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空間中，四面體\(A-BCD\)，\(\overline{AB}=\overline{CD}=6\)，\(\overline{AC}=\overline{AD}=\overline{BC}=5\)，\(\overline{BD}=7\)，求四面體\(A-BCD\)的體積為 。
[解答]
我發現用內積也滿好算的!
令 A(0,0,0), B(6,0,0), C(3,4,0), D(x,y,z)
則有 \(\displaystyle x^2+y^2+z^2=25 ...(1)\)
\(\displaystyle 18=\overrightarrow{DA}\cdot\overrightarrow{DC}=(-x,-y,-z)\cdot(3-x,4-y,-z)\)
\(\displaystyle 19=\overrightarrow{DA}\cdot\overrightarrow{DB}=(-x,-y,-z)\cdot(6-x,-y,-z)\)
\(\displaystyle x(x-3)+y(y-4)+z^2=18...(2)\)
\(\displaystyle x(x-6)+y^2+z^2=19...(3)\)
(1) 代入(3) 可得 \(\displaystyle x(x-6)+25-x^2=19\), 解出 \(\displaystyle x=1\) 代回
可得 \(\displaystyle y^2+z^2=24\) 與 \(\displaystyle y(y-4)+z^2=20\), 再解出 \(\displaystyle y=1\)
於是 \(\displaystyle z=\pm \sqrt{23}\)
可取 \(\displaystyle \overrightarrow{AD}=(1,1,\sqrt{23})\)
所以四面體 A-BCD 體積為 \(\displaystyle \frac{1}{6}|\left |
\begin {array} {clr}
1 & 1 & \sqrt{23} \\
6 & 0 & 0 \\
3 & 4 & 0 \\
\end {array} \right | |=4\sqrt{23}
\)
