題目3.:如圖有一平行六面體,若\(\displaystyle\overline{AE}=3, \overline{EH}=5, \overline{EF}=6, \cos \angle AEH=\frac{4}{5}, \cos \angle AEF=\frac{1}{2}, \cos \angle HEF=\frac{3}{5}\),求\(\overline{AG}=\)?
解答:
\(\vec{AG} = -\vec{EA}+\vec{EH}+\vec{EF}\)
\(\displaystyle\overline{AG}^2 = \vec{AG}\cdot\vec{AG}\)
\(= \overline{EA}^2+\overline{EH}^2+\overline{EF}^2-2\vec{EA}\cdot{EH}-2\vec{EA}\cdot{EF}+2\vec{EH}\cdot{EF}\)
\(\displaystyle= 9+25+36-2\cdot3\cdot5\cdot\frac{4}{5}-2\cdot3\cdot6\cdot\frac{1}{2}+2\cdot5\cdot6\cdot\frac{3}{5}\)
\(=64\)
\(\displaystyle\Rightarrow \overline{AG}=8\)