Cho hình lập phương ABCD.A'B'C'D'. M, N lần lượt thuộc các đoạn AD, A'C sao cho

a) Chứng minh $\left({\mathrm{AB}}^{\text{'}}{\mathrm{D}}^{\text{'}}\right)\text{//}\left({\mathrm{BC}}^{\text{'}}\mathrm{D}\right)$.

Từ giả thiết ta có $\left.\begin{array}{l}\mathrm{BD}\text{//}{\mathrm{B}}^{\text{'}}{\mathrm{D}}^{\text{'}}\\ {\mathrm{AB}}^{\text{'}}\text{//}{\mathrm{DC}}^{\text{'}}\end{array}\right\}\Rightarrow \left({\mathrm{AB}}^{\text{'}}{\mathrm{D}}^{\text{'}}\right)\text{//}\left({\mathrm{BC}}^{\text{'}}\mathrm{D}\right)$

b) Chứng minh ${\mathrm{AC}}^{\text{'}}\perp {\mathrm{A}}^{\text{'}}\mathrm{B}$

Ta có: $\overrightarrow{{\mathrm{AC}}^{\text{'}}}.\overrightarrow{{\mathrm{A}}^{\text{'}}\mathrm{B}}=\left(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{AD}}+\overrightarrow{{\mathrm{AA}}^{\text{'}}}\right).$$\left(\overrightarrow{\mathrm{AB}}-\overrightarrow{{\mathrm{AA}}^{\text{'}}}\right)$

$$\begin{gathered} ={\overrightarrow{\mathrm{AB}}}^2+\overrightarrow{\mathrm{AD}}.\overrightarrow{\mathrm{AB}}+\overrightarrow{{\mathrm{AA}}^{\text{'}}}.\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{AB}}.\overrightarrow{{\mathrm{AA}}^{\text{'}}} \\ -\overrightarrow{\mathrm{AD}}.\overrightarrow{{\mathrm{AA}}^{\text{'}}}-{\overrightarrow{{\mathrm{AA}}^{\text{'}}}}^2 \\ \Rightarrow \overrightarrow{{\mathrm{AC}}^{\text{'}}}\perp \overrightarrow{{\mathrm{A}}^{\text{'}}\mathrm{B}}\Rightarrow {\mathrm{AC}}^{\text{'}}\perp {\mathrm{A}}^{\text{'}}\mathrm{B} \end{gathered}$$.

c) Chứng minh $\mathrm{MN}\text{//}\left({\mathrm{AB}}^{\text{'}}{\mathrm{D}}^{\text{'}}\right)$

Dễ thấy M, N không thuộc (AB'D').

$\mathrm{MN}\text{//}\left({\mathrm{AB}}^{\text{'}}{\mathrm{D}}^{\text{'}}\right)\iff \overrightarrow{\mathrm{MN}},\overrightarrow{{\mathrm{AB}}^{\text{'}}},\overrightarrow{{\mathrm{AD}}^{\text{'}}}$ đồng phẳng $\iff \exists \mathrm{m},\mathrm{n}:\overrightarrow{\mathrm{MN}}=\mathrm{m}\overrightarrow{{\mathrm{AB}}^{\text{'}}}+\mathrm{n}\overrightarrow{{\mathrm{AD}}^{\text{'}}}$

Đặt $\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{b}},\overrightarrow{{\mathrm{AA}}^{\text{'}}}=\overrightarrow{\mathrm{c}}$

 $$\begin{gathered} \overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{AN}}-\overrightarrow{\mathrm{AM}}=\overrightarrow{{\mathrm{AA}}^{\text{'}}}+\overrightarrow{{\mathrm{A}}^{\text{'}}\mathrm{N}}-\frac{1}{5}\overrightarrow{\mathrm{AD}} \\ =\overrightarrow{\mathrm{c}}+\frac{2}{5}\overrightarrow{{\mathrm{A}}^{\text{'}}\mathrm{C}}-\frac{1}{5}\overrightarrow{\mathrm{b}} \\ =\overrightarrow{\mathrm{c}}+\frac{2}{5}\left(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}\right)-\frac{1}{5}\overrightarrow{\mathrm{b}} \\ =\frac{2}{5}\overrightarrow{\mathrm{a}}+\frac{1}{5}\overrightarrow{\mathrm{b}}+\frac{3}{5}\overrightarrow{\mathrm{c}} \end{gathered}$$

$\overrightarrow{{\mathrm{AB}}^{\text{'}}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}$, $\overrightarrow{{\mathrm{AD}}^{\text{'}}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$

$\overrightarrow{\mathrm{MN}}=\mathrm{m}\overrightarrow{{\mathrm{AB}}^{\text{'}}}+\mathrm{n}\overrightarrow{{\mathrm{AD}}^{\text{'}}}$$\iff \frac{2}{5}\overrightarrow{\mathrm{a}}+\frac{1}{5}\overrightarrow{\mathrm{b}}+\frac{3}{5}\overrightarrow{\mathrm{c}}$$=\mathrm{m}\overrightarrow{\mathrm{a}}+\mathrm{n}\overrightarrow{\mathrm{b}}+\left(\mathrm{m}+\mathrm{n}\right)\overrightarrow{\mathrm{c}}$$\iff \left\{\begin{array}{l}\mathrm{m}=\frac{2}{5}\\ \mathrm{n}=\frac{1}{5}\end{array}\right.$$\Rightarrow \overrightarrow{\mathrm{MN}}=\frac{2}{5}\overrightarrow{{\mathrm{AB}}^{\text{'}}}+\frac{1}{5}\overrightarrow{{\mathrm{AD}}^{\text{'}}}$

Vậy MN // (AB'D').