Cho tứ diện ABCD. Gọi M, N lần lượt là trung điểm của BC và AD, biết AB = CD = a

* Hướng dẫn giải

Cách 1.

Gọi I là trung điểm của AC. Ta có $\left\{\begin{array}{l}\mathrm{IM}\parallel \mathrm{AB}\\ \mathrm{IN}\parallel \mathrm{CD}\end{array}\right.\Rightarrow$$\widehat{\left(\mathrm{AB},\mathrm{CD}\right)}\text{=}\widehat{\left(\mathrm{IM},\mathrm{IN}\right)}$

Đặt $\widehat{\mathrm{MIN}}=\mathrm{\alpha }$, xét tam giác IMN có $\mathrm{IM}=\frac{\mathrm{AB}}{2}=\frac{\mathrm{a}}{2},$$\mathrm{IN}=\frac{\mathrm{CD}}{2}=\frac{\mathrm{a}}{2},$$\mathrm{MN}=\frac{\mathrm{a}\sqrt{3}}{2}$.

Theo định lí côsin, ta có

$\mathrm{cos\alpha }=\frac{{\mathrm{IM}}^2+{\mathrm{IN}}^2-{\mathrm{MN}}^2}{2\mathrm{IM}.\mathrm{IN}}$$=\frac{{\left(\frac{\mathrm{a}}{2}\right)}^2+{\left(\frac{\mathrm{a}}{2}\right)}^2-{\left(\frac{\mathrm{a}\sqrt{3}}{2}\right)}^2}{2.\frac{\mathrm{a}}{2}.\frac{\mathrm{a}}{2}}$$=-\frac{1}{2}<0$

$\Rightarrow \widehat{\mathrm{MIN}}={120}^0$ suy ra $\widehat{\left(\mathrm{AB},\mathrm{CD}\right)}{\text{=60}}^0$.

Cách 2.

$\cos \widehat{\left(\mathrm{AB},\mathrm{CD}\right)}=\cos \widehat{\left(\mathrm{IM},\mathrm{IN}\right)}$$=\frac{\left|\overrightarrow{\mathrm{IM}}.\overrightarrow{\mathrm{IN}}\right|}{\left|\overrightarrow{\mathrm{IM}}\right|\left|\overrightarrow{\mathrm{IN}}\right|}$

$\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{IN}}-\overrightarrow{\mathrm{IM}}\Rightarrow$${\overrightarrow{\mathrm{MN}}}^2={\left(\overrightarrow{\mathrm{IN}}-\overrightarrow{\mathrm{IM}}\right)}^2$$={\mathrm{IM}}^2+{\mathrm{IN}}^2-2\overrightarrow{\mathrm{IN}}.\overrightarrow{\mathrm{IM}}$

$\overrightarrow{\mathrm{IN}}.\overrightarrow{\mathrm{IM}}=\frac{{\mathrm{IM}}^2+{\mathrm{IN}}^2-{\mathrm{MN}}^2}{2}=-\frac{{\mathrm{a}}^2}{8}$

$\cos \widehat{\left(\mathrm{AB},\mathrm{CD}\right)}=\left|\cos \widehat{\left(\mathrm{IM},\mathrm{IN}\right)}\right|$$=\frac{\left|\overrightarrow{\mathrm{IM}}.\overrightarrow{\mathrm{IN}}\right|}{\left|\overrightarrow{\mathrm{IM}}\right|\left|\overrightarrow{\mathrm{IN}}\right|}=\frac{1}{2}$

Vậy $\widehat{\left(\mathrm{AB},\mathrm{CD}\right)}{\text{=60}}^0$.