Cho ba vectơ
Cho ba vectơ \(\overrightarrow u (1;2;3),\overrightarrow v (2;2; – 1),\overrightarrow {\rm{w}} \left( {4;0; – 4} \right)\). Tìm tọa độ của vectơ \(\overrightarrow x \), biết
\(\eqalign{ & a)\overrightarrow x = \overrightarrow u – \overrightarrow v ; \cr & b)\overrightarrow x = \overrightarrow u – \overrightarrow v + 2\overrightarrow {\rm{w}} ; \cr & c)\overrightarrow x = 2\overrightarrow u + 4\overrightarrow v – \overrightarrow {\rm{w}} ; \cr & d)\overrightarrow x = 5\overrightarrow u – 3\overrightarrow v – {1 \over 2}\overrightarrow {\rm{w}} . \cr & e)2\overrightarrow x – 3\overrightarrow u = \overrightarrow {\rm{w}} ; \cr & g)2\overrightarrow u + \overrightarrow v – \overrightarrow {\rm{w}} + 3\overrightarrow x = \overrightarrow 0 ; \cr} \)
Giải
\(\eqalign{ & a)\overrightarrow x = (1 – 2;2 – 2;3 + 1) = \left( { – 1;0;4} \right). \cr & b)\overrightarrow x = ( – 1 + 8;0 + 0;4 – 8) = (7;0; – 4). \cr & c)\overrightarrow x = (2 + 8 – 4;4 + 8 – 0;6 – 4 + 4) \cr&\;\;\;\;\;\;\;\;= (6;12;6). \cr & d)\overrightarrow x = (5 – 6 – 2;10 – 6 + 0;15 + 3 + 2) \cr&\;\;\;\;\;\;\;\;= ( – 3;4;20). \cr & e)2\overrightarrow x = 3\overrightarrow u + \overrightarrow {\rm{w}} \Rightarrow \overrightarrow x = {3 \over 2}\overrightarrow u + {1 \over 2}\overrightarrow {\rm{w}} . \cr & \Rightarrow \overrightarrow x = \left( {{3 \over 2} + 2;3 + 0;{9 \over 2} – 0} \right) = \left( {{7 \over 2};3;{5 \over 2}} \right), \cr & g)\;3\overrightarrow x = – 2\overrightarrow u – \overrightarrow v + \overrightarrow {\rm{w}} \cr&\;\;\;\;\;\;\;\;\;\;\;= ( – 2 – 2 + 4; – 4 – 2 + 0; – 6 + 1 – 4) \cr & \Rightarrow 3\overrightarrow x = (0; – 6; – 9) \Rightarrow \overrightarrow x = (0, – 2; – 3). \cr} \)
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