Cho vecto a ( -3j) vecto b ( 3i+4j) vecto c ( 5j-3i) Tính / a+b-2c/ ( / là dấu GTTĐ nha) Tìm vecto z cùng phương vecto i sao cho 2z + b và a-c cùng phương

$\begin{array}{l} \overrightarrow a = - 3\overrightarrow j = 0\overrightarrow i - 3\overrightarrow j \Rightarrow \overrightarrow a = \left( {0; - 3} \right)\\ \overrightarrow b = 3\overrightarrow i + 4\overrightarrow j \Rightarrow \overrightarrow b = \left( {3;4} \right)\\ \overrightarrow c = 5\overrightarrow j - 3\overrightarrow i = - 3\overrightarrow i + 5\overrightarrow j \Rightarrow \overrightarrow c = \left( { - 3;5} \right)\\ \Rightarrow \overrightarrow a + \overrightarrow b - 2\overrightarrow c = \left( {9; - 9} \right) \Rightarrow \left| {\overrightarrow a + \overrightarrow b - 2\overrightarrow c } \right| = \sqrt {{9^2} + {{\left( { - 9} \right)}^2}} = 9\sqrt 2 \\ \overrightarrow z = \left( {x;0} \right) \Rightarrow 2\overrightarrow z + \overrightarrow b = \left( {2x + 3;4} \right);\overrightarrow a - \overrightarrow c = \left( {3; - 8} \right)\\ 2\overrightarrow z + \overrightarrow b \,cung\,phuong\,\overrightarrow a - \overrightarrow c \Leftrightarrow \frac{{2x + 3}}{3} = \frac{4}{{ - 8}}\\ \Leftrightarrow - 8\left( {2x + 3} \right) = 12 \Leftrightarrow - 16x = 36 \Leftrightarrow x = - \frac{9}{4}\\ \Rightarrow \overrightarrow z = \left( { - \frac{9}{4};0} \right) \end{array}$