Nếu hai trong ba vectơ đó cùng hướng thì ba vectơ đồng phẳng; nếu hai trong ba vectơ đó

* Hướng dẫn giải

Đặt $\overrightarrow{\mathrm{BA}}=\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{BB}\text{'}}=\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{c}}$ thì $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ là ba vec tơ không đồng phẳng và $\overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}};$$\overrightarrow{\mathrm{BC}\text{'}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}},\overrightarrow{\mathrm{BA}\text{'}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}$

Ta có $\overrightarrow{\mathrm{MA}}=-\frac{1}{4}\overrightarrow{\mathrm{MD}}$$\Rightarrow \overrightarrow{\mathrm{BA}}-\overrightarrow{\mathrm{BM}}=-\frac{1}{4}\left(\overrightarrow{\mathrm{BD}}-\overrightarrow{\mathrm{BM}}\right)$$\Rightarrow \frac{5}{4}\overrightarrow{\mathrm{BM}}=\overrightarrow{\mathrm{BA}}+\frac{1}{4}\overrightarrow{\mathrm{BD}}$

$\Rightarrow \overrightarrow{\mathrm{BM}}=\frac{4\overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{BD}}}{5}=\frac{4\overrightarrow{\mathrm{a}}+\left(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}\right)}{5}$$=\frac{5\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}}{5}$

Tương tự $\overrightarrow{\mathrm{BN}}=\frac{3\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{b}}+2\overrightarrow{\mathrm{c}}}{5}$

$\overrightarrow{\mathrm{MN}}=\overrightarrow{\mathrm{BN}}-\overrightarrow{\mathrm{BM}}=\frac{-2\overrightarrow{\mathrm{a}}+3\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}}{5}$$=-\frac{2}{5}\left(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}}\right)+\frac{3}{5}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})$$=-\frac{2}{5}\overrightarrow{\mathrm{BD}}+\frac{3}{5}\overrightarrow{\mathrm{BC}\text{'}}$

Suy ra $\overrightarrow{\mathrm{MN}},\overrightarrow{\mathrm{DB}},\overrightarrow{\mathrm{BC}\text{'}}$ đồng phẳng mà $\mathrm{N}\notin \left(\mathrm{BC}\text{'}\mathrm{D}\right)$$\Rightarrow \mathrm{MN}\text{//}\left(\mathrm{BC}\text{'}\mathrm{D}\right)$

Chọn đáp án B.