1650

若互不相等的单位空间向量 ${\displaystyle \overrightarrow{O{P}_i}}$ (${\displaystyle i=1,2,3,4}$) 满足 ${\displaystyle |\overrightarrow{P_1{P}_2}|=|\overrightarrow{P_2{P}_3}|=|\overrightarrow{P_3{P}_4}|=|\overrightarrow{P_4{P}_1}|=1}$, 则 ${\displaystyle |\overrightarrow{P_1{P}_3}{|}^2+|\overrightarrow{P_2{P}_4}{|}^2}$ 的取值范围为_____.

如图, 取 ${\displaystyle P_1{P}_3}$ 中点 ${\displaystyle M}$. 整个图形关于平面 ${\displaystyle P_{1}O{P}_3}$ 和平面 ${\displaystyle O{P}_2{P}_4}$ 对称. 设 ${\displaystyle \mathrm{\angle }{P}_{1}OM=\theta }$, ${\displaystyle \mathrm{\angle }{P}_{2}MO=\alpha }$. 则在 ${\displaystyle \mathrm{\Delta }O{P}_{2}M}$ 中由余弦定理 $$\cos \alpha =\frac{2{\cos }^2\theta -1}{2{\cos }^2\theta }\ge 0\Rightarrow {\cos }^2\theta \ge \frac{1}{2},$$ 另一方面 ${\displaystyle {\cos }^2\theta <1}$ (这里不能取等, 否则 ${\displaystyle P_1,P_3}$ 会重合). 这样 $$\begin{array}{rl}& |\overrightarrow{P_1{P}_3}{|}^2+|\overrightarrow{P_2{P}_4}{|}^2=(2\sin \theta {)}^2+(2\cos \theta \sin \alpha {)}^2\\ =& 4{\sin }^2\theta +4{\cos }^2\theta (1-{\cos }^2\alpha )=8-(\frac{1}{{\cos }^2\theta }+4{\cos }^2\theta )\in (3,4].\end{array}$$