Cho hai véctơ: $\overrightarrow{a}\left(x_a;\,y_a\right),\,\,\overrightarrow{b}\left(x_b;\,y_b\right)$ và $\overrightarrow{a}\times\overrightarrow{b}=x_ay_b-x_by_a.$ Chứng minh rằng: $$\left|\overrightarrow{a}\times\overrightarrow{b} \right|=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\sin\left(\overrightarrow{a};\,\overrightarrow{b}\right)$$
Ta có:$\overrightarrow a\centerdot \overrightarrow b=x_ax_b+y_ay_b$ và $|\overrightarrow a\centerdot \overrightarrow b|=|\overrightarrow a||\overrightarrow b|\cos(\overrightarrow a;\overrightarrow b)$
Lại có:
     $(\overrightarrow a\times\overrightarrow b)^2+(\overrightarrow a\centerdot \overrightarrow b)^2$
$=(x_ay_b-x_by_a)^2+(x_ax_b+y_ay_b)^2$
$=x_a^2y_b^2+x_b^2y_a^2+x_a^2x_b^2+y_a^2y_b^2$
$=(x_a^2+y_a^2)(x_b^2+y_b^2)$
$=|\overrightarrow a|^2|\overrightarrow b|^2$
Suy ra: $|\overrightarrow a\times \overrightarrow b|^2=|\overrightarrow a|^2|\overrightarrow b|^2\sin^2(\overrightarrow a;\overrightarrow b)$
$\Rightarrow |\overrightarrow a\times \overrightarrow b|=|\overrightarrow a||\overrightarrow b|\sin(\overrightarrow a;\overrightarrow b)$, do $\sin(\overrightarrow a;\overrightarrow b)\ge0.$

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