Cho ba vectơ \(\overrightarrow m = \left( {1;1} \right),\overrightarrow n = \left( {2;2} \right),\overrightarrow p = \left( { - 1; - 1} \right)\). Tìm tọa độ của các vectơ
a) \(\overrightarrow m + 2\overrightarrow n - 3\overrightarrow p \);
b) \(\left( {\overrightarrow n .\overrightarrow p } \right)\overrightarrow m \)
Cho hai vectơ \(\overrightarrow a = \left( {{a_1},{a_2}} \right),\overrightarrow b = \left( {{b_1},{b_2}} \right)\), ta có:
+ \(\overrightarrow a \pm \overrightarrow b = \left( {{a_1} \pm {b_1},{a_2} \pm {b_2}} \right)\)
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+ \(k\overrightarrow a = \left( {k{a_1},k{a_2}} \right)\)
+ \(\overrightarrow a .\overrightarrow b = {a_1}{b_1} + {a_2}{b_2}\)
a) \(2\overrightarrow n = \left( {4;4} \right),3\overrightarrow p = \left( { - 3; - 3} \right)\)
\( \Rightarrow \overrightarrow m + 2\overrightarrow n - 3\overrightarrow p = \left( {1;1} \right) + \left( {4;4} \right) - \left( { - 3; - 3} \right) = \left( {8;8} \right)\)
b) \(\overrightarrow n .\overrightarrow p = 2\left( { - 1} \right) + 2\left( { - 1} \right) = - 4 \Rightarrow \left( {\overrightarrow n .\overrightarrow p } \right)\overrightarrow m = - 4\left( {1;1} \right) = \left( { - 4; - 4} \right)\)