共通テスト数学ⅡBC｜空間ベクトルの基本問題10問【PDF・解説つき】

151 数学ⅡBC【解答】

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\(\overrightarrow{\rm AB}=(-2~,~4~,~-2)\)

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\(\overrightarrow{\rm AC}=(1~,~1~,~-2)\)

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\(\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}=6\)

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\(|\,\overrightarrow{\rm AB}\,|=2\sqrt{\,6\,}~,~|\,\overrightarrow{\rm AC}\,|=\sqrt{\,6\,}\)

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\(\triangle{\rm ABC}=3\sqrt{\,3\,}\)

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【より詳しい解説】
点 \({\rm O}\) を原点として、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}&=&\overrightarrow{\rm OB}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}-2\\[2pt]3\\[2pt]-1\end{array}\,\right)-\left(\,\begin{array}{c}0\\[2pt]-1\\[2pt]1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-2-0\\[2pt]3+1\\[2pt]-1-1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-2\\[2pt]4\\[2pt]-2\end{array}\,\right) \end{eqnarray}\)

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

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

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

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

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AB}\,|&=&\sqrt{\,(-2)^2+4^2+(-2)^2\,}
\\[3pt]~~~&=&\sqrt{\,4+16+4\,}
\\[3pt]~~~&=&\sqrt{\,24\,}
\\[3pt]~~~&=&2\sqrt{\,6\,}\end{eqnarray}\)

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

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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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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\cos\theta
\\[5pt]~~~6&=&2\sqrt{\,6\,}\cdot\sqrt{\,6\,}\cdot\cos\theta
\\[5pt]~~~12\cos\theta&=&6
\\[5pt]~~~\cos\theta&=&\displaystyle \frac{\,6\,}{\,12\,}
\\[5pt]~~~\cos\theta&=&\displaystyle \frac{\,1\,}{\,2\,}\end{eqnarray}\)

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\(0^{\circ}{\small ~≦~}\theta{\small ~≦~}180^{\circ}\) より、\(\theta=60^{\circ}\)

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よって、\(\triangle{\rm ABC}\) の面積は、

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\(\begin{eqnarray}~~~\triangle{\rm ABC}&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot 2\sqrt{\,6\,}\cdot\sqrt{\,6\,}\cdot\sin 60^{\circ}
\\[5pt]~~~&=&6\cdot \displaystyle \frac{\,\sqrt{\,3\,}\,}{\,2\,}
\\[5pt]~~~&=&3\sqrt{\,3\,}\end{eqnarray}\)