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 {\displaystyle R^{3}} のベクトル a → = ( a b c ) {\displaystyle {\vec {a}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)} , b → = ( x y z ) {\displaystyle {\vec {b}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)} に関して、 a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} を次で定める.
a → × b → = ( b z − c y c x − a z a y − b x ) {\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。
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. a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} で a → {\displaystyle {\vec {a}}} と b → {\displaystyle {\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) a → × b → = − ( b → × a → ) {\displaystyle {\vec {a}}\times {\vec {b}}=-({\vec {b}}\times {\vec {a}})}
(2) k ( a → × b → ) = ( k a → ) × b → = a → × ( k b → ) {\displaystyle k({\vec {a}}\times {\vec {b}})=(k{\vec {a}})\times {\vec {b}}={\vec {a}}\times (k{\vec {b}})}
(3) a → × ( b → + c → ) = a → × b → + a → × c → {\displaystyle {\vec {a}}\times ({\vec {b}}+{\vec {c}})={\vec {a}}\times {\vec {b}}+{\vec {a}}\times {\vec {c}}}
(4) ( a → + b → ) × c → = a → × c → + b → × c → {\displaystyle ({\vec {a}}+{\vec {b}})\times {\vec {c}}={\vec {a}}\times {\vec {c}}+{\vec {b}}\times {\vec {c}}}
証明
(1)
a → = ( a b c ) {\displaystyle {\vec {a}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)} , b → = ( x y z ) {\displaystyle {\vec {b}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)} にて b → × a → = ( y c − z b z a − x c x b − y a ) = ( − ( b z − c y ) − ( c x − a z ) − ( a y − b x ) ) = − ( b z − c y c x − a z a y − b x ) = − a → × b → {\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)
k ( a → × b → ) = ( k ( b z − c y ) k ( c x − a z ) k ( a y − b x ) ) = ( k b z − k c y k c x − k a z k a y − k b x ) = ( ( k b ) z − ( k c ) y ( k c ) x − ( k a ) z ( k a ) y − ( k b ) x ) = ( k a → ) × b → {\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}}} .
同様に k ( a → × b → ) = ( k b z − k c y k c x − k a z k a y − k b x ) = ( b ( k z ) − c ( k y ) c ( k x ) − a ( k z ) a ( k y ) − b ( k x ) ) = a → × ( k b → ) {\displaystyle 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)
a → = ( a 1 a 2 a 3 ) {\displaystyle {\vec {a}}=\left({\begin{array}{c}a_{1}\\a_{2}\\a_{3}\end{array}}\right)} , b → = ( b 1 b 2 b 3 ) {\displaystyle {\vec {b}}=\left({\begin{array}{c}b_{1}\\b_{2}\\b_{3}\end{array}}\right)} , c → = ( c 1 c 2 c 3 ) {\displaystyle {\vec {c}}=\left({\begin{array}{c}c_{1}\\c_{2}\\c_{3}\end{array}}\right)} にて
a → × ( b → + c → ) = ( a 1 a 2 a 3 ) × ( b 1 + c 1 b 2 + c 2 b 3 + c 3 ) = ( 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 ) ) = ( 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 ) {\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)}
= ( 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 ) = ( a 2 b 3 − a 3 b 2 a 3 b 1 − a 1 b 3 a 1 b 2 − a 2 b 1 ) + ( a 2 c 3 − a 3 c 2 a 3 c 1 − a 1 c 3 a 1 c 2 − a 2 c 1 ) = a → × b → + a → × c → {\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)
( a → + b → ) × c → = − c → × ( a → + b → ) = − ( c → × a → + c → × b → ) = − ( − a → × c → − b → × c → ) = a → × c → + b → × c → {\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.
定理5 ベクトル積の意味
(1) a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} は, a → {\displaystyle {\vec {a}}} , b → {\displaystyle {\vec {b}}} の両方と直交する.
(2) a → {\displaystyle {\vec {a}}} と b → {\displaystyle {\vec {b}}} が張る平行四辺形の面積 S {\displaystyle S} は, S = | a → × b → | {\displaystyle S=|{\vec {a}}\times {\vec {b}}|}
(3) a → {\displaystyle {\vec {a}}} , b → {\displaystyle {\vec {b}}} , c → {\displaystyle {\vec {c}}} が張る平行六面体の体積 V {\displaystyle V} は, V = | ( a → × b → ) ⋅ c → | {\displaystyle V=|({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}|}
証明
a → = O A → = ( a b c ) {\displaystyle {\vec {a}}={\vec {OA}}=\left({\begin{array}{c}a\\b\\c\end{array}}\right)} , b → = O B → = ( x y z ) {\displaystyle {\vec {b}}={\vec {OB}}=\left({\begin{array}{c}x\\y\\z\end{array}}\right)} とする.
(1)
a → {\displaystyle {\vec {a}}} と a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} の内積をとる.
a → ⋅ ( a → × b → ) = ( a b c ) ⋅ ( b z − c y c x − a z a y − b x ) = a ( b z − c y ) + b ( c x − a z ) + c ( a y − b x ) = a b z − a c y + b c x − a b z + a c y − b c x {\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} = a b z − a c y + b c x − a b z + a c y − b c x = 0 {\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} .
同様に
b → ⋅ ( a → × b → ) = ( x y z ) ⋅ ( b z − c y c x − a z a y − b x ) = x ( b z − c y ) + y ( c x − a z ) + z ( a y − b x ) = b x z − c x y + c x y − a y z + a y z − b x z {\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} = b x z − c x y + c x y − a y z + a y z − b x z = 0 {\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, a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} は a → {\displaystyle {\vec {a}}} , b → {\displaystyle {\vec {b}}} の両方と直交する.
(2)
a → {\displaystyle {\vec {a}}} , b → {\displaystyle {\vec {b}}} のなす角を θ {\displaystyle \theta } とすると,内積の性質より,
| a → | | b → | cos θ = a → ⋅ b → = a x + b y + c z {\displaystyle |{\vec {a}}||{\vec {b}}|\cos \theta ={\vec {a}}\cdot {\vec {b}}=ax+by+cz}
O A {\displaystyle \mathrm {OA} } を底辺としたときの B {\displaystyle \mathrm {B} } の高さを h {\displaystyle h} とすると,
S = O A × h = | a → | | b → | sin θ ( ∵ h = | b → | sin θ ) {\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,
S 2 = | a → | 2 | b → | 2 sin 2 θ = | a → | 2 | b → | 2 ( 1 − cos 2 θ ) = | a → | 2 | b → | 2 − | a → | 2 | b → | 2 cos 2 θ {\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 }
| a → | 2 | b → | 2 cos 2 θ = ( a → ⋅ b → ) 2 = ( a x + b y + c z ) 2 {\displaystyle |{\vec {a}}|^{2}|{\vec {b}}|^{2}\cos ^{2}\theta =({\vec {a}}\cdot {\vec {b}})^{2}=(ax+by+cz)^{2}} ,また | a → | 2 | b → | 2 = ( a 2 + b 2 + c 2 ) ( x 2 + y 2 + z 2 ) {\displaystyle |{\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 ) − ( a x + b y + c z ) 2 {\displaystyle 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 + a b x y + a c x z + a b x y + b 2 y 2 + b c y z + a c x z + b c y z + c 2 z 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})}
= 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 + a b x y + a c x z + a b x y + b 2 y 2 + b c y z + a c x z + b c y z + 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}}})}
= a 2 y 2 + a 2 z 2 + b 2 x 2 + b 2 z 2 + c 2 x 2 + c 2 y 2 − ( a b x y + a c x z + a b x y + b c y z + a c x z + b c y z ) {\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})}
= ( b 2 z 2 − 2 b c y z + c 2 y 2 ) + ( c 2 x 2 − 2 a c x z + a 2 z 2 ) + ( a 2 y 2 − 2 a b x y + b 2 x 2 ) {\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})}
= ( b z − c y ) 2 + ( c x − a z ) 2 + ( a y − b x ) 2 {\displaystyle =(bz-cy)^{2}+(cx-az)^{2}+(ay-bx)^{2}}
= | a → × b → | 2 {\displaystyle =|{\vec {a}}\times {\vec {b}}|^{2}}
なぜならば | a → × b → | 2 = ( a → × b → ) ⋅ ( a → × b → ) = ( b z − c y c x − a z a y − b x ) ⋅ ( b z − c y c x − a z a y − b x ) = ( b z − c y ) 2 + ( c x − a z ) 2 + ( a y − b x ) 2 {\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, S = | a → × b → | {\displaystyle S=|{\vec {a}}\times {\vec {b}}|}
(3) c → = O C {\displaystyle {\vec {c}}=\mathrm {OC} } とする. a → {\displaystyle {\vec {a}}} , b → {\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 C {\displaystyle \mathrm {C} } の高さを l {\displaystyle l} , a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} と c → {\displaystyle {\vec {c}}} のなす角を φ {\displaystyle \varphi } とすると,
a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} は a → {\displaystyle {\vec {a}}} , b → {\displaystyle {\vec {b}}} が張る平行四辺形に垂直なので,
l = | O C cos φ | = | c → | ⋅ | cos φ | {\displaystyle l=|\mathrm {OC} \cos \varphi |=|{\vec {c}}|\cdot |\cos \varphi |} だから,
V = S l = | a → × b → | ⋅ | c → | ⋅ | cos φ | = | | a → × b → | ⋅ | c → | ⋅ cos φ | = | ( a → × b → ) ⋅ c → | {\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 }
演習1. {\displaystyle \quad }
a → = ( 1 − 3 2 ) {\displaystyle {\vec {a}}=\left({\begin{array}{c}1\\-3\\2\end{array}}\right)} , b → = ( 2 − 2 3 ) {\displaystyle {\vec {b}}=\left({\begin{array}{c}2\\-2\\3\end{array}}\right)} , c → = ( − 1 − 1 0 ) {\displaystyle {\vec {c}}=\left({\begin{array}{c}-1\\-1\\0\end{array}}\right)} のとき,
(1) a → × b → , ( a → × b → ) ⋅ c → {\displaystyle {\vec {a}}\times {\vec {b}},({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}} を求めよ.
(2) a → , b → {\displaystyle {\vec {a}},{\vec {b}}} の両方と直交する単位ベクトルを求めよ.
(3) O A → = a → , O B → = b → , O C → = c → {\displaystyle {\vec {\mathrm {OA} }}={\vec {a}},{\vec {\mathrm {OB} }}={\vec {b}},{\vec {\mathrm {OC} }}={\vec {c}}} とするとき,三角錐 O − A B C {\displaystyle \mathrm {O-ABC} } の体積を求めよ.
解答例
(1) a → × b → = ( 1 − 3 2 ) × ( 2 − 2 3 ) = ( ( − 3 ) ⋅ 3 − 2 ⋅ ( − 2 ) 2 ⋅ 2 − 1 ⋅ 3 1 ⋅ ( − 2 ) − ( − 3 ) ⋅ 2 ) = ( − 5 1 4 ) {\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)} .
( a → × b → ) ⋅ c → = ( − 5 1 4 ) ⋅ ( − 1 − 1 0 ) = − 5 ⋅ ( − 1 ) + 1 ⋅ ( − 1 ) + 4 ⋅ 0 = 4 {\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) a → × b → {\displaystyle {\vec {a}}\times {\vec {b}}} は a → {\displaystyle {\vec {a}}} と b → {\displaystyle {\vec {b}}} に垂直なので, d → = a → × b → {\displaystyle {\vec {d}}={\vec {a}}\times {\vec {b}}} を単位化する.
± 1 | d → | d → = ± 1 ( − 5 ) 2 + 1 2 + 4 2 {\displaystyle \pm {\frac {1}{|{\vec {d}}|}}{\vec {d}}=\pm {\frac {1}{\sqrt {(-5)^{2}+1^{2}+4^{2}}}}} ( − 5 1 4 ) = ± 1 42 ( − 5 1 4 ) {\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) a → , b → , c → {\displaystyle {\vec {a}},{\vec {b}},{\vec {c}}} が張る平行六面体の体積 V {\displaystyle V} は,
V = | ( a → × b → ) ⋅ c → = | 4 | = 4 {\displaystyle V=|({\vec {a}}\times {\vec {b}})\cdot {\vec {c}}=|4|=4}
a → , b → {\displaystyle {\vec {a}},{\vec {b}}} が張る平行四辺形の面積を S {\displaystyle S} , a → , b → {\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 h {\displaystyle h} とすると V = S h {\displaystyle V=Sh} であり,
(三角錐 O − A B C {\displaystyle \mathrm {O-ABC} } の体積) = 1 3 ( 1 2 S ) h = 1 6 V = 4 6 = 2 3 {\displaystyle ={\frac {1}{3}}\left({\frac {1}{2}}S\right)h={\frac {1}{6}}V={\frac {4}{6}}={\frac {2}{3}}} .
◼ {\displaystyle \blacksquare }