Bài 22 trang 23 Sách giáo khoa (SGK) Hình học 10 Nâng cao

Lời giải chi tiết

Ta có
\(\eqalign{
& \overrightarrow {OM} = {1 \over 2}\overrightarrow {OA} = {1 \over 2}\overrightarrow {OA} + 0.\overrightarrow {OB} \cr&\Rightarrow m = {1 \over 2}, n = 0. \cr
& \overrightarrow {MN} = \overrightarrow {ON} - \overrightarrow {OM} \cr&= {1 \over 2}\overrightarrow {OB} - {1 \over 2}\overrightarrow {OA} \cr&= \left({ - {1 \over 2}} \right)\overrightarrow {OA} + {1 \over 2}\overrightarrow {OB} \cr& \Rightarrow m = - {1 \over 2}, n = {1 \over 2}. \cr
& \overrightarrow {AN} = \overrightarrow {ON} - \overrightarrow {OA} \cr&= {1 \over 2}\overrightarrow {OB} - \overrightarrow {OA}\cr& = \left({ - 1} \right)\overrightarrow {OA} + {1 \over 2}\overrightarrow {OB}\cr&\Rightarrow m = - 1, n = {1 \over 2}. \cr
& \overrightarrow {MB} = \overrightarrow {OB} - \overrightarrow {OM} \cr&= \overrightarrow {OB} - {1 \over 2}\overrightarrow {OA} \cr& = \left({ - {1 \over 2}} \right)\overrightarrow {OA} + \overrightarrow {OB}\cr&\Rightarrow m = - {1 \over 2}, n = 1. \cr} \)
Cách khác:
$\overrightarrow{\mathrm{OM}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}} \Leftrightarrow \frac{1}{2} \overrightarrow{\mathrm{OA}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}$
(Vì M là trung điểm OA)
$\Leftrightarrow\left(\frac{1}{2}-m\right) \overrightarrow{\mathrm{OA}}=n \overrightarrow{\mathrm{OB}}(1)$
Mà $\overrightarrow{\mathrm{OA}}$ và $\overrightarrow{\mathrm{OB}}$ không cùng phương do đó (1) xảy ra $\Leftrightarrow$
$\left\{\begin{array}{l}\left(\frac{1}{2}-\mathrm{m}\right) \overrightarrow{\mathrm{OA}}=\overrightarrow{0} \\ \mathrm{n} \overrightarrow{\mathrm{OB}}=\overrightarrow{0}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{1}{2}-\mathrm{m}=0 \\ \mathrm{n}=0\end{array} \Leftrightarrow\left\{\begin{array}{l}\mathrm{m}=\frac{1}{2} \\ \mathrm{n}=0\end{array}\right.\right.\right.$
Vậy $\overrightarrow{\mathrm{OM}}=\frac{1}{\mathrm{OA}}+0 \cdot \overrightarrow{\mathrm{OB}}$
$+\overrightarrow{\mathrm{OM}}=m \overrightarrow{\mathrm{OA}}+n \overrightarrow{\mathrm{OB}}$
$\Leftrightarrow \frac{1}{2} \overrightarrow{\mathrm{AB}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}$
(do $\mathrm{M}, \mathrm{N}$ lần lượt là trung điềm $\mathrm{OA}$ và $\mathrm{OB}$ )
$\Leftrightarrow \frac{1}{2}(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}})=m \overrightarrow{\mathrm{OA}}+n \overrightarrow{\mathrm{OB}}$
$\Leftrightarrow\left(\frac{1}{2}-\mathrm{n}\right) \overrightarrow{\mathrm{OB}}=\left(\frac{1}{2}+\mathrm{n}\right) \overrightarrow{\mathrm{OA}}(2)$
Mà OA và $\overrightarrow{\text { OB }}$ không cùng phương do đó (2) xày ra $\Leftrightarrow$
$\left\{\begin{array}{l}\left(\frac{1}{2}-n\right) \overrightarrow{O B}=\overrightarrow{0} \\ \left(\frac{1}{2}+m\right) \overrightarrow{O A}=\overrightarrow{0}\end{array} \Rightarrow\left\{\begin{array}{l}\frac{1}{2}-n=0 \\ \frac{1}{2}+m=0\end{array} \Leftrightarrow\left\{\begin{array}{l}n=\frac{1}{2} \\ m=-\frac{1}{2}\end{array}\right.\right.\right.$
Vậy $\overrightarrow{\mathrm{MN}}=\frac{1}{2} \overrightarrow{\mathrm{OA}}+\frac{1}{2} \cdot \overrightarrow{\mathrm{OB}}$
$\begin{aligned}
&+\overrightarrow{\mathrm{AN}}=m \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}\\
&\Leftrightarrow \overrightarrow{\mathrm{ON}}-\overrightarrow{\mathrm{OA}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}\\
&\Leftrightarrow \frac{1}{2} \overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}\\
&\Leftrightarrow\left(\frac{1}{2}-n\right) \overrightarrow{\mathrm{OB}}=(\mathrm{m}+1) \overrightarrow{\mathrm{OA}}(3)\\
&\text { Mà } \overrightarrow{\mathrm{OA}} \text { và } \overrightarrow{\mathrm{OB}} \text { không cùng phương do đó }(3) \text { xày } \mathrm{ra} \Leftrightarrow\\
&\left\{\begin{array}{l}
\left(\frac{1}{2}-n\right) \overrightarrow{O B}=\overrightarrow{0} \\
(m+1) \overrightarrow{O A}=\overrightarrow{0}
\end{array} \Rightarrow\left\{\begin{array}{l}
\frac{1}{2}-n=0 \\
m+1=0
\end{array} \Leftrightarrow\left\{\begin{array}{l}
n=\frac{1}{2} \\
m=-1
\end{array}\right.\right.\right.\\
&\text { Vậy } \overrightarrow{\mathrm{AN}}=-1 \cdot \overrightarrow{\mathrm{OA}}+\frac{1}{2} \overrightarrow{\mathrm{OB}}
\end{aligned}$
$+\overrightarrow{\mathrm{MB}}=m \overrightarrow{\mathrm{OA}}+n \overrightarrow{\mathrm{OB}}$
$\Leftrightarrow \overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OM}}=m \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}$
$\Leftrightarrow \overrightarrow{\mathrm{OB}}-\frac{1}{2} \overrightarrow{\mathrm{OA}}=\mathrm{m} \overrightarrow{\mathrm{OA}}+\mathrm{n} \overrightarrow{\mathrm{OB}}$
$\Leftrightarrow(1-\mathrm{n}) \overrightarrow{\mathrm{OB}}=\left(\frac{1}{2}+\mathrm{m}\right) \overrightarrow{\mathrm{OA}}(4)$
Mà $\overrightarrow{\mathrm{OA}}$ và $\overrightarrow{\mathrm{OB}}$ không cùng phương do đó (4) xày $\mathrm{ra} \Leftrightarrow$ $\left\{\begin{array}{l}(1-\mathrm{n}) \overrightarrow{\mathrm{OB}}=\overrightarrow{0} \\ \left(\frac{1}{2}+\mathrm{m}\right) \overrightarrow{\mathrm{OA}}=\overrightarrow{0}\end{array} \Rightarrow\left\{\begin{array}{l}1-\mathrm{n}=0 \\ \frac{1}{2}+\mathrm{m}=0\end{array} \Leftrightarrow\left\{\begin{array}{l}\mathrm{n}=1 \\ \mathrm{~m}=-\frac{1}{2}\end{array}\right.\right.\right.$
Vậy $\overrightarrow{\mathrm{MB}}=-\frac{1}{2} \overrightarrow{\mathrm{OA}}+1 \cdot \overrightarrow{\mathrm{OB}}$