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Câu hỏi: 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
\(\eqalign{ & a)\overrightarrow x = \overrightarrow u - \overrightarrow v ; \cr & b)\overrightarrow x = \overrightarrow u - \overrightarrow v + 2\overrightarrow {\rm{w}} ; \cr & c)\overrightarrow x = 2\overrightarrow u + 4\overrightarrow v - \overrightarrow {\rm{w}} ; \cr & d)\overrightarrow x = 5\overrightarrow u - 3\overrightarrow v - {1 \over 2}\overrightarrow {\rm{w}} . \cr & e)2\overrightarrow x - 3\overrightarrow u = \overrightarrow {\rm{w}} ; \cr & g)2\overrightarrow u + \overrightarrow v - \overrightarrow {\rm{w}} + 3\overrightarrow x = \overrightarrow 0 ; \cr} \)
\(\eqalign{ & a)\overrightarrow x = \overrightarrow u - \overrightarrow v ; \cr & b)\overrightarrow x = \overrightarrow u - \overrightarrow v + 2\overrightarrow {\rm{w}} ; \cr & c)\overrightarrow x = 2\overrightarrow u + 4\overrightarrow v - \overrightarrow {\rm{w}} ; \cr & d)\overrightarrow x = 5\overrightarrow u - 3\overrightarrow v - {1 \over 2}\overrightarrow {\rm{w}} . \cr & e)2\overrightarrow x - 3\overrightarrow u = \overrightarrow {\rm{w}} ; \cr & g)2\overrightarrow u + \overrightarrow v - \overrightarrow {\rm{w}} + 3\overrightarrow x = \overrightarrow 0 ; \cr} \)
Lời giải chi tiết
\(\eqalign{ & a)\overrightarrow x = (1 - 2; 2 - 2; 3 + 1) = \left({ - 1; 0; 4} \right). \cr & b)\overrightarrow x = (- 1 + 8; 0 + 0; 4 - 8) = (7; 0; - 4). \cr & c)\overrightarrow x = (2 + 8 - 4; 4 + 8 - 0; 6 - 4 + 4) \cr& = (6; 12; 6). \cr & d)\overrightarrow x = (5 - 6 - 2; 10 - 6 + 0; 15 + 3 + 2) \cr& = (- 3; 4; 20). \cr & e)2\overrightarrow x = 3\overrightarrow u + \overrightarrow {\rm{w}} \Rightarrow \overrightarrow x = {3 \over 2}\overrightarrow u + {1 \over 2}\overrightarrow {\rm{w}} . \cr & \Rightarrow \overrightarrow x = \left({{3 \over 2} + 2; 3 + 0;{9 \over 2} - 0} \right) = \left({{7 \over 2}; 3;{5 \over 2}} \right), \cr & g) 3\overrightarrow x = - 2\overrightarrow u - \overrightarrow v + \overrightarrow {\rm{w}} \cr& = (- 2 - 2 + 4; - 4 - 2 + 0; - 6 + 1 - 4) \cr & \Rightarrow 3\overrightarrow x = (0; - 6; - 9) \Rightarrow \overrightarrow x = (0, - 2; - 3). \cr} \)
\(\eqalign{ & a)\overrightarrow x = (1 - 2; 2 - 2; 3 + 1) = \left({ - 1; 0; 4} \right). \cr & b)\overrightarrow x = (- 1 + 8; 0 + 0; 4 - 8) = (7; 0; - 4). \cr & c)\overrightarrow x = (2 + 8 - 4; 4 + 8 - 0; 6 - 4 + 4) \cr& = (6; 12; 6). \cr & d)\overrightarrow x = (5 - 6 - 2; 10 - 6 + 0; 15 + 3 + 2) \cr& = (- 3; 4; 20). \cr & e)2\overrightarrow x = 3\overrightarrow u + \overrightarrow {\rm{w}} \Rightarrow \overrightarrow x = {3 \over 2}\overrightarrow u + {1 \over 2}\overrightarrow {\rm{w}} . \cr & \Rightarrow \overrightarrow x = \left({{3 \over 2} + 2; 3 + 0;{9 \over 2} - 0} \right) = \left({{7 \over 2}; 3;{5 \over 2}} \right), \cr & g) 3\overrightarrow x = - 2\overrightarrow u - \overrightarrow v + \overrightarrow {\rm{w}} \cr& = (- 2 - 2 + 4; - 4 - 2 + 0; - 6 + 1 - 4) \cr & \Rightarrow 3\overrightarrow x = (0; - 6; - 9) \Rightarrow \overrightarrow x = (0, - 2; - 3). \cr} \)