}_{3}-{b}_{3}{c}_{2}\right)\right)$

$=\left({a}_{2}{b}_{1}{c}_{2}-{a}_{2}{b}_{2}{c}_{1}-{a}_{3}{b}_{3}{c}_{1}+{a}_{3}{b}_{1}{c}_{3},{a}_{3}{b}_{2}{c}_{3}-{a}_{3}{b}_{3}{c}_{2}-{a}_{1}{b}_{1}{c}_{2}+{a}_{1}{b}_{2}{c}_{1},{a}_{1}{b}_{3}{c}_{1}-{a}_{1}{b}_{1}{c}_{3}-{a}_{2}{b}_{2}{c}_{3}+{a}_{2}{b}_{3}{c}_{2}\right)$

$=\left(\left({a}_{2}{b}_{1}{c}_{2}+{a}_{3}{b}_{1}{c}_{3}\right)-\left({a}_{2}{b}_{2}{c}_{1}+{a}_{3}{b}_{3}{c}_{1}\right),\left({a}_{3}{b}_{2}{c}_{3}+{a}_{1}{b}_{2}{c}_{1}\right)-\left({a}_{3}{b}_{3}{c}_{2}+{a}_{1}{b}_{1}{c}_{2}\right),\left({a}_{1}{b}_{3}{c}_{1}+{a}_{2}{b}_{3}{c}_{2}\right)-\left({a}_{1}{b}_{1}{c}_{3}+{a}_{2}{b}_{2}{c}_{3}\right)\right)$

$=\left(\left({a}_{2}{c}_{2}+{a}_{3}{c}_{3}\right){b}_{1}-\left({a}_{2}{b}_{2}+{a}_{3}{b}_{3}\right){c}_{1},\left({a}_{3}{c}_{3}+{a}_{1}{c}_{1}\right){b}_{2}-\left({a}_{3}{b}_{3}+{a}_{1}{b}_{1}\right){c}_{2},\left({a}_{1}{c}_{1}+{a}_{2}{c}_{2}\right){b}_{3}-\left({a}_{1}{b}_{1}+{a}_{2}{b}_{2}\right){c}_{3}\right)$

$=\left(\left({a}_{2}{c}_{2}+{a}_{3}{c}_{3}\right){b}_{1}-\left({a}_{2}{b}_{2}+{a}_{3}{b}_{3}\right){c}_{1}+{a}_{1}{b}_{1}{c}_{1}-{a}_{1}{b}_{1}{c}_{1},\left({a}_{3}{c}_{3}+{a}_{1}{c}_{1}\right){b}_{2}-\left({a}_{3}{b}_{3}+{a}_{1}{b}_{1}\right){c}_{2}+{a}_{2}{b}_{2}{c}_{2}-{a}_{2}{b}_{2}{c}_{2},\left({a}_{1}{c}_{1}+{a}_{2}{c}_{2}\right){b}_{3}-\left({a}_{1}{b}_{1}+{a}_{2}{b}_{2}\right){c}_{3}+{a}_{3}{b}_{3}{c}_{3}-{a}_{3}{b}_{3}{c}_{3}\right)$

$=\left(\left({a}_{2}{c}_{2}+{a}_{3}{c}_{3}+{a}_{1}{c}_{1}\right){b}_{1}-\left({a}_{2}{b}_{2}+{a}_{3}{b}_{3}+{a}_{1}{b}_{1}\right){c}_{1},\left({a}_{3}{c}_{3}+{a}_{1}{c}_{1}+{a}_{2}{c}_{2}\right){b}_{2}-\left({a}_{3}{b}_{3}+{a}_{1}{b}_{1}+{a}_{2}{b}_{2}\right){c}_{2},\left({a}_{1}{c}_{1}+{a}_{2}{c}_{2}+{a}_{3}{c}_{3}\right){b}_{3}-\left({a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}\right){c}_{3}\right)$

$=\left(\left({a}_{2}{c}_{2}+{a}_{3}{c}_{3}+{a}_{1}{c}_{1}\right){b}_{1},\left({a}_{3}{c}_{3}+{a}_{1}{c}_{1}+{a}_{2}{c}_{2}\right){b}_{2},\left({a}_{1}{c}_{1}+{a}_{2}{c}_{2}+{a}_{3}{c}_{3}\right){b}_{3}\right)-\left(\left({a}_{2}{b}_{2}+{a}_{3}{b}_{3}+{a}_{1}{b}_{1}\right){c}_{1},\left({a}_{3}{b}_{3}+{a}_{1}{b}_{1}+{a}_{2}{b}_{2}\right){c}_{2},\left({a}_{1}{b}_{1}+{a}_{2}{b}_{2}+{a}_{3}{b}_{3}\right){c}_{3}\right)$

$=\left({a}_{2}{c}_{2}+{a}_{3}{c}_{3}+{a}_{1}{c}_{1}\right)\left({b}_{1},{b}_{2},{b}_{3}\right)-\left({a}_{2}{b}_{2}+{a}_{3}{b}_{3}+{a}_{1}{b}_{1}\right)\left({c}_{1},{c}_{2},{c}_{3}\right)$

$=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)\stackrel{\to }{c}$

よって

$\stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$$=\left(\stackrel{\to }{a}·\stackrel{\to }{c}\right)\stackrel{\to }{b}-\left(\stackrel{\to }{a}·\stackrel{\to }{b}\right)\stackrel{\to }{c}$　･･････(2)

### (1)，(2)より

$\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}\ne \stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$

ホーム>>カテゴリー分類>>ベクトル>>外積>> $\left(\stackrel{\to }{a}×\stackrel{\to }{b}\right)×\stackrel{\to }{c}$$\ne \stackrel{\to }{a}×\left(\stackrel{\to }{b}×\stackrel{\to }{c}\right)$