Đáp án:
\(\begin{array}{l}
a)\,\,\,\overrightarrow u = \left( {1;\,\,1} \right).\\
b)\,\,\,\overrightarrow c = \frac{{23}}{{10}}\overrightarrow a + \frac{9}{{10}}\overrightarrow b .
\end{array}\)
Giải thích các bước giải:
Cho \(\overrightarrow a = \left( {1; - 2} \right),\,\,\,\overrightarrow b = \left( {3;\,\,4} \right),\,\,\,\overrightarrow c = \left( {5; - 1} \right)\)
\(\begin{array}{l}a)\,\,\,\overrightarrow u = 2\overrightarrow a + \overrightarrow b - \overrightarrow c = 2\left( {1; - 2} \right) + \left( {3;\,\,4} \right) - \left( {5; - 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2; - 4} \right) + \left( {4;\,\,4} \right) + \left( { - 5;\,\,1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {2 + 4 - 5; - 4 + 4 + 1} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {1;\,\,1} \right).\end{array}\)
b) Phân tích \(\overrightarrow c \) theo \(\overrightarrow a ,\,\,\,\overrightarrow b .\)
Gọi \(m,\,\,n \in \mathbb{R}\) thỏa mãn:\(\overrightarrow c = m\overrightarrow a + n\overrightarrow b \)
\(\begin{array}{l} \Leftrightarrow \left( {5; - 1} \right) = m\left( {1; - 2} \right) + n\left( {3;\,\,4} \right)\\ \Leftrightarrow \left( {5; - 1} \right) = \left( {m + 3n;\,\, - 2m + 4n} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}m + 3n = 5\\ - 2m + 4n = - 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m = \frac{{23}}{{10}}\\n = \frac{9}{{10}}\end{array} \right..\\ \Rightarrow \overrightarrow c = \frac{{23}}{{10}}\overrightarrow a + \frac{9}{{10}}\overrightarrow b .\end{array}\)
