Giải câu 7 bài biểu thức tọa độ của các phép toán vectơ

$\overrightarrow{F}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{2}}}+\overrightarrow{{{F}_{3}}}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{23}}}$

$\overrightarrow{{{F}_{23}}}=\overrightarrow{{{F}_{2}}}+\overrightarrow{{{F}_{3}}}$

$ \Rightarrow {{F}_{23}}^{2}={{F}_{2}}^{2}+{{F}_{3}}^{2}+2{{F}_{2}}.{{F}_{3}}.\cos \left(\overrightarrow{{{F}_{2}}},\overrightarrow{{{F}_{3}}} \right)$

$={{600}^{2}}+{{800}^{2}}+2.600.800.\cos {{75}^{o}}$

$=1248466,283$ (N)

$\Rightarrow {{F}_{23}}=\sqrt{1248466,283}=1117,347879$ $\approx 1117,35$ (N)

$cos\left( \overrightarrow{{{F}_{23}}},\overrightarrow{{{F}_{3}}}\right)=\frac{{{F}_{23}}^{2}+{{F}_{3}}^{2}-{{F}_{2}}^{2}}{2.{{F}_{23}}.{{F}_{3}}}=\frac{1248466,283+{{800}^{2}}-{{600}^{2}}}{2.1117,35.800}\approx 0,855$

$\widehat{\left( \overrightarrow{{{F}_{23}}},\overrightarrow{{{F}_{3}}} \right)}\approx {{31}^{o}}$ => $\widehat{\left(\overrightarrow{{{F}_{23}}},\overrightarrow{{{F}_{1}}} \right)}$ = 45^o -31^o =14^o

$\overrightarrow{F}=\overrightarrow{{{F}_{1}}}+\overrightarrow{{{F}_{23}}}$

$ \Rightarrow {{F}}^{2}={{F}_{23}}^{2}+{{F}_{1}}^{2}+2{{F}_{23}}.{{F}_{1}}.\cos \left(\overrightarrow{{{F}_{23}}},\overrightarrow{{{F}_{1}}} \right)$

$={1248466,283}+{{1500}^{2}}+2.1117,35.1500.\cos {{14}^{o}}$

$=6750946,072$

 $\Rightarrow {{F}}=\sqrt{6750946,072}=2598,258277$ $\approx 2598$ (N)