空間における三角形の面積

数研出版｜数学Ｃ[708] p.81 演習問題A 2

点 \({\rm O}\) を原点として、

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\(\begin{eqnarray} \overrightarrow{\rm AB}&=&\overrightarrow{\rm OB}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}-1\\[2pt]-1\\[2pt]2\end{array}\,\right)-\left(\,\begin{array}{c}0\\[2pt]1\\[2pt]1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-1-0\\[2pt]-1-1\\[2pt]2-1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-1\\[2pt]-2\\[2pt]1\end{array}\,\right) \end{eqnarray}\)

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\(\begin{eqnarray} \overrightarrow{\rm AC}&=&\overrightarrow{\rm OC}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}2\\[2pt]3\\[2pt]1\end{array}\,\right)-\left(\,\begin{array}{c}0\\[2pt]1\\[2pt]1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}2-0\\[2pt]3-1\\[2pt]1-1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}2\\[2pt]2\\[2pt]0\end{array}\,\right) \end{eqnarray}\)

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\( \overrightarrow{\rm AB} \) と \( \overrightarrow{\rm AC} \) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&-1\cdot 2+(-2)\cdot 2+1\cdot 0
\\[3pt]~~~&=&-2-4+0
\\[3pt]~~~&=&-6\end{eqnarray}\)

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\( \overrightarrow{\rm AB} \) と \( \overrightarrow{\rm AC} \) の大きさは、

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AB}\,|&=&\sqrt{(-1)^2+(-2)^2+1^2}
\\[3pt]~~~&=&\sqrt{1+4+1}
\\[3pt]~~~&=&\sqrt{6}\end{eqnarray}\)

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AC}\,|&=&\sqrt{2^2+2^2+0^2}
\\[3pt]~~~&=&\sqrt{4+4+0}
\\[3pt]~~~&=&\sqrt{8}
\\[3pt]~~~&=&2\sqrt{2}\end{eqnarray}\)

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\( \overrightarrow{\rm AB} \) と \( \overrightarrow{\rm AC} \) のなす角を \( \theta \) とし、内積の定義の式より、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\cos\theta
\\[5pt]~-6&=&\sqrt{6}\cdot 2\sqrt{2}\cdot\cos\theta
\\[5pt]~2\sqrt{12}\cos\theta&=&-6
\\[5pt]~4\sqrt{3}\cos\theta&=&-6
\\[5pt]~\cos\theta&=&-\displaystyle \frac{\,\sqrt{3}\,}{\,2\,}\end{eqnarray}\)

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\( 0^\circ {\small ~≦~}\theta{\small ~≦~}180^\circ \) より、\(\theta=150^\circ \)

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よって、\({\small (1)}~\) \( \angle {\rm BAC}=150^\circ \)

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よって、\( \triangle {\rm ABC} \) の面積は、

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\(\begin{eqnarray}~\triangle {\rm ABC}&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot \sqrt{6}\cdot 2\sqrt{2}\cdot\sin 150^\circ
\\[5pt]~~~&=&\sqrt{12}\cdot \displaystyle \frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&2\sqrt{3}\cdot \displaystyle \frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{3}\end{eqnarray}\)

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よって、\({\small (2)}~\) \( \triangle {\rm ABC}=\sqrt{3} \)

数研出版｜高等学校数学Ｃ[709] p.74 章末問題B 9
数研出版｜新編数学Ｃ[710] p.73 章末問題B 6

点 \({\rm O}\) は原点なので、

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\(\begin{eqnarray} \overrightarrow{\rm OA}&=&\left(\,\begin{array}{c}-1\\[2pt]-2\\[2pt]1\end{array}\,\right) \end{eqnarray}\)

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\(\begin{eqnarray} \overrightarrow{\rm OB}&=&\left(\,\begin{array}{c}2\\[2pt]2\\[2pt]0\end{array}\,\right) \end{eqnarray}\)

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\( \overrightarrow{\rm OA} \) と \( \overrightarrow{\rm OB} \) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm OA}\cdot\overrightarrow{\rm OB}&=&(-1)\cdot 2+(-2)\cdot 2+1\cdot 0
\\[3pt]~~~&=&-2-4+0
\\[3pt]~~~&=&-6\end{eqnarray}\)

---

\( \overrightarrow{\rm OA} \) と \( \overrightarrow{\rm OB} \) の大きさは、

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm OA}\,|&=&\sqrt{(-1)^2+(-2)^2+1^2}
\\[3pt]~~~&=&\sqrt{1+4+1}
\\[3pt]~~~&=&\sqrt{6}\end{eqnarray}\)

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm OB}\,|&=&\sqrt{2^2+2^2+0^2}
\\[3pt]~~~&=&\sqrt{4+4+0}
\\[3pt]~~~&=&\sqrt{8}
\\[3pt]~~~&=&2\sqrt{2}\end{eqnarray}\)

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\( \overrightarrow{\rm OA} \) と \( \overrightarrow{\rm OB} \) のなす角を \( \theta \) とし、内積の定義の式より、

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\(\begin{eqnarray}~~~\overrightarrow{\rm OA}\cdot\overrightarrow{\rm OB}&=&|\,\overrightarrow{\rm OA}\,|\cdot|\,\overrightarrow{\rm OB}\,|\cdot\cos\theta
\\[5pt]~-6&=&\sqrt{6}\cdot 2\sqrt{2}\cdot\cos\theta
\\[5pt]~2\sqrt{12}\cos\theta&=&-6
\\[5pt]~4\sqrt{3}\cos\theta&=&-6
\\[5pt]~\cos\theta&=&-\displaystyle \frac{\,\sqrt{3}\,}{\,2\,}\end{eqnarray}\)

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\( 0^\circ {\small ~≦~}\theta{\small ~≦~}180^\circ \) より、\(\theta=150^\circ \)

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よって、\({\small (1)}~\) \( \angle {\rm AOB}=150^\circ \)

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よって、\( \triangle {\rm OAB} \) の面積 \( S \)は、

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\(\begin{eqnarray}~S&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm OA}\,|\cdot|\,\overrightarrow{\rm OB}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot \sqrt{6}\cdot 2\sqrt{2}\cdot\sin 150^\circ
\\[5pt]~~~&=&\sqrt{12}\cdot \displaystyle \frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&2\sqrt{3}\cdot \displaystyle \frac{\,1\,}{\,2\,}
\\[5pt]~~~&=&\sqrt{3}\end{eqnarray}\)

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よって、\({\small (2)}~\) \( S=\sqrt{3} \)