Bài 7:
Giả sử $\overrightarrow{c}= m\overrightarrow{a} + n\overrightarrow{b}$
Khi đó:
$\quad \begin{cases}5 = m.2 + n.1\\0 = m.(-2) + n.4\end{cases}$
$\to \begin{cases}2m + n = 5\\-2m + 4n = 0\end{cases}$
$\to \begin{cases}m = 2\\n = 1\end{cases}$
Vậy $\overrightarrow{c}= 2\overrightarrow{a} + \overrightarrow{b}$
8) Ta có:
$3\overrightarrow{AB} - 7\overrightarrow{CD}+ 2\overrightarrow{DA}= 4\overrightarrow{DB} + 3\overrightarrow{AC}$
$\to - 7\overrightarrow{CD}= 4\overrightarrow{DB}+ 3(\overrightarrow{AC}-\overrightarrow{AB})$
$\to 7\overrightarrow{DC}= 4\overrightarrow{DB}+ 3\overrightarrow{BC}$
$\to 7\overrightarrow{DC} = 4(\overrightarrow{DB} +\overrightarrow{BC})- \overrightarrow{BC}$
$\to 7\overrightarrow{DC} = 4\overrightarrow{DC}-\overrightarrow{BC}$
$\to 3\overrightarrow{DC} = \overrightarrow{CB}$
$\to 2\overrightarrow{DC}=\dfrac23\overrightarrow{CB}$
$\to 2\overrightarrow{DC}=\dfrac23(-14;14)$
$\to 2\overrightarrow{DC} =\left(-\dfrac{28}{3};\dfrac{28}{3}\right)$
$\to P = \left|2\overrightarrow{DC}\right| =\sqrt{\left(-\dfrac{28}{3}\right)^2 + \left(\dfrac{28}{3}\right)^2}$
$\to P = \dfrac{28\sqrt2}{3}$
