解答・解説[3](2)

\begin{eqnarray*} \overrightarrow{\mathrm{BH}}・\overrightarrow{\mathrm{AP}} &=& 0 \\ (\overrightarrow{\mathrm{AH}} - \overrightarrow{\mathrm{AB}})・\overrightarrow{\mathrm{AP}} &=& 0 \\ (\overrightarrow{k\mathrm{AP}} - \overrightarrow{\mathrm{AB}})・\overrightarrow{\mathrm{AP}} &=& 0 \\ k|\overrightarrow{\mathrm{AP}}|^2 - \overrightarrow{\mathrm{AB}} ・ \overrightarrow{\mathrm{AP}} &=& 0 …①　\\ \end{eqnarray*}

\begin{eqnarray*} \overrightarrow{\mathrm{AP}} &=& \frac{3\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{4} より \\[3mm] |\overrightarrow{\mathrm{AP}}|^2 &=& \frac{9|\overrightarrow{\mathrm{AB}}|^2 + 6\overrightarrow{\mathrm{AB}}・\overrightarrow{\mathrm{AC}} + |\overrightarrow{\mathrm{AC}}|^2}{16} \\[3mm] &=& \frac{1}{16} \{9・3^2+6・(-3)+7^2\} \\[3mm] &=& \frac{1}{16} (81-18+49) \\[3mm] &=& \frac{1}{16} ・ 112 \\[3mm] &=& 7 \\ |\overrightarrow{\mathrm{AP}}|&=& \sqrt{7} \end{eqnarray*}

\begin{eqnarray*} \overrightarrow{\mathrm{AB}}・\overrightarrow{\mathrm{AP}} &=& \overrightarrow{\mathrm{AB}}\left(\frac{3\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AC}}}{4}\right) \\[3mm] &=& \frac{3}{4}|\overrightarrow{\mathrm{AB}}|^2 + \frac{\overrightarrow{\mathrm{AB}}・\overrightarrow{\mathrm{AC}}}{4} \\[3mm] &=& \frac{3}{4}・3^2-\frac{3}{4} \\[3mm] &=& \frac{27}{4}-\frac{3}{4} \\[3mm] &=& \frac{24}{4} \\[3mm] &=& 6 \end{eqnarray*}

\begin{eqnarray*} k|\overrightarrow{\mathrm{AP}}|^2 - \overrightarrow{\mathrm{AB}} ・ \overrightarrow{\mathrm{AP}} &=& 0 \\[3mm] k・\sqrt{7}^2 - 6 &=& 0 \\[3mm] 7k &=& 6 \\[3mm] k &=& \frac{6}{7} \end{eqnarray*}