Giải bài 7 trang 98 – SGK môn Hình học lớp 11

Ta có: ${{S}_{ABC}}=\dfrac{1}{2}AB.AC.\sin \widehat{A}=\dfrac{1}{2}.AB.AC.\sqrt{1-{{\cos }^{2}}\widehat{A}}$

Ta lại có: $\cos \widehat{A}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|}\Rightarrow {{\cos }^{2}}\widehat{A}=\dfrac{{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}{|\overrightarrow{AB}{{|}^{2}}|\overrightarrow{AC}{{|}^{2}}}$

Nên

$\begin{aligned} & 1-{{\cos }^{2}}\widehat{A}=1-\dfrac{{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}}=\dfrac{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}-{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}} \\ & \Rightarrow \sqrt{1-{{\cos }^{2}}\widehat{A}}=\sqrt{\dfrac{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}-{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}}} \\ \end{aligned}$

Do đó:

$\begin{aligned} & {{S}_{ABC}}=\dfrac{1}{2}AB.AC.\sqrt{\dfrac{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}-{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}}} \\ & =\dfrac{1}{2}.AB.AC.\dfrac{\sqrt{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}-{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}}}{AB.AC} \\ & =\dfrac{1}{2}\sqrt{{{\left| \overrightarrow{AB} \right|}^{2}}{{\left| \overrightarrow{AC} \right|}^{2}}-{{\left( \overrightarrow{AB}.\overrightarrow{AC} \right)}^{2}}} \\ \end{aligned}$

Ghi nhớ:   

${{S}_{ABC}}=\dfrac{1}{2}AB.AC.\sin \widehat{A}$