向量法解一道动态立体几何难题

题目:如图,直线$l\perp$平面$\alpha$,垂足为$O$.正四面体$ABCD$的棱长为$1$.点$C$在平面$\alpha$内,点$B$是直线$l$上的动点,则点$O$到直线$AD$的距离的最大值和最小值分别为?

此图像的alt属性为空;文件名为20190123.png

解:设平面$\alpha$的一个单位法向量为$\bm{n}$.则
$$
\ov{OA}=\ov{BA}-\ov{BO}=\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n},
$$
设点$O$在直线$AD$上的射影是点$P$,则
\begin{align*}
\ov{OP}&=\ov{OA}+\ov{AP}\\&=\ov{OA}-\left(\ov{OA}\cdot\ov{AD}\right)\ov{AD}
\\&=\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n}-\left[\left(\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n}\right)\cdot\ov{AD}\right]\ov{AD}
\\&=\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n}-\left[\ov{BA}\cdot\ov{AD}-\left(\ov{BC}\cdot\bm{n}\right)\left(\ov{AD}\cdot\bm{n}\right)\right]\ov{AD}
\\&=\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n}-\left[-\frac{1}{2}-\left(\ov{BC}\cdot\bm{n}\right)\left(\ov{AD}\cdot\bm{n}\right)\right]\ov{AD}
\\&=\ov{BA}-\left(\ov{BC}\cdot\bm{n}\right)\bm{n}+\left(\ov{BC}\cdot\bm{n}\right)\left(\ov{AD}\cdot\bm{n}\right)\ov{AD}+\frac{1}{2}\ov{AD},
\end{align*}
因此
\begin{align*}
\ov{OP}^2&=\ov{BA}^2+(\ov{BC}\cdot\bm{n})^2+(\ov{BC}\cdot\bm{n})^{2}(\ov{AD}\cdot\bm{n})^2+\frac{1}{4}\ov{AD}^2-2(\ov{BC}\cdot\bm{n})(\ov{BA}\cdot\bm{n})\\&+2\ov{BA}\cdot\ov{AD}(\ov{BC}\cdot\bm{n})(\ov{AD}\cdot\bm{n})+\ov{BA}\cdot\ov{AD}-2(\ov{BC}\cdot\bm{n})^2(\ov{AD}\cdot\bm{n})^2-(\ov{BC}\cdot\bm{n})(\ov{AD}\cdot\bm{n})+(\ov{BC}\cdot\bm{n})(\ov{AD}\cdot\bm{n})\ov{AD}^2
\\&=\frac{3}{4}+(\ov{BC}\cdot\bm{n})^{2}-(\ov{BC}\cdot\bm{n})^2(\ov{AD}\cdot\bm{n})^2-2(\ov{BC}\cdot\bm{n})(\ov{BA}\cdot\bm{n})-(\ov{BC}\cdot\bm{n})(\ov{AD}\cdot\bm{n})
\\&=\frac{3}{4}+(\ov{BC}\cdot\bm{n})^2-(\ov{BC}\cdot\bm{n})^2(\ov{AD}\cdot\bm{n})^2-2(\ov{BC}\cdot\bm{n})(\ov{BA}+\frac{1}{2}\ov{AD})\cdot\bm{n}
\\&=\frac{3}{4}+2(\ov{BC}\cdot\bm{n})(\frac{1}{2}\ov{BC}-\ov{BA}-\frac{1}{2}\ov{AD})\cdot\bm{n}-(\ov{BC}\cdot\bm{n})^{2}(\ov{AD}\cdot\bm{n})^{2}
\end{align*}
记线段$AD$的中点是$M$,线段$BC$的中点是$N$.则
$$
\ov{OP}^2=\frac{3}{4}+2(\ov{BC}\cdot\bm{n})(\ov{MN}\cdot\bm{n})-(\ov{BC}\cdot\bm{n})^2(\ov{AD}\cdot\bm{n})^2=\frac{3}{4}+\sqrt{2}(\ov{BC}\cdot\bm{n})(\sqrt{2}\ov{MN}\cdot\bm{n})-(\ov{BC}\cdot\bm{n})^2(\ov{AD}\cdot\bm{n})^2
$$
$\{\ov{BC},\ov{AD},\sqrt{2}\ov{MN}\}$构成$\mathbf{R}^3$的一组单位正交基底,设
$$
\bm{n}\cdot\ov{BC}=x,\bm{n}\cdot\ov{AD}=y,\bm{n}\cdot (\sqrt{2}\ov{MN})=z,
$$

$$
x^2+y^2+z^2=\bm{n}^2=1,
$$
于是
$$
\ov{OP}^2=\frac{3}{4}+\sqrt{2}xz-x^2y^2\leq
\frac{3}{4}+\frac{\sqrt{2}}{2}(2xz)\leq
\frac{3}{4}+\frac{\sqrt{2}}{2}(x^2+z^2)\leq \frac{3}{4}+\frac{\sqrt{2}}{2},
$$
当且仅当$y=0$且$x=z=\pm\frac{\sqrt{2}}{2}$时,$\ov{OP}^2$取到最大值$\frac{3}{4}+\frac{\sqrt{2}}{2}=\left(\frac{\sqrt{2}+1}{2}\right)^{2}$.且
\begin{align*}
\ov{OP}^2&=\frac{3}{4}+\sqrt{2}xz-x^2y^2
\\&=\frac{3}{4}+\sqrt{2}xz-x^2(1-x^2-z^2)
\\&=(x^2-\frac{1}{2})^2+(xz+\frac{\sqrt{2}}{2})^2,
\end{align*}
因为
$$
xz\geq -\frac{x^2+z^2}{2}\geq -\frac{1}{2}>-\frac{\sqrt{2}}{2},
$$
所以
$$
\ov{OP}^2\geq (x^2-\frac{1}{2})^2+(-\frac{1}{2}+\frac{\sqrt{2}}{2})^2\geq (\frac{\sqrt{2}-1}{2})^2.
$$
等号当且仅当$y=0$,$x=-z=\pm \frac{\sqrt{2}}{2}$时成立.综上,
$$
\frac{\sqrt{2}-1}{2}\leq |\ov{OP}|\leq \frac{\sqrt{2}+1}{2},
$$
即点$O$到直线$AD$距离的最大值为$\frac{\sqrt{2}+1}{2}$,最小值为$\frac{\sqrt{2}-1}{2}$.

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