1.4 平行四辺形の面積

定理 1.17 (平行四辺形の面積) 点

からなる平行四辺形

の面積は

$\displaystyle S=\mathrm{abs}\left( \begin{vmatrix}a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\right)$

で与えられる．

（証明）まず，

$\displaystyle \theta=\angle AOB, \quad \vec{a}= \begin{bmatrix}a_{1} \\ a_{2} \end{bmatrix}\,, \quad \vec{b}= \begin{bmatrix}b_{1} \\ b_{2} \end{bmatrix}$

とおく．すると，

$\displaystyle S$	$\displaystyle = \Vert\vec{a}\Vert\,\Vert\vec{b}\Vert\sin\theta= \Vert\vec{a}\Ve... ...ert\vec{b}\Vert^2- \left(\Vert\vec{a}\Vert\Vert\vec{b}\Vert\cos\theta\right)^2}$	(1)
	$\displaystyle = \sqrt{ (\vec{a}\cdot\vec{a})(\vec{b}\cdot\vec{b})- (\vec{a}\cdo... ...= \sqrt{ (a_{1}{}^2+a_{2}{}^2)(b_{1}{}^2+b_{2}{}^2)- (a_{1}b_{1}+a_{2}b_{2})^2}$	(2)
	$\displaystyle = \sqrt{ a_{1}{}^2b_{2}{}^2-2a_{1}a_{2}b_{1}b_{2}+ a_{2}{}^2b_{1}... ...= \sqrt{(a_{1}b_{2}-a_{2}b_{1})^2}= \left\vert a_{1}b_{2}-a_{2}b_{1}\right\vert$	(3)
	$\displaystyle = \mathrm{abs}\left( \begin{vmatrix}a_{1} & b_{1} \\ a_{2} & b_{2} \end{vmatrix}\right)$	(4)

と得られる．