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# 線型代数学/行列と行列式/第三類/外積

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

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

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。

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． ${\displaystyle {\vec {a}}\times {\vec {b}}}$${\displaystyle {\vec {a}}}$${\displaystyle {\vec {b}}}$ を入れ替えると，符号が逆になる．

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

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

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

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

(1)

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

(2)

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

(3)

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

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

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

(4)

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

${\displaystyle \blacksquare }$

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．

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

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

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

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

(1)

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

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

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

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

(2)

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

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

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

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

と表されるので，

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

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

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

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

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

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

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

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

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

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

よって，${\displaystyle S=|{\vec {a}}\times {\vec {b}}|}$

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

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

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

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

${\displaystyle \blacksquare }$

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

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

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

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

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

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

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

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

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

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

${\displaystyle {\vec {a}},{\vec {b}}}$ が張る平行四辺形の面積を ${\displaystyle S}$${\displaystyle {\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 ${\displaystyle h}$ とすると ${\displaystyle V=Sh}$ であり，

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

${\displaystyle \blacksquare }$