3点がつくる平面上に点がある条件

平面 \( {\rm ABC} \) 上に点 \( {\rm P} \) があるので

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\( \overrightarrow{\rm AP}=s\overrightarrow{\rm AB}+t\overrightarrow{\rm AC} \) となる実数 \( s~,~t \) がある

 

原点を \( {\rm O} \) とすると、

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\(\begin{eqnarray}~~~\overrightarrow{\rm AP}&=&\overrightarrow{\rm OP}-\overrightarrow{\rm OA}
\\[5pt]~~~&=&\left(\,\begin{array}{c}x\\[2pt]-5\\[2pt]0\end{array}\,\right)-\left(\,\begin{array}{c}1\\[2pt]2\\[2pt]3\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}x-1\\[2pt]-5-2\\[2pt]0-3\end{array}\,\right)=\left(\,\begin{array}{c}x-1\\[2pt]-7\\[2pt]-3\end{array}\,\right)
\end{eqnarray}\)

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

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

 

\( \overrightarrow{\rm AP}=s\overrightarrow{\rm AB}+t\overrightarrow{\rm AC} \) より、

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\(\begin{eqnarray}~~~\left(\,\begin{array}{c}x-1\\[2pt]-7\\[2pt]-3\end{array}\,\right)&=&s\left(\,\begin{array}{c}3\\[2pt]-3\\[2pt]-1\end{array}\,\right)+t\left(\,\begin{array}{c}1\\[2pt]1\\[2pt]1\end{array}\,\right)
\\[5pt]~~~&=&\left(\,\begin{array}{c}3s+t\\[2pt]-3s+t\\[2pt]-s+t\end{array}\,\right)
\end{eqnarray}\)

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それぞれの成分が等しいので、

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\(\begin{eqnarray}~~~ \left\{~\begin{array}{l}
x-1=3s+t\\ -7=-3s+t \\ -3=-s+t
\end{array}\right.\end{eqnarray}\)

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\(\begin{eqnarray}~~~~\Leftrightarrow ~ \left\{~\begin{array}{l}
x=3s+t+1~ ~ ~ \cdots {\small [\,1\,]}\\ 3s-t=7~ ~ ~\hspace{18pt} \cdots {\small [\,2\,]} \\ s-t=3~ ~ ~\hspace{18pt} \cdots {\small [\,3\,]}
\end{array}\right.
\end{eqnarray}\)

 

\({\small [\,2\,]}-{\small [\,3\,]}\) より、

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\(\begin{eqnarray}~~~~~
3s-t&=&7 \\~~
-\big{)}~~~~s-t&=&3\\
\hline 2s&=&4
\\[2pt] s&=&2\end{eqnarray}\)

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\(\small [\,3\,]\) に代入すると、

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\(\begin{eqnarray}~~~2-t&=&3
\\[3pt]~~~-t&=&1
\\[3pt]~~~t&=&-1\end{eqnarray}\)

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\( s=2~,~t=-1 \) を \(\small [\,1\,]\) に代入すると、

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\(\begin{eqnarray}~~~x&=&3\cdot 2+(-1)+1
\\[3pt]~~~x&=&6\end{eqnarray}\)

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したがって、\( x=6 \) となる