수선의 교점과 내적

삼각형 $\mathrm{ABC}$에서 코사인법칙에 의해 $\displaystyle \cos A = \frac{9 ^{2} + 7 ^{2} - 8 ^{2}}{2 \times 9 \times 7} = \frac{66}{126} = \frac{11}{21}$ 따라서 $\displaystyle \mathrm{\overline{AP}} = 7 \times \frac{11}{21} = \frac{11}{3}$이고 $\displaystyle \mathrm{\overline{AQ}} = 9 \times \frac{11}{21} = \frac{33}{7}$이다. ${\vec{\mathrm{AR}}} \bullet \left( {\vec{\mathrm{AB}}} + {\vec{\mathrm{AC}}} \right)$$= {\vec{\mathrm{AR}}} \bullet {\vec{\mathrm{AB}}} + {\vec{\mathrm{AR}}} \bullet {\vec{\mathrm{AC}}}$ $= \mathrm{\left| {\vec{AP}} \right|} \times \left| {\vec{AB}} \right| + \left| {\vec{AQ}} \right| \times \left| {\vec{AC}} \right|$ $\displaystyle = \frac{11}{3} \times 9 + \frac{33}{7} \times 7$ $= 33 + 33 = 66$ [다른 풀이] 삼각형 $\mathrm{ABC}$에서 $\left| \mathrm{\vec{BC}} \right| ^{2}$$= \left| \mathrm{\vec{AC}} - \vec{AB} \right| ^{2}$ $= \left| \mathrm{\vec{AC}} \right| ^{2} - 2 \vec{AB} \bullet \vec{AC} + \left| \vec{AB} \right| ^{2}$ $\left| \mathrm{\vec{AB}} \right| = 9$, $\left| \mathrm{\vec{AC}} \right| = 7$, $\left| \mathrm{\vec{BC}} \right| = 8$이므로 $8 ^{2} = 7 ^{2} - 2 \mathrm{\vec{AB}} \bullet \vec{AC} + 9 ^{2}$ $\therefore$ $\mathrm{\vec{AB}} \bullet \vec{AC} = 33$ $\mathrm{\vec{AR}} \bullet \left( \vec{AB} + \vec{AC} \right) = \vec{AR} \bullet \vec{AB} + \vec{AR} \bullet \vec{AC}$에서 $\mathrm{\vec{AR}} \bullet \vec{AB}$$\mathrm{=} \left( \vec{AC} + \vec{CR} \right) \bullet \vec{AB}$ $= \mathrm{\vec{AB}} \bullet \vec{AC} + \vec{AB} \bullet \vec{CR}$ $= \mathrm{\vec{AB}} \bullet \vec{AC}$ ($\because$ $\mathrm{\vec{AB}} \perp \vec{CR}$) $\mathrm{\vec{AR}} \bullet \vec{AC}$$= \left( \mathrm{\vec{AB}} + \vec{BR} \right) \bullet \vec{AC}$ $= \mathrm{\vec{AB}} \bullet \vec{AC} + \vec{AC} \bullet \vec{BR}$ $= \mathrm{\vec{AB}} \bullet \vec{AC}$ ($\because$ $\mathrm{\vec{AC}} \perp \vec{BR}$) $\therefore$ $\mathrm{\vec{AR}} \bullet \left( \vec{AB} + \vec{AC} \right) = 2 \vec{AB} \bullet \vec{AC} = 66$