令和05年度(2023)久留米大学医学部推薦入試数学の過去問と解答（３／４）

 \[
\vec{p}\cdot \vec{q}=|\vec{p}||\vec{q}|\cos \frac{2\pi}{3}=1 \cdot 1 \cdot \left( -\frac12 \right)=-\frac12. \quad \cdots \text{（答）} 
\]
\( \vec{p} \perp \vec{r},~ \vec{q} \perp \vec{r}\)なので
\[
\vec{p} \cdot \vec{r}=\vec{q} \cdot \vec{r}=0. \quad \cdots \text{（答）} 
\]

\begin{align*}
\overrightarrow{\textrm{AK}}= & \overrightarrow{\textrm{AF}}+\overrightarrow{\textrm{FE}}+\overrightarrow{\textrm{EK}} \\
= & \overrightarrow{\textrm{AF}}+\left( \overrightarrow{\textrm{AB}}+\overrightarrow{\textrm{AF}} \right) +\overrightarrow{\textrm{AG}} \\
= & \vec{p}+2\vec{q}+\vec{r} \quad \therefore 4. \quad \cdots \text{（答）} 
\end{align*}

\begin{align*}
\overrightarrow{\textrm{AI}}= & \overrightarrow{\textrm{AB}}+\overrightarrow{\textrm{BC}}+\overrightarrow{\textrm{CI}} \\
= & \overrightarrow{\textrm{AB}}+\left( \overrightarrow{\textrm{AB}}+\overrightarrow{\textrm{AF}} \right)+\overrightarrow{\textrm{AG}} \\
= & 2\vec{p}+\vec{q}+\vec{r} \quad \therefore 3. \quad \cdots \text{（答）} 
\end{align*}

\begin{align*}
\overrightarrow{\textrm{AP}}= & s\overrightarrow{\textrm{AK}}+t\overrightarrow{\textrm{AI}} \\
= & s \left( \vec{p}+ 2\vec{q}+\vec{r} \right) + t \left( 2\vec{p}+ \vec{q}+\vec{r} \right)\\
= & (s+2t)\vec{p}+(2s+t)\vec{q}+(s+t)\vec{r} \quad \cdots [1]
\end{align*}
よって \( \qquad 4,~2,~0 \quad \cdots \text{（答）}  \) 

また一方，
\begin{align*}
\overrightarrow{\textrm{AL}}= & \overrightarrow{\textrm{AF}}+\overrightarrow{\textrm{FL}} \\
= & \overrightarrow{\textrm{AF}}+\overrightarrow{\textrm{AG}} \\
= & \vec{q}+\vec{r} \quad \cdots \text{（答）} 
\end{align*}
\begin{align*}
\overrightarrow{\textrm{LC}}= & \overrightarrow{\textrm{LI}}+\overrightarrow{\textrm{IC}} \\
= & 2\overrightarrow{\textrm{AB}}+\overrightarrow{\textrm{GA}} \\
= & 2\vec{p}-\vec{r} \quad \cdots \text{（答）} 
\end{align*}
点\(\textrm{P}\)は直線\(\textrm{LC}\)上にあるので，実数\(u\)を用いて
\begin{align*}
\overrightarrow{\textrm{AP}}= & \overrightarrow{\textrm{AL}}+u\overrightarrow{\textrm{LC}} \\
= & \left( \vec{q}+\vec{r} \right) +u \left( 2\vec{p}-\vec{r} \right) \\
= & 2u\vec{p}+ \vec{q}+(1-u)\vec{r} \quad \cdots [2]
\end{align*}

\(\vec{p} ,~\vec{q},~\vec{r}~\)は互いに一次独立であるから，\([1],~[2]\)より
\begin{align*}
\begin{cases}
s+2t= 2u \\
2s+t= 1 \\
s+t= 1-u
\end{cases}
\end{align*}
これを解いて\( s=\frac25,~t=\frac15,~u=\frac25. \)したがって
\[
\overrightarrow{\textrm{AP}}=\frac25 \overrightarrow{\textrm{AK}}+\frac15 \overrightarrow{\textrm{AI}}  \quad \cdots \text{（答）} 
\]

また，3点\(~\textrm{A,P,Q}~\)は一直線上に並んでいるので，
\begin{align*}
\overrightarrow{\textrm{AQ}}= & k\overrightarrow{\textrm{AP}} \\
= & \frac{2k}{5}\overrightarrow{\textrm{AK}}+\frac{k}{5}\overrightarrow{\textrm{AI}}
\end{align*}
を満たす実数\(~k~\)が存在する。
点\(\textrm{Q}~\)は，線分\(\textrm{KI}~\)上にあるから
\begin{align*}
\frac{2k}{5}+\frac{k}{5}= & 1 \\
k= & \frac{5}{3}
\end{align*}
したがって
\begin{align*}
\overrightarrow{\textrm{AQ}}= & \frac25 \cdot \frac53 \overrightarrow{\textrm{AK}}+\frac15 \cdot \frac53 \overrightarrow{\textrm{AI}} \\
= & \frac23 \overrightarrow{\textrm{AK}} +\frac13 \overrightarrow{\textrm{AI}}
\end{align*}

よって点\(~\textrm{Q}~\)は，線分\(~\textrm{KI}~\)を\(1:2\)に内分する点である。\(\therefore 2. \quad \cdots \text{（答）} \)