Cho tam giác \(OAB\). Gọi \(M\) và \(N\) lần lượt là trung điểm của \(OA\) và \(OB\). Tìm các số \(m, n\) sao cho:
a) \(\overrightarrow {OM} = m\overrightarrow {OA} + n\overrightarrow {OB} \)
b) \(\overrightarrow {AN} = m\overrightarrow {OA} + n\overrightarrow {OB} \)
c) \(\overrightarrow {MN} = m\overrightarrow {OA} + n\overrightarrow {OB} \)
d) \(\overrightarrow {MB} = m\overrightarrow {OA} + n\overrightarrow {OB} \)
a) Ta có: \(\overrightarrow {OM} = {1 \over 2}\overrightarrow {OA} \)
\(\begin{array}{l}
\overrightarrow {OM} = n\overrightarrow {OA} + n\overrightarrow {OB}\\ \Rightarrow m\overrightarrow {OA} + n\overrightarrow {OB} = \frac{1}{2}\overrightarrow {OA} \\
\Leftrightarrow \left( {m - \frac{1}{2}} \right)\overrightarrow {OA} + n\overrightarrow {OB} = \overrightarrow 0 \\
\Leftrightarrow \left( {m - \frac{1}{2}} \right)\overrightarrow {OA} + n\overrightarrow {OB} = \overrightarrow 0 \\
\Leftrightarrow \left\{ \begin{array}{l}
m - \frac{1}{2} = 0\\
n = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m = \frac{1}{2}\\
n = 0
\end{array} \right..
\end{array}\)
Vậy \(m = {1 \over 2}; \, \, n = 0.\)
b) Ta có: vì \(N\) là trung điểm \(OB\)
\(\eqalign{
& 2\overrightarrow {AN} = \overrightarrow {AO} + \overrightarrow {AB} \cr
& \Rightarrow 2\overrightarrow {AN} = \overrightarrow {AO} + \overrightarrow {AO} + \overrightarrow {OB} \cr
& \Rightarrow 2\overrightarrow {AN} = 2\overrightarrow {AO} + \overrightarrow {OB}\cr& \Rightarrow \overrightarrow {AN} = - \overrightarrow {OA} + {1 \over 2}\overrightarrow {OB} \cr} \)
\(\begin{array}{l} \Leftrightarrow \left( {m + 1} \right)\overrightarrow {OA} + \left( {n - \frac{1}{2}} \right)\overrightarrow {OB} = \overrightarrow 0 \\ \Leftrightarrow \left\{ \begin{array}{l}m + 1 = 0\\n - \frac{1}{2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = - 1\\n = \frac{1}{2}\end{array} \right..\end{array}\)
Vậy \(m = - 1; \, \, n = {1 \over 2}.\)
\(\eqalign{ c) \, \,& \overrightarrow {MN} = {1 \over 2}\overrightarrow {AB}\cr& \Rightarrow \overrightarrow {MN} = {1 \over 2}(\overrightarrow {AO} + \overrightarrow {OB} ) \cr & \Rightarrow \overrightarrow {MN} = - {1 \over 2}\overrightarrow {OA} + {1 \over 2}\overrightarrow {OB} \cr} \)
\(\begin{array}{l} \Leftrightarrow \left( {m + \frac{1}{2}} \right)\overrightarrow {OA} + \left( {n - \frac{1}{2}} \right)\overrightarrow {OB} = \overrightarrow 0 \\ \Leftrightarrow \left\{ \begin{array}{l}m + \frac{1}{2} = 0\\n - \frac{1}{2} = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = - \frac{1}{2}\\n = \frac{1}{2}\end{array} \right..\end{array}\)
Vậy \(m = - {1 \over 2}, \, \, n = {1 \over 2}.\)
d) Ta có:
\(\eqalign{
& 2\overrightarrow {BM} = \overrightarrow {BA} + \overrightarrow {BO}\cr& \Rightarrow 2\overrightarrow {BM} = \overrightarrow {BO} + \overrightarrow {OA} + \overrightarrow {BO} \cr
& \Rightarrow 2\overrightarrow {BM} = 2\overrightarrow {BO} + \overrightarrow {OA}\cr& \Rightarrow 2\overrightarrow {MB} = - \overrightarrow {OA} + 2\overrightarrow {OB} \cr
& \Rightarrow \overrightarrow {MB} = - {1 \over 2}\overrightarrow {OA} + \overrightarrow {OB} \cr} \)
\(\begin{array}{l} \Leftrightarrow \left( {m + 1} \right)\overrightarrow {OA} + \left( {n - 1} \right)\overrightarrow {OB} = \overrightarrow 0 \\ \Leftrightarrow \left\{ \begin{array}{l}m + \frac{1}{2} = 0\\n - 1 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = - \frac{1}{2}\\n = 1\end{array} \right..\end{array}\)
Vậy \(m = - {1 \over 2}, \, \, n = 1.\)