ベクトル解析

ベクトル3重積

2種類のベクトル3重積

ベクトル \(\pmb{a}\)、\(\pmb{b}\)、\(\pmb{c}\) に対して

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\cdot\cdot\cdot\left(a\right)$$

または

$$\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}\cdot\cdot\cdot\left(b\right)$$

ベクトル \(\pmb{a}\)、\(\pmb{b}\)、\(\pmb{c}\) の ベクトル3重積 と呼ぶ。外積 は計算順序によって違う結果を生じるので基本的には

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\neq\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}$$

である。(もちろん特定のベクトルにおいて等しくなるパターンは存在する。)

ベクトルの外積外積の計算方法、性質の紹介と共に外積がどんなところで役立っているかを解説する。...

ベクトル3重積 \(\left(a\right)\)、\(\left(b\right)\) はそれぞれ次の関係式 \(\left(A\right)\)、\(\left(B\right)\) が成り立つ。

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{a}\cdot\pmb{b}\right)\pmb{c}\cdot\cdot\cdot\left(A\right)$$

$$\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{b}\cdot\pmb{c}\right)\pmb{a} \cdot\cdot\cdot\left(B\right) $$

関係式 \(\left(A\right)\) の証明をするために、\(\pmb{a}=\left(a_{1},a_{2},a_{3}\right)\)、\(\pmb{b}=\left(b_{1},b_{2},b_{3}\right)\)、\(\pmb{c}=\left(c_{1},c_{2},c_{3}\right)\) と成分を設定して ベクトル3重積 を計算してみよう。

関係式 \(\left(B\right)\) の証明は例題で用意した

ベクトル3重積(A) の計算

ここでは、 ベクトル3重積(A) \(\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\) のみ計算する。

No.1 : 外積 \(\pmb{b}\times\pmb{c}\) を成分計算

\begin{eqnarray}&&\pmb{b}\times\pmb{c}\\\\&=&\begin{vmatrix}\pmb{i} & \pmb{j} & \pmb{k}\\b_{1} & b_{2} & b_{3}\\c_{1} & c_{2} & c_{3}\end{vmatrix}\\\\&=&\left(b_{2}c_{3}-b_{3}c_{2}\right)\pmb{i}+\left(b_{3}c_{1}-b_{1}c_{3}\right)\pmb{j}+\left(b_{1}c_{2}-b_{2}c_{1}\right)\pmb{k}\end{eqnarray}

ここで、外積 \(\pmb{b}\times\pmb{c}\) の各成分が長ったらしいので \(d_{1}\)、\(d_{2}\)、\(d_{3}\) と置く。すなわち

\begin{cases}d_{1}=b_{2}c_{3}-b_{3}c_{2}\\\\d_{2}=b_{3}c_{1}-b_{1}c_{3}\\\\d_{3}=b_{1}c_{2}-b_{2}c_{1}\end{cases}

とすると、

$$\pmb{b}\times\pmb{c}=d_{1}\pmb{i}+d_{2}\pmb{j}+d_{3}\pmb{k}$$

とスッキリさせることができた。

No.2 : ベクトル3重積 \(\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\) を成分計算

いよいよ ベクトル3重積

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)=\left(a_{1},a_{2},a_{3}\right)\times\left(d_{1},d_{2},d_{3}\right)$$

を計算してみよう。

\begin{eqnarray}&&\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\\\\&=&\begin{vmatrix}\pmb{i} & \pmb{j} & \pmb{k}\\a_{1} & a_{2} & a_{3}\\d_{1} & d_{2} & d_{3}\end{vmatrix}\\\\&=&\left(a_{2}d_{3}-a_{3}d_{2}\right)\pmb{i}+\left(a_{3}d_{1}-a_{1}d_{3}\right)\pmb{j}+\left(a_{1}d_{2}-a_{2}d_{1}\right)\pmb{k}\end{eqnarray}

No.3 : \(x\) 成分 「\(a_{2}d_{3}-a_{3}d_{2}\)」 をひとまず代入計算

ここで No.1 で置き換えた \(d\) を元に戻す。一気にやると計算が長ったらしくなるので、\(x\) 方向成分 \(a_{2}d_{3}-a_{3}d_{2}\) のみ計算してみよう。

\begin{eqnarray}a_{2}d_{3}-a_{3}d_{2}&=&a_{2}\left(\color{blue}{b_{1}}c_{2}-b_{2}\color{green}{c_{1}}\right)-a_{3}\left(b_{3}\color{green}{c_{1}}-\color{blue}{b_{1}}c_{3}\right)\\\\&=&a_{2}\color{blue}{b_{1}}c_{2}-a_{2}b_{2}\color{green}{c_{1}}-a_{3}b_{3}\color{green}{c_{1}}+a_{3}\color{blue}{b_{1}}c_{3}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{1}}\end{eqnarray}

No.4 : 式変形のための項「\(a_{1}b_{1}c_{1}-a_{1}b_{1}c_{1}\)」を加える。

No.3 で求めた式において、\(b_{1}\) の係数 「\(a_{2}c_{2}+a_{3}c_{3}\)」と \(c_{1}\) の係数 「\(a_{2}b_{2}+a_{3}b_{3}\)」はどうにも

内積

\begin{cases}\pmb{a}\cdot\pmb{c}=a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\\\\\pmb{a}\cdot\pmb{b}=a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\end{cases}

を彷彿とさせる。

内積変形のために、打ち消し項 \(a_{1}\color{blue}{b_{1}}c_{1}-a_{1}b_{1}\color{green}{c_{1}}\left(=0\right)\) を加えると、

\begin{eqnarray}&&a_{2}d_{3}-a_{3}d_{2}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{1}}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{1}}+a_{1}\color{blue}{b_{1}}c_{1}-a_{1}b_{1}\color{green}{c_{1}}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{1}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\end{eqnarray}

No.5 : \(y\) 成分 、\(z\) 成分も同様に求める。

\(y\) 成分について

\begin{eqnarray}&&a_{3}d_{1}-a_{1}d_{3}\\\\&=&a_{3}\left(\color{blue}{b_{2}}c_{3}-b_{3}\color{green}{c_{2}}\right)-a_{1}\left(b_{1}\color{green}{c_{2}}-\color{blue}{b_{2}}c_{1}\right)\\\\&=&a_{3}\color{blue}{b_{2}}c_{3}-a_{3}b_{3}\color{green}{c_{2}}-a_{1}b_{1}\color{green}{c_{2}}+a_{1}\color{blue}{b_{2}}c_{1}\\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(a_{1}b_{1}+a_{3}b_{3}\right)\color{green}{c_{2}}\end{eqnarray}

内積変形のために、打ち消し項 \(a_{2}\color{blue}{b_{2}}c_{2}-a_{2}b_{2}\color{green}{c_{2}}\left(=0\right)\) を加えると、

\begin{eqnarray}&&a_{3}d_{1}-a_{1}d_{3}\\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(a_{1}b_{1}+a_{3}b_{3}\right)\color{green}{c_{2}}\\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(a_{1}b_{1}+a_{3}b_{3}\right)\color{green}{c_{2}}+a_{2}\color{blue}{b_{2}}c_{2}-a_{2}b_{2}\color{green}{c_{2}}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{2}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\end{eqnarray}

と求まる。


一方、\(z\) 成分について

\begin{eqnarray}&&a_{1}d_{2}-a_{2}d_{1}\\\\&=&a_{1}\left(\color{blue}{b_{3}}c_{1}-b_{1}\color{green}{c_{3}}\right)-a_{2}\left(b_{2}\color{green}{c_{3}}-\color{blue}{b_{3}}c_{2}\right)\\\\&=&a_{1}\color{blue}{b_{3}}c_{1}-a_{1}b_{1}\color{green}{c_{3}}-a_{2}b_{2}\color{green}{c_{3}}+a_{2}\color{blue}{b_{3}}c_{2}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(a_{1}b_{1}+a_{2}b_{2}\right)\color{green}{c_{3}}\end{eqnarray}

内積変形のために、打ち消し項 \(a_{3}\color{blue}{b_{3}}c_{3}-a_{3}b_{3}\color{green}{c_{3}}\left(=0\right)\) を加えると、

\begin{eqnarray}&&a_{1}d_{2}-a_{2}d_{1}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(a_{1}b_{1}+a_{2}b_{2}\right)\color{green}{c_{3}}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(a_{1}b_{1}+a_{2}b_{2}\right)\color{green}{c_{3}}+a_{3}\color{blue}{b_{3}}c_{3}-a_{3}b_{3}\color{green}{c_{3}}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{3}}-\left(a_{1}b_{1}+a_{2}b_{2}+a_{3}b_{3}\right)\color{green}{c_{3}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\end{eqnarray}

No.6 : No.4 と No.5 の結果からベクトル3重積を求める

No.4 と No.5 の結果をまとめると

\begin{cases}a_{2}d_{3}-a_{3}d_{2}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\\\\a_{3}d_{1}-a_{1}d_{3}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\\\\a_{1}d_{2}-a_{2}d_{1}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\end{cases}

であるので、ベクトル3重積は

\begin{eqnarray}&&\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\\\\&=&\left(a_{2}d_{3}-a_{3}d_{2}\right)\color{red}{\pmb{i}}+\left(a_{3}d_{1}-a_{1}d_{3}\right)\color{blue}{\pmb{j}}+\left(a_{1}d_{2}-a_{2}d_{1}\right)\color{green}{\pmb{k}}\\\\&=&\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\}\color{red}{\pmb{i}}+\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\}\color{blue}{\pmb{j}}+\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\}\color{green}{\pmb{k}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}\color{red}{\pmb{i}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\color{red}{\pmb{i}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}\color{blue}{\pmb{j}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\color{blue}{\pmb{j}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}\color{green}{\pmb{k}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\color{green}{\pmb{k}}\\\\&=&\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}\color{red}{\pmb{i}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}\color{green}{\pmb{k}}\}-\{\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\color{red}{\pmb{i}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\color{blue}{\pmb{j}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\color{green}{\pmb{k}}\}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\{\color{blue}{b_{1}}\color{red}{\pmb{i}}+\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\color{blue}{b_{3}}\color{green}{\pmb{k}}\}-\left(\pmb{a}\cdot\pmb{b}\right)\{\color{green}{c_{1}}\color{red}{\pmb{i}}+\color{green}{c_{2}}\color{blue}{\pmb{j}}+\color{green}{c_{3}}\color{green}{\pmb{k}}\}\end{eqnarray}

となる。\(\pmb{b}=\color{blue}{b_{1}}\color{red}{\pmb{i}}+\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\color{blue}{b_{3}}\color{green}{\pmb{k}}\)、\(\pmb{c}=\color{green}{c_{1}}\color{red}{\pmb{i}}+\color{green}{c_{2}}\color{blue}{\pmb{j}}+\color{green}{c_{3}}\color{green}{\pmb{k}}\) であるので

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{a}\cdot\pmb{b}\right)\pmb{c}$$

例題

ベクトル3重積(B)

$$\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{b}\cdot\pmb{c}\right)\pmb{a}$$

を示せ。

例題の解答

No.1 : 外積 \(\pmb{a}\times\pmb{b}\) を成分計算

\begin{eqnarray}&&\pmb{a}\times\pmb{b}\\\\&=&\begin{vmatrix}\pmb{i} & \pmb{j} & \pmb{k}\\a_{1} & a_{2} & a_{3}\\b_{1} & b_{2} & b_{3}\end{vmatrix}\\\\&=&\left(a_{2}b_{3}-a_{3}b_{2}\right)\pmb{i}+\left(a_{3}b_{1}-a_{1}b_{3}\right)\pmb{j}+\left(a_{1}b_{2}-a_{2}b_{1}\right)\pmb{k}\end{eqnarray}

ここで、外積 \(\pmb{a}\times\pmb{b}\) の各成分が長ったらしいので \(e_{1}\)、\(e_{2}\)、\(e_{3}\) と置く。すなわち

\begin{cases}e_{1}=a_{2}b_{3}-a_{3}b_{2}\\\\e_{2}=a_{3}b_{1}-a_{1}b_{3}\\\\e_{3}=a_{1}b_{2}-a_{2}b_{1}\end{cases}

とすると、

$$\pmb{a}\times\pmb{b}=e_{1}\pmb{i}+e_{2}\pmb{j}+e_{3}\pmb{k}$$

とスッキリさせることができた。

No.2 : ベクトル3重積 \(\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c} \) を成分計算

ベクトル3重積$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)=\left(a_{1},a_{2},a_{3}\right)\times\left(d_{1},d_{2},d_{3}\right)$$を計算する。

\begin{eqnarray}&&\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}\\\\&=&\begin{vmatrix}\pmb{i} & \pmb{j} & \pmb{k}\\e_{1} & e_{2} & e_{3}\\c_{1} & c_{2} & c_{3}\end{vmatrix}\\\\&=&\left(e_{2}c_{3}-e_{3}c_{2}\right)\pmb{i}+\left(e_{3}c_{1}-e_{1}c_{3}\right)\pmb{j}+\left(e_{1}c_{2}-e_{2}c_{1}\right)\pmb{k}\end{eqnarray}

No.3 : \(x\) 成分 「\(e_{2}c_{3}-e_{3}c_{2}\)」 を代入計算

No.1 で置き換えた \(e\) を元の \(a\) と \(b\) の式に戻す。式の対称性から計算結果も対称的になるので、ひとまず\(x\) 方向成分 \(e_{2}c_{3}-e_{3}c_{2}\) のみ計算する。

\begin{eqnarray}e_{2}c_{3}-e_{3}c_{2}&=&\left(a_{3}\color{blue}{b_{1}}-\color{red}{a_{1}}b_{3}\right)c_{3}-\left(\color{red}{a_{1}}b_{2}-a_{2}\color{blue}{b_{1}}\right)c_{2}\\\\&=&a_{3}\color{blue}{b_{1}}c_{3}-\color{red}{a_{1}}b_{3}c_{3}-\color{red}{a_{1}}b_{2}c_{2}+a_{2}\color{blue}{b_{1}}c_{2}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{1}}\end{eqnarray}

No.4 : 式変形のための項「\(a_{1}b_{1}c_{1}-a_{1}b_{1}c_{1}\)」を加える。

No.3 で求めた式において、\(b_{1}\) の係数 「\(a_{2}c_{2}+a_{3}c_{3}\)」と \(a_{1}\) の係数 「\(b_{2}c_{2}+b_{3}c_{3}\)」は

内積

\begin{cases}\pmb{a}\cdot\pmb{c}=a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\\\\\pmb{b}\cdot\pmb{c}=b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}\end{cases}

を彷彿とさせる。

内積変形のために、打ち消し項 \(a_{1}\color{blue}{b_{1}}c_{1}-\color{red}{a_{1}}b_{1}c_{1}\left(=0\right)\) を加えると、

\begin{eqnarray}&&e_{2}c_{3}-e_{3}c_{2}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{1}}\\\\&=&\left(a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{1}}+a_{1}\color{blue}{b_{1}}c_{1}-\color{red}{a_{1}}b_{1}c_{1}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{1}}-\left(b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{1}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{1}}\end{eqnarray}

No.5 : \(y\) 成分 、\(z\) 成分も同様に求める。

\(y\) 成分について

\begin{eqnarray}&& e_{3}c_{1}-e_{1}c_{3} \\\\&=&\left(a_{1} \color{blue}{ b_{2}}- \color{red}{a_{2}}b_{1}\right)c_{1}-\left( \color{red}{a_{2}}b_{3}-a_{3}\color{blue}{b_{2}}\right) c_{3} \\\\&=& a_{1} \color{blue}{ b_{2}} c_{1} – \color{red}{ a_{2}}b_{1}c_{1} – \color{red}{a_{2}}b_{3} c_{3} +a_{3} \color{blue}{ b_{2}} c_{3} \\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(b_{1}c_{1}+b_{3}c_{3}\right)\color{red}{a_{2}}\end{eqnarray}

内積変形のために、打ち消し項 \(a_{2}\color{blue}{b_{2}}c_{2}-\color{red}{a_{2}}b_{2}c_{2}\left(=0\right)\) を加えると、

\begin{eqnarray}&&e_{3}c_{1}-e_{1}c_{3} \\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(b_{1}c_{1}+b_{3}c_{3}\right)\color{red}{a_{2}}\\\\&=&\left(a_{1}c_{1}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(b_{1}c_{1}+b_{3}c_{3}\right)\color{red}{a_{2}}+a_{2}\color{blue}{b_{2}}c_{2}-\color{red}{a_{2}}b_{2}c_{2}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{2}}-\left(b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{2}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{2}}\end{eqnarray}


\(z\) 成分について

\begin{eqnarray}&&e_{1}c_{2}-e_{2}c_{1} \\\\&=&\left(a_{2} \color{blue}{ b_{3}}- \color{red}{a_{3}}b_{2}\right)c_{2}-\left( \color{red}{a_{3}}b_{1}-a_{1}\color{blue}{b_{3}}\right) c_{1} \\\\&=& a_{2} \color{blue}{ b_{3}} c_{2} – \color{red}{ a_{3}}b_{2}c_{2} – \color{red}{a_{3}}b_{1} c_{1} +a_{1} \color{blue}{ b_{3}} c_{1} \\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(b_{1}c_{1}+b_{2}c_{2}\right)\color{red}{a_{3}}\end{eqnarray}

内積変形のために、打ち消し項 \(a_{3}\color{blue}{b_{3}}c_{3}-\color{red}{a_{3}}b_{3}c_{3}\left(=0\right)\) を加えると、

\begin{eqnarray}&&e_{1}c_{2}-e_{2}c_{1} \\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(b_{1}c_{1}+b_{2}c_{2}\right)\color{red}{a_{3}}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}\right)\color{blue}{b_{3}}-\left(b_{1}c_{1}+b_{2}c_{2}\right)\color{red}{a_{3}}+a_{3}\color{blue}{b_{3}}c_{3}-\color{red}{a_{3}}b_{3}c_{3}\\\\&=&\left(a_{1}c_{1}+a_{2}c_{2}+a_{3}c_{3}\right)\color{blue}{b_{3}}-\left(b_{1}c_{1}+b_{2}c_{2}+b_{3}c_{3}\right)\color{red}{a_{3}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{3}}\end{eqnarray}

No.6 : No.4 と No.5 の結果からベクトル3重積を求める

No.4 と No.5 の結果をまとめると

\begin{cases}e_{2}c_{3}-e_{3}c_{2}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{1}}\\\\e_{3}c_{1}-e_{1}c_{3}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{2}}\\\\e_{1}c_{2}-e_{2}c_{1}=\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{b}\cdot\pmb{c}\right)\color{red}{a_{3}}\end{cases}

であるので、ベクトル3重積は

\begin{eqnarray}&&\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)\\\\&=&\left(a_{2}d_{3}-a_{3}d_{2}\right)\color{red}{\pmb{i}}+\left(a_{3}d_{1}-a_{1}d_{3}\right)\color{blue}{\pmb{j}}+\left(a_{1}d_{2}-a_{2}d_{1}\right)\color{green}{\pmb{k}}\\\\&=&\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\}\color{red}{\pmb{i}}+\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\}\color{blue}{\pmb{j}}+\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\}\color{green}{\pmb{k}}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}\color{red}{\pmb{i}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\color{red}{\pmb{i}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}\color{blue}{\pmb{j}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\color{blue}{\pmb{j}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}\color{green}{\pmb{k}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\color{green}{\pmb{k}}\\\\&=&\{\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{1}}\color{red}{\pmb{i}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\left(\pmb{a}\cdot\pmb{c}\right)\color{blue}{b_{3}}\color{green}{\pmb{k}}\}-\{\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{1}}\color{red}{\pmb{i}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{2}}\color{blue}{\pmb{j}}-\left(\pmb{a}\cdot\pmb{b}\right)\color{green}{c_{3}}\color{green}{\pmb{k}}\}\\\\&=&\left(\pmb{a}\cdot\pmb{c}\right)\{\color{blue}{b_{1}}\color{red}{\pmb{i}}+\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\color{blue}{b_{3}}\color{green}{\pmb{k}}\}-\left(\pmb{a}\cdot\pmb{b}\right)\{\color{green}{c_{1}}\color{red}{\pmb{i}}+\color{green}{c_{2}}\color{blue}{\pmb{j}}+\color{green}{c_{3}}\color{green}{\pmb{k}}\}\end{eqnarray}

となる。\(\pmb{b}=\color{blue}{b_{1}}\color{red}{\pmb{i}}+\color{blue}{b_{2}}\color{blue}{\pmb{j}}+\color{blue}{b_{3}}\color{green}{\pmb{k}}\)、\(\pmb{a}=\color{red}{a_{1}}\color{red}{\pmb{i}}+\color{red}{a_{2}}\color{blue}{\pmb{j}}+\color{red}{a_{3}}\color{green}{\pmb{k}}\) であるので

$$\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{b}\cdot\pmb{c}\right)\pmb{a}$$

練習問題

\(\left(\pmb{a}\times\pmb{b}\right)\cdot\left(\pmb{c}\times\pmb{d}\right)=\left(\pmb{a}\cdot\pmb{c}\right)\left(\pmb{b}\cdot\pmb{d}\right)-\left(\pmb{b}\cdot\pmb{c}\right)\left(\pmb{a}\cdot\pmb{d}\right)\) を示せ。

まとめ

ベクトル \(\pmb{a}\)、\(\pmb{b}\)、\(\pmb{c}\) の ベクトル3重積 は2種類あり、

$$\pmb{a}\times\left(\pmb{b}\times\pmb{c}\right)=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{a}\cdot\pmb{b}\right)\pmb{c}\cdot\cdot\cdot\left(A\right)$$

$$\left(\pmb{a}\times\pmb{b}\right)\times\pmb{c}=\left(\pmb{a}\cdot\pmb{c}\right)\pmb{b}-\left(\pmb{b}\cdot\pmb{c}\right)\pmb{a} \cdot\cdot\cdot\left(B\right) $$

といった塩梅で、外積順序に応じて結果も異なる。(交換律が不成立)