線型代数学/行列と行列式/第三類/外積

内積があるのなら外積があってもいいのでは，と思っている人もいることだろう． 単に「外積」と呼ばれることもある，3次元実数ベクトルについての外積，すなわち「ベクトル積」を紹介する．

定義3 外積

$R^{3}$のベクトル ${\vec {a}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)$ ， ${\vec {b}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)$ に関して、${\vec {a}}\times {\vec {b}}$ を次で定める．

${\vec {a}}\times {\vec {b}}=\left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)$

ベクトル積は 3 次元ベクトルの場合のみについて定義される演算である。

定義のとおりだが、実際の計算は図に示したように第 1 成分を下に付け加え， ×の形に積を取り，使っていない成分に押し込むという感じで技化しておくとよい．

ベクトル積に関して次の計算法則が成り立つ． 交換法則が成り立たないことに注意する． ${\vec {a}}\times {\vec {b}}$ で ${\vec {a}}$ と ${\vec {b}}$ を入れ替えると，符号が逆になる．

定理4 ベクトル積の計算法則

(1) ${\vec {a}}\times {\vec {b}}=-({\vec {b}}\times {\vec {a}})$

(2) $k({\vec {a}}\times {\vec {b}})=(k{\vec {a}})\times {\vec {b}}={\vec {a}}\times (k{\vec {b}})$

(3) ${\vec {a}}\times ({\vec {b}}+{\vec {c}})={\vec {a}}\times {\vec {b}}+{\vec {a}}\times {\vec {c}}$

(4) $({\vec {a}}+{\vec {b}})\times {\vec {c}}={\vec {a}}\times {\vec {c}}+{\vec {b}}\times {\vec {c}}$

証明

(1)

${\vec {a}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)$ ， ${\vec {b}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)$ にて ${\vec {b}}\times {\vec {a}}=\left({\begin{array}{c}yc-zb\\za-xc\\xb-ya\end{array}}\right)=\left({\begin{array}{c}-(bz-cy)\\-(cx-az)\\-(ay-bx)\end{array}}\right)=-\left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)=-{\vec {a}}\times {\vec {b}}$．

(2)

$k({\vec {a}}\times {\vec {b}})=\left({\begin{array}{c}k(bz-cy)\\k(cx-az)\\k(ay-bx)\end{array}}\right)=\left({\begin{array}{c}kbz-kcy\\kcx-kaz\\kay-kbx\end{array}}\right)=\left({\begin{array}{c}(kb)z-(kc)y\\(kc)x-(ka)z\\(ka)y-(kb)x\end{array}}\right)=(k{\vec {a}})\times {\vec {b}}$．

同様に $k({\vec {a}}\times {\vec {b}})=\left({\begin{array}{c}kbz-kcy\\kcx-kaz\\kay-kbx\end{array}}\right)=\left({\begin{array}{c}b(kz)-c(ky)\\c(kx)-a(kz)\\a(ky)-b(kx)\end{array}}\right)={\vec {a}}\times (k{\vec {b}})$．

(3)

${\vec {a}}=\left({\begin{array}{c}a_{1}\\a_{2}\\a_{3}\end{array}}\right)$ ， ${\vec {b}}=\left({\begin{array}{c}b_{1}\\b_{2}\\b_{3}\end{array}}\right)$ ， ${\vec {c}}=\left({\begin{array}{c}c_{1}\\c_{2}\\c_{3}\end{array}}\right)$ にて

${\vec {a}}\times ({\vec {b}}+{\vec {c}})=\left({\begin{array}{c}a_{1}\\a_{2}\\a_{3}\end{array}}\right)\times \left({\begin{array}{c}b_{1}+c_{1}\\b_{2}+c_{2}\\b_{3}+c_{3}\end{array}}\right)=\left({\begin{array}{c}a_{2}(b_{3}+c_{3})-a_{3}(b_{2}+c_{2})\\a_{3}(b_{1}+c_{1})-a_{1}(b_{3}+c_{3})\\a_{1}(b_{2}+c_{2})-a_{2}(b_{1}+c_{1})\end{array}}\right)=\left({\begin{array}{c}a_{2}b_{3}+a_{2}c_{3}-a_{3}b_{2}-a_{3}c_{2}\\a_{3}b_{1}+a_{3}c_{1}-a_{1}b_{3}-a_{1}c_{3}\\a_{1}b_{2}+a_{1}c_{2}-a_{2}b_{1}-a_{2}c_{1}\end{array}}\right)$

$=\left({\begin{array}{c}a_{2}b_{3}-a_{3}b_{2}+a_{2}c_{3}-a_{3}c_{2}\\a_{3}b_{1}-a_{1}b_{3}+a_{3}c_{1}-a_{1}c_{3}\\a_{1}b_{2}-a_{2}b_{1}+a_{1}c_{2}-a_{2}c_{1}\end{array}}\right)=\left({\begin{array}{c}a_{2}b_{3}-a_{3}b_{2}\\a_{3}b_{1}-a_{1}b_{3}\\a_{1}b_{2}-a_{2}b_{1}\end{array}}\right)+\left({\begin{array}{c}a_{2}c_{3}-a_{3}c_{2}\\a_{3}c_{1}-a_{1}c_{3}\\a_{1}c_{2}-a_{2}c_{1}\end{array}}\right)={\vec {a}}\times {\vec {b}}+{\vec {a}}\times {\vec {c}}$．

(4)

$({\vec {a}}+{\vec {b}})\times {\vec {c}}=-{\vec {c}}\times ({\vec {a}}+{\vec {b}})=-({\vec {c}}\times {\vec {a}}+{\vec {c}}\times {\vec {b}})=-(-{\vec {a}}\times {\vec {c}}-{\vec {b}}\times {\vec {c}})={\vec {a}}\times {\vec {c}}+{\vec {b}}\times {\vec {c}}$

$\blacksquare$

2次元ベクトル，3次元ベクトルの内積は，図形的な解釈が可能であった． ベクトル積が図形的には何を表しているかを紹介する．

定理5 ベクトル積の意味

(1) ${\vec {a}}\times {\vec {b}}$ は，${\vec {a}}$，${\vec {b}}$ の両方と直交する．

(2) ${\vec {a}}$ と ${\vec {b}}$ が張る平行四辺形の面積 $S$ は，$S=|{\vec {a}}\times {\vec {b}}|$

(3) ${\vec {a}}$，${\vec {b}}$，${\vec {c}}$ が張る平行六面体の体積 $V$ は，$V=|({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}|$

証明

${\vec {a}}={\vec {OA}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)$， ${\vec {b}}={\vec {OB}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)$ とする．

(1)

${\vec {a}}$ と ${\vec {a}}\times {\vec {b}}$ の内積をとる．

${\vec {a}}\cdot ({\vec {a}}\times {\vec {b}})=\left({\begin{array}{c}a\\b\\c\end{array}}\right)\cdot \left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)=a(bz-cy)+b(cx-az)+c(ay-bx)=abz-acy+bcx-abz+acy-bcx$
$={\cancel {\color {red}abz}}-{\cancel {\color {blue}acy}}+{\cancel {\color {green}bcx}}-{\cancel {\color {red}abz}}+{\cancel {\color {blue}acy}}-{\cancel {\color {green}bcx}}=0$．

同様に

${\vec {b}}\cdot ({\vec {a}}\times {\vec {b}})=\left({\begin{array}{c}x\\y\\z\end{array}}\right)\cdot \left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)=x(bz-cy)+y(cx-az)+z(ay-bx)=bxz-cxy+cxy-ayz+ayz-bxz$
$={\cancel {\color {red}bxz}}-{\cancel {\color {blue}cxy}}+{\cancel {\color {blue}cxy}}-{\cancel {\color {green}ayz}}+{\cancel {\color {green}ayz}}-{\cancel {\color {red}bxz}}=0$．

よって，${\vec {a}}\times {\vec {b}}$ は ${\vec {a}}$，${\vec {b}}$ の両方と直交する．

(2)

${\vec {a}}$，${\vec {b}}$ のなす角を $\theta$ とすると，内積の性質より，

$|{\vec {a}}||{\vec {b}}|\cos \theta ={\vec {a}}\cdot {\vec {b}}=ax+by+cz$

$\mathrm {OA}$ を底辺としたときの $\mathrm {B}$ の高さを $h$ とすると，

$S=\mathrm {OA} \times h=|{\vec {a}}||{\vec {b}}|\sin \theta (\because h=|{\vec {b}}|\sin \theta )$

と表されるので，

$S^{2}=|{\vec {a}}|^{2}|{\vec {b}}|^{2}\sin ^{2}\theta =|{\vec {a}}|^{2}|{\vec {b}}|^{2}(1-\cos ^{2}\theta )=|{\vec {a}}|^{2}|{\vec {b}}|^{2}-|{\vec {a}}|^{2}|{\vec {b}}|^{2}\cos ^{2}\theta$

$|{\vec {a}}|^{2}|{\vec {b}}|^{2}\cos ^{2}\theta =({\vec {a}}\cdot {\vec {b}})^{2}=(ax+by+cz)^{2}$，また $|{\vec {a}}|^{2}|{\vec {b}}|^{2}=(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})$ だから，

$S^{2}=(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})-(ax+by+cz)^{2}$

$=a^{2}x^{2}+a^{2}y^{2}+a^{2}z^{2}+b^{2}x^{2}+b^{2}y^{2}+b^{2}z^{2}+c^{2}x^{2}+c^{2}y^{2}+c^{2}z^{2}-(a^{2}x^{2}+abxy+acxz+abxy+b^{2}y^{2}+bcyz+acxz+bcyz+c^{2}z^{2})$

$={\cancel {\color {red}a^{2}x^{2}}}+a^{2}y^{2}+a^{2}z^{2}+b^{2}x^{2}+{\cancel {\color {blue}b^{2}y^{2}}}+b^{2}z^{2}+c^{2}x^{2}+c^{2}y^{2}+{\cancel {\color {green}c^{2}z^{2}}}-({\cancel {\color {red}a^{2}x^{2}}}+abxy+acxz+abxy+{\cancel {\color {blue}b^{2}y^{2}}}+bcyz+acxz+bcyz+{\cancel {\color {green}c^{2}z^{2}}})$

$={\color {red}a^{2}y^{2}}+{\color {blue}a^{2}z^{2}}+{\color {red}b^{2}x^{2}}+{\color {green}b^{2}z^{2}}+{\color {blue}c^{2}x^{2}}+{\color {green}c^{2}y^{2}}-({\color {red}abxy}+{\color {blue}acxz}+{\color {red}abxy}+{\color {green}bcyz}+{\color {blue}acxz}+{\color {green}bcyz})$

$=(b^{2}z^{2}-2bcyz+c^{2}y^{2})+(c^{2}x^{2}-2acxz+a^{2}z^{2})+(a^{2}y^{2}-2abxy+b^{2}x^{2})$

$=(bz-cy)^{2}+(cx-az)^{2}+(ay-bx)^{2}$

$=|{\vec {a}}\times {\vec {b}}|^{2}$

なぜならば $|{\vec {a}}\times {\vec {b}}|^{2}=({\vec {a}}\times {\vec {b}})\cdot ({\vec {a}}\times {\vec {b}})=\left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)\cdot \left({\begin{array}{c}bz-cy\\cx-az\\ay-bx\end{array}}\right)=(bz-cy)^{2}+(cx-az)^{2}+(ay-bx)^{2}$

よって，$S=|{\vec {a}}\times {\vec {b}}|$

(3) ${\vec {c}}=\mathrm {OC}$ とする． ${\vec {a}}$，${\vec {b}}$ が張る平面を平行四辺形の底面としてみたときの $\mathrm {C}$ の高さを $l$， ${\vec {a}}\times {\vec {b}}$ と ${\vec {c}}$ のなす角を $\varphi$ とすると，

${\vec {a}}\times {\vec {b}}$ は ${\vec {a}}$，${\vec {b}}$ が張る平行四辺形に垂直なので，

$l=|\mathrm {OC} \cos \varphi |=|{\vec {c}}|\cdot |\cos \varphi |$ だから，

$V=Sl=|{\vec {a}}\times {\vec {b}}|\cdot |{\vec {c}}|\cdot |\cos \varphi |=\left||{\vec {a}}\times {\vec {b}}|\cdot |{\vec {c}}|\cdot \cos \varphi \right|=\left|({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}\right|$．

$\blacksquare$

演習1.$\quad$

${\vec {a}}=\left({\begin{array}{c}1\\-3\\2\end{array}}\right)$ ， ${\vec {b}}=\left({\begin{array}{c}2\\-2\\3\end{array}}\right)$ ， ${\vec {c}}=\left({\begin{array}{c}-1\\-1\\0\end{array}}\right)$ のとき，

(1) ${\vec {a}}\times {\vec {b}},({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}$ を求めよ．

(2) ${\vec {a}},{\vec {b}}$ の両方と直交する単位ベクトルを求めよ．

(3) ${\vec {\mathrm {OA} }}={\vec {a}},{\vec {\mathrm {OB} }}={\vec {b}},{\vec {\mathrm {OC} }}={\vec {c}}$ とするとき，三角錐 $\mathrm {O-ABC}$ の体積を求めよ．

解答例

(1) ${\vec {a}}\times {\vec {b}}=\left({\begin{array}{c}1\\-3\\2\end{array}}\right)\times \left({\begin{array}{c}2\\-2\\3\end{array}}\right)=\left({\begin{array}{c}(-3)\cdot 3-2\cdot (-2)\\2\cdot 2-1\cdot 3\\1\cdot (-2)-(-3)\cdot 2\end{array}}\right)=\left({\begin{array}{c}-5\\1\\4\end{array}}\right)$．

$({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}=\left({\begin{array}{c}-5\\1\\4\end{array}}\right)\cdot \left({\begin{array}{c}-1\\-1\\0\end{array}}\right)=-5\cdot (-1)+1\cdot (-1)+4\cdot 0=4$.

(2) ${\vec {a}}\times {\vec {b}}$ は ${\vec {a}}$ と ${\vec {b}}$ に垂直なので，${\vec {d}}={\vec {a}}\times {\vec {b}}$ を単位化する.

$\pm {\frac {1}{|{\vec {d}}|}}{\vec {d}}=\pm {\frac {1}{\sqrt {(-5)^{2}+1^{2}+4^{2}}}}$ $\left({\begin{array}{c}-5\\1\\4\end{array}}\right)=\pm {\frac {1}{\sqrt {42}}}\left({\begin{array}{c}-5\\1\\4\end{array}}\right)$．(解となるベクトルは二つ）

(3) ${\vec {a}},{\vec {b}},{\vec {c}}$ が張る平行六面体の体積 $V$ は，

$V=|({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}=|4|=4$

${\vec {a}},{\vec {b}}$ が張る平行四辺形の面積を $S$，${\vec {a}},{\vec {b}}$ が張る平行四辺形を底面として見たときの平行六面体の高さを $h$ とすると $V=Sh$ であり，

（三角錐 $\mathrm {O-ABC}$ の体積）$={\frac {1}{3}}\left({\frac {1}{2}}S\right)h={\frac {1}{6}}V={\frac {4}{6}}={\frac {2}{3}}$．

$\blacksquare$