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Câu hỏi: Tìm tọa độ của các vectơ sau trong mặt phẳng tọa độ
\(\eqalign{
& \overrightarrow a = - \overrightarrow i ; \overrightarrow b = 5\overrightarrow j ; \overrightarrow c = 3\overrightarrow i - 4\overrightarrow j \cr
& \overrightarrow d = {1 \over 2}(\overrightarrow j - \overrightarrow i) ; \overrightarrow e = 0,15\overrightarrow i + 1,3\overrightarrow {j} \cr&\overrightarrow f = \pi \overrightarrow i - (\cos {24^0})\overrightarrow {j}\cr} \)
\(\eqalign{
& \overrightarrow a = - \overrightarrow i ; \overrightarrow b = 5\overrightarrow j ; \overrightarrow c = 3\overrightarrow i - 4\overrightarrow j \cr
& \overrightarrow d = {1 \over 2}(\overrightarrow j - \overrightarrow i) ; \overrightarrow e = 0,15\overrightarrow i + 1,3\overrightarrow {j} \cr&\overrightarrow f = \pi \overrightarrow i - (\cos {24^0})\overrightarrow {j}\cr} \)
Phương pháp giải
Sử dụng lí thuyết: \(\overrightarrow a = (x, y) \Rightarrow \overrightarrow a = x\overrightarrow i + y\overrightarrow j \)
Lời giải chi tiết
\(\begin{array}{l}
\overrightarrow a = - \overrightarrow i = \left({ - 1} \right)\overrightarrow i + 0\overrightarrow j \\
\Rightarrow \overrightarrow a = \left({ - 1; 0} \right)\\
\overrightarrow b = 5\overrightarrow j = 0\overrightarrow i + 5\overrightarrow j \\
\Rightarrow \overrightarrow b = \left({0; 5} \right)\\
\overrightarrow c = 3\overrightarrow i - 4\overrightarrow j = 3\overrightarrow i + \left({ - 4} \right)\overrightarrow j \\
\Rightarrow \overrightarrow c = \left({3; - 4} \right)\\
\overrightarrow d = \frac{1}{2}\left({\overrightarrow j - \overrightarrow i } \right) = \frac{1}{2}\overrightarrow j - \frac{1}{2}\overrightarrow i \\
= \left({ - \frac{1}{2}} \right)\overrightarrow i + \frac{1}{2}\overrightarrow j \\
\Rightarrow \overrightarrow d = \left({ - \frac{1}{2};\frac{1}{2}} \right)\\
\overrightarrow e = 0,15\overrightarrow i + 1,3\overrightarrow j \\
\Rightarrow \overrightarrow e = \left({0,15; 1,3} \right)\\
\overrightarrow f = \pi \overrightarrow i - \cos {24^0}\overrightarrow j \\
= \pi \overrightarrow i + \left({ - \cos {{24}^0}} \right)\overrightarrow j \\
\Rightarrow \overrightarrow f = \left({\pi ; - \cos {{24}^0}} \right)
\end{array}\)
Sử dụng lí thuyết: \(\overrightarrow a = (x, y) \Rightarrow \overrightarrow a = x\overrightarrow i + y\overrightarrow j \)
Lời giải chi tiết
\(\begin{array}{l}
\overrightarrow a = - \overrightarrow i = \left({ - 1} \right)\overrightarrow i + 0\overrightarrow j \\
\Rightarrow \overrightarrow a = \left({ - 1; 0} \right)\\
\overrightarrow b = 5\overrightarrow j = 0\overrightarrow i + 5\overrightarrow j \\
\Rightarrow \overrightarrow b = \left({0; 5} \right)\\
\overrightarrow c = 3\overrightarrow i - 4\overrightarrow j = 3\overrightarrow i + \left({ - 4} \right)\overrightarrow j \\
\Rightarrow \overrightarrow c = \left({3; - 4} \right)\\
\overrightarrow d = \frac{1}{2}\left({\overrightarrow j - \overrightarrow i } \right) = \frac{1}{2}\overrightarrow j - \frac{1}{2}\overrightarrow i \\
= \left({ - \frac{1}{2}} \right)\overrightarrow i + \frac{1}{2}\overrightarrow j \\
\Rightarrow \overrightarrow d = \left({ - \frac{1}{2};\frac{1}{2}} \right)\\
\overrightarrow e = 0,15\overrightarrow i + 1,3\overrightarrow j \\
\Rightarrow \overrightarrow e = \left({0,15; 1,3} \right)\\
\overrightarrow f = \pi \overrightarrow i - \cos {24^0}\overrightarrow j \\
= \pi \overrightarrow i + \left({ - \cos {{24}^0}} \right)\overrightarrow j \\
\Rightarrow \overrightarrow f = \left({\pi ; - \cos {{24}^0}} \right)
\end{array}\)