四面体の垂直条件と四面体の体積

数研出版｜数学Ｃ[708] p.81 演習問題B 4

原点を \( \rm O \) として、

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\({\small (1)}~\) 辺 \( \rm CD \) の中点 \( \rm M \) は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm OM}&=&\displaystyle \frac{\,\overrightarrow{\rm OC}+\overrightarrow{\rm OD}\,}{\,2\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\left\{\left(\,\begin{array}{c}5\\[2pt]1\\[2pt]8\end{array}\,\right)+\left(\,\begin{array}{c}3\\[2pt]-3\\[2pt]6\end{array}\,\right)\right\}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\left(\,\begin{array}{c}8\\[2pt]-2\\[2pt]14\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}4\\[2pt]-1\\[2pt]7\end{array}\,\right)\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\overrightarrow{\rm BM}&=&\overrightarrow{\rm OM}-\overrightarrow{\rm OB}
\\[5pt]~~~&=&\left(\,\begin{array}{c}4\\[2pt]-1\\[2pt]7\end{array}\,\right)-\left(\,\begin{array}{c}1\\[2pt]3\\[2pt]2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}4-1\\[2pt]-1-3\\[2pt]7-2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}3\\[2pt]-4\\[2pt]5\end{array}\,\right)\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\overrightarrow{\rm CD}&=&\overrightarrow{\rm OD}-\overrightarrow{\rm OC}
\\[5pt]~~~&=&\left(\,\begin{array}{c}3\\[2pt]-3\\[2pt]6\end{array}\,\right)-\left(\,\begin{array}{c}5\\[2pt]1\\[2pt]8\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}3-5\\[2pt]-3-1\\[2pt]6-8\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-2\\[2pt]-4\\[2pt]-2\end{array}\,\right)\end{eqnarray}\)

---

ここで、\(\overrightarrow{\rm BM}\) と \(\overrightarrow{\rm CD}\) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm BM}\cdot\overrightarrow{\rm CD}&=&3\cdot(-2)+(-4)\cdot(-4)+5\cdot(-2)
\\[3pt]~~~&=&-6+16-10
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm BM}\neq\overrightarrow{0}~,~\overrightarrow{\rm CD}\neq\overrightarrow{0}\) より、\( {\rm BM}\perp{\rm CD} \)

 

\({\small (2)}~\) ここで、\(\triangle {\rm BCD}\) の面積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm BC}&=&\overrightarrow{\rm OC}-\overrightarrow{\rm OB}
\\[5pt]~~~&=&\left(\,\begin{array}{c}5\\[2pt]1\\[2pt]8\end{array}\,\right)-\left(\,\begin{array}{c}1\\[2pt]3\\[2pt]2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}5-1\\[2pt]1-3\\[2pt]8-2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}4\\[2pt]-2\\[2pt]6\end{array}\,\right)\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\overrightarrow{\rm BD}&=&\overrightarrow{\rm OD}-\overrightarrow{\rm OB}
\\[5pt]~~~&=&\left(\,\begin{array}{c}3\\[2pt]-3\\[2pt]6\end{array}\,\right)-\left(\,\begin{array}{c}1\\[2pt]3\\[2pt]2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}3-1\\[2pt]-3-3\\[2pt]6-2\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}2\\[2pt]-6\\[2pt]4\end{array}\,\right)\end{eqnarray}\)

---

\( \overrightarrow{\rm BC} \) と \( \overrightarrow{\rm BD} \) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm BC}\cdot\overrightarrow{\rm BD}&=&4\cdot 2+(-2)\cdot (-6)+6\cdot 4
\\[3pt]~~~&=&8+12+24
\\[3pt]~~~&=&44\end{eqnarray}\)

---

\( \overrightarrow{\rm BC} \) と \( \overrightarrow{\rm BD} \) の大きさは、

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm BC}\,|&=&\sqrt{4^2+(-2)^2+6^2}
\\[3pt]~~~&=&\sqrt{16+4+36}
\\[3pt]~~~&=&\sqrt{56}
\\[3pt]~~~&=&2\sqrt{14}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm BD}\,|&=&\sqrt{2^2+(-6)^2+4^2}
\\[3pt]~~~&=&\sqrt{4+36+16}
\\[3pt]~~~&=&\sqrt{56}
\\[3pt]~~~&=&2\sqrt{14}\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm BC}\cdot\overrightarrow{\rm BD}&=&|\,\overrightarrow{\rm BC}\,|\cdot|\,\overrightarrow{\rm BD}\,|\cdot\cos\theta
\\[5pt]~44&=&2\sqrt{14}\cdot 2\sqrt{14}\cdot\cos\theta
\\[5pt]~56\cos\theta&=&44
\\[5pt]~\cos\theta&=&\displaystyle \frac{\,11\,}{\,14\,}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\sin^2\theta&=&1-\cos^2\theta
\\[5pt]~~~&=&1-\displaystyle \frac{\,121\,}{\,196\,}
\\[5pt]~~~&=&\displaystyle \frac{\,75\,}{\,196\,}\end{eqnarray}\)

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\( 0^\circ {\small ~≦~}\theta{\small ~≦~}180^\circ \) より、\(\sin\theta=\displaystyle \frac{\,5\sqrt{3}\,}{\,14\,} \)

---

よって、\( \triangle {\rm BCD} \) の面積は、

---

\(\begin{eqnarray}~\triangle {\rm BCD}&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm BC}\,|\cdot|\,\overrightarrow{\rm BD}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot 2\sqrt{14}\cdot 2\sqrt{14}\cdot\displaystyle \frac{\,5\sqrt{3}\,}{\,14\,}
\\[5pt]~~~&=&28\cdot \displaystyle \frac{\,5\sqrt{3}\,}{\,14\,}
\\[5pt]~~~&=&10\sqrt{3}\end{eqnarray}\)

 

\({\small (3)}~\) \(\overrightarrow{\rm AB}\) は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}&=&\overrightarrow{\rm OB}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}1\\[2pt]3\\[2pt]2\end{array}\,\right)-\left(\,\begin{array}{c}8\\[2pt]2\\[2pt]-3\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}1-8\\[2pt]3-2\\[2pt]2+3\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}-7\\[2pt]1\\[2pt]5\end{array}\,\right)\end{eqnarray}\)

---

ここで、\(\overrightarrow{\rm AB}\) と \(\overrightarrow{\rm BC}\) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm BC}&=&(-7)\cdot4+1\cdot(-2)+5\cdot6
\\[3pt]~~~&=&-28-2+30
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AB}\neq\overrightarrow{0}~,~\overrightarrow{\rm BC}\neq\overrightarrow{0}\) より、\( {\rm AB}\perp{\rm BC} \)

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm BD}&=&(-7)\cdot2+1\cdot(-6)+5\cdot4
\\[3pt]~~~&=&-14-6+20
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AB}\neq\overrightarrow{0}~,~\overrightarrow{\rm BD}\neq\overrightarrow{0}\) より、\( {\rm AB}\perp{\rm BD} \)

 

\({\small (4)}~\) \({\rm AB}\perp{\rm BC}~,~{\rm AB}\perp{\rm BD}\) より、四面体 \( \rm ABCD \) の体積は \( \triangle \rm BCD \) が底面、\( \rm AB \) が高さとなるので、

---

\(\overrightarrow{\rm AB}\) の大きさは、

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AB}\,|&=&\sqrt{(-7)^2+1^2+5^2}
\\[3pt]~~~&=&\sqrt{49+1+25}
\\[3pt]~~~&=&\sqrt{75}
\\[3pt]~~~&=&5\sqrt{3}\end{eqnarray}\)

---

したがって、四面体 \( \rm ABCD \) の体積 \( \rm V \) は、

---

\(\begin{eqnarray}~~~V&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot\triangle \rm BCD\cdot|\,\overrightarrow{\rm AB}\,|
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot10\sqrt{3}\cdot5\sqrt{3}
\\[5pt]~~~&=&50\end{eqnarray}\)

東京書籍｜Advanced数学Ｃ[701] p.69 練習問題B 10

原点を \( \rm O \) として、

---

\({\small (1)}~\)

---

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

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AD}&=&\overrightarrow{\rm OD}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}5\\[2pt]-4\\[2pt]8\end{array}\,\right)-\left(\,\begin{array}{c}1\\[2pt]-1\\[2pt]3\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}5-1\\[2pt]-4+1\\[2pt]8-3\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}4\\[2pt]-3\\[2pt]5\end{array}\,\right)\end{eqnarray}\)

---

ここで、\(\overrightarrow{\rm AD}\) と \(\overrightarrow{\rm AB}\) の内積は、

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AD}\cdot\overrightarrow{\rm AB}&=&4\cdot3+(-3)\cdot4+5\cdot0
\\[3pt]~~~&=&12-12+0
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AD}\neq\overrightarrow{0}~,~\overrightarrow{\rm AB}\neq\overrightarrow{0}\) より、\( {\rm AD}\perp{\rm AB} \)

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AD}\cdot\overrightarrow{\rm AC}&=&4\cdot(-1)+(-3)\cdot2+5\cdot2
\\[3pt]~~~&=&-4-6+10
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AD}\neq\overrightarrow{0}~,~\overrightarrow{\rm AC}\neq\overrightarrow{0}\) より、\( {\rm AD}\perp{\rm AC} \)

 

\({\small (2)}~\) \( \overrightarrow{\rm AB} \) と \( \overrightarrow{\rm AC} \) の内積は、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&3\cdot(-1)+4\cdot 2+0\cdot 2
\\[3pt]~~~&=&-3+8+0
\\[3pt]~~~&=&5\end{eqnarray}\)

---

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

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\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AB}\,|&=&\sqrt{3^2+4^2+0^2}
\\[3pt]~~~&=&\sqrt{9+16+0}
\\[3pt]~~~&=&\sqrt{25}
\\[3pt]~~~&=&5\end{eqnarray}\)

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AC}\,|&=&\sqrt{(-1)^2+2^2+2^2}
\\[3pt]~~~&=&\sqrt{1+4+4}
\\[3pt]~~~&=&\sqrt{9}
\\[3pt]~~~&=&3\end{eqnarray}\)

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内積の定義の式より、

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\cos\theta
\\[5pt]~5&=&5\cdot 3\cdot\cos\theta
\\[5pt]~15\cos\theta&=&5
\\[5pt]~\cos\theta&=&\displaystyle \frac{\,1\,}{\,3\,}\end{eqnarray}\)

 

\({\small (3)}~\) ここで、\(\triangle {\rm ABC}\) の面積は、

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\(\begin{eqnarray}~~~\sin^2\theta&=&1-\cos^2\theta
\\[5pt]~~~&=&1-\displaystyle \frac{\,1\,}{\,9\,}
\\[5pt]~~~&=&\displaystyle \frac{\,8\,}{\,9\,}\end{eqnarray}\)

---

\( 0^\circ {\small ~≦~}\theta{\small ~≦~}180^\circ \) より、\(\sin\theta=\displaystyle \frac{\,2\sqrt{2}\,}{\,3\,} \)

---

よって、\( \triangle {\rm ABC} \) の面積は、

---

\(\begin{eqnarray}~\triangle {\rm ABC}&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot 5\cdot 3\cdot\displaystyle \frac{\,2\sqrt{2}\,}{\,3\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot 10\sqrt{2}
\\[5pt]~~~&=&5\sqrt{2}\end{eqnarray}\)

 

\({\small (4)}~\) \({\rm AD}\perp{\rm AB}~,~{\rm AD}\perp{\rm AC}\) より、四面体 \( \rm ABCD \) の体積は \( \triangle \rm ABC \) が底面、\( \rm AD \) が高さとなるので、

---

\(\overrightarrow{\rm AD}\) の大きさは、

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AD}\,|&=&\sqrt{4^2+(-3)^2+5^2}
\\[3pt]~~~&=&\sqrt{16+9+25}
\\[3pt]~~~&=&\sqrt{50}
\\[3pt]~~~&=&5\sqrt{2}\end{eqnarray}\)

---

したがって、四面体 \( \rm ABCD \) の体積 \( \rm V \) は、

---

\(\begin{eqnarray}~~~V&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot\triangle \rm ABC\cdot|\,\overrightarrow{\rm AD}\,|
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot5\sqrt{2}\cdot5\sqrt{2}
\\[5pt]~~~&=&\displaystyle \frac{\,50\,}{\,3\,}\end{eqnarray}\)

東京書籍｜Standard数学Ｃ[702] p.69 Level Up 9

原点を \( \rm O \) として、

---

\({\small (1)}~\)

---

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

---

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

---

ここで、\(\triangle {\rm ABC}\) の面積は、

---

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

---

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

---

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

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AB}\,|&=&\sqrt{2^2+0^2+1^2}
\\[3pt]~~~&=&\sqrt{4+0+1}
\\[3pt]~~~&=&\sqrt{5}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AC}\,|&=&\sqrt{1^2+3^2+(-4)^2}
\\[3pt]~~~&=&\sqrt{1+9+16}
\\[3pt]~~~&=&\sqrt{26}\end{eqnarray}\)

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AB}\cdot\overrightarrow{\rm AC}&=&|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\cos\theta
\\[5pt]~-2&=&\sqrt{5}\cdot\sqrt{26}\cdot\cos\theta
\\[5pt]~\sqrt{130}\cos\theta&=&-2
\\[5pt]~\cos\theta&=&-\displaystyle \frac{\,2\,}{\,\sqrt{130}\,}\end{eqnarray}\)

---

\(\begin{eqnarray}~~~\sin^2\theta&=&1-\cos^2\theta
\\[5pt]~~~&=&1-\displaystyle \frac{\,4\,}{\,130\,}
\\[5pt]~~~&=&\displaystyle \frac{\,126\,}{\,130\,}
\\[5pt]~~~&=&\displaystyle \frac{\,63\,}{\,65\,}\end{eqnarray}\)

---

\( 0^\circ {\small ~≦~}\theta{\small ~≦~}180^\circ \) より、\(\sin\theta=\displaystyle \frac{\,\sqrt{63}\,}{\,\sqrt{65}\,}=\displaystyle \frac{\,3\sqrt{7}\,}{\,\sqrt{65}\,} \)

---

よって、\( \triangle {\rm ABC} \) の面積は、

---

\(\begin{eqnarray}~\triangle {\rm ABC}&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot|\,\overrightarrow{\rm AB}\,|\cdot|\,\overrightarrow{\rm AC}\,|\cdot\sin\theta
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot \sqrt{5}\cdot\sqrt{26}\cdot\displaystyle \frac{\,3\sqrt{7}\,}{\,\sqrt{65}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot\displaystyle \frac{\,3\sqrt{7}\cdot\sqrt{130}\,}{\,\sqrt{65}\,}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,2\,}\cdot 3\sqrt{7}\cdot\sqrt{2}
\\[5pt]~~~&=&\displaystyle \frac{\,3\sqrt{14}\,}{\,2\,}\end{eqnarray}\)

 

\({\small (2)}~\) \(\overrightarrow{\rm AD}\) は、

---

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

---

ここで、\(\overrightarrow{\rm AD}\) と \(\overrightarrow{\rm AB}\) の内積は、

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AD}\cdot\overrightarrow{\rm AB}&=&(-1)\cdot2+3\cdot0+2\cdot1
\\[3pt]~~~&=&-2+0+2
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AD}\neq\overrightarrow{0}~,~\overrightarrow{\rm AB}\neq\overrightarrow{0}\) より、\( {\rm AD}\perp{\rm AB} \)

---

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

---

\(\begin{eqnarray}~~~\overrightarrow{\rm AD}\cdot\overrightarrow{\rm AC}&=&(-1)\cdot1+3\cdot3+2\cdot(-4)
\\[3pt]~~~&=&-1+9-8
\\[3pt]~~~&=&0\end{eqnarray}\)

---

\(\overrightarrow{\rm AD}\neq\overrightarrow{0}~,~\overrightarrow{\rm AC}\neq\overrightarrow{0}\) より、\( {\rm AD}\perp{\rm AC} \)

---

\({\rm AD}\perp{\rm AB}~,~{\rm AD}\perp{\rm AC}\) より、直線 \( \rm AD \) は平面 \( \rm ABC \) に垂直である

 

\({\small (3)}~\) 直線 \( \rm AD \) は平面 \( \rm ABC \) に垂直であるから、四面体 \( \rm ABCD \) の体積は \( \triangle \rm ABC \) が底面、\( \rm AD \) が高さとなるので、

---

\(\overrightarrow{\rm AD}\) の大きさは、

---

\(\begin{eqnarray}~~~|\,\overrightarrow{\rm AD}\,|&=&\sqrt{(-1)^2+3^2+2^2}
\\[3pt]~~~&=&\sqrt{1+9+4}
\\[3pt]~~~&=&\sqrt{14}\end{eqnarray}\)

---

したがって、四面体 \( \rm ABCD \) の体積 \( V \) は、

---

\(\begin{eqnarray}~~~V&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot\triangle \rm ABC\cdot|\,\overrightarrow{\rm AD}\,|
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot\displaystyle \frac{\,3\sqrt{14}\,}{\,2\,}\cdot\sqrt{14}
\\[5pt]~~~&=&\displaystyle \frac{\,1\,}{\,3\,}\cdot\displaystyle \frac{\,3\cdot14\,}{\,2\,}
\\[5pt]~~~&=&7\end{eqnarray}\)