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Câu hỏi: Cho \(S\) là diện tích tam giác \(ABC\). Chứng minh rằng:
\(S=\dfrac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}.\)
\(S=\dfrac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}.\)
Phương pháp giải
Sử dụng các công thức:
\(\begin{array}{l}{S_{ABC}} = \dfrac{1}{2}AB. AC.\sin A\\\sin A = \sqrt {1 - {{\cos }^2}A} \\\cos A = \dfrac{{\overrightarrow {AB} .\overrightarrow {AC} }}{{\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|}}\end{array}\)
Lời giải chi tiết
\(S_{ABC}=\dfrac{1}{2}AB. AC.\sin A\) \(=\dfrac{1}{2}AB. AC.\sqrt{1-\cos^{2}A}\)
\(=\dfrac{1}{2}AB. AC.\sqrt{1-\left(\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|} \right)^{2}}\)
\(= \dfrac{1}{2}\sqrt {A{B^2}. A{C^2} - A{B^2}A{C^2}.\dfrac{{{{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}}}{{{{\left| {\overrightarrow {AB} } \right|}^2}.{{\left| {\overrightarrow {AC} } \right|}^2}}}} \)
\(= \dfrac{1}{2}\sqrt {{{\overrightarrow {AB} }^2}.{{\overrightarrow {AC} }^2} - A{B^2}. A{C^2}.\dfrac{{{{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}}}{{A{B^2}. A{C^2}}}} \)
\(=\dfrac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}.\)
Sử dụng các công thức:
\(\begin{array}{l}{S_{ABC}} = \dfrac{1}{2}AB. AC.\sin A\\\sin A = \sqrt {1 - {{\cos }^2}A} \\\cos A = \dfrac{{\overrightarrow {AB} .\overrightarrow {AC} }}{{\left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|}}\end{array}\)
Lời giải chi tiết
\(S_{ABC}=\dfrac{1}{2}AB. AC.\sin A\) \(=\dfrac{1}{2}AB. AC.\sqrt{1-\cos^{2}A}\)
\(=\dfrac{1}{2}AB. AC.\sqrt{1-\left(\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{|\overrightarrow{AB}|.|\overrightarrow{AC}|} \right)^{2}}\)
\(= \dfrac{1}{2}\sqrt {A{B^2}. A{C^2} - A{B^2}A{C^2}.\dfrac{{{{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}}}{{{{\left| {\overrightarrow {AB} } \right|}^2}.{{\left| {\overrightarrow {AC} } \right|}^2}}}} \)
\(= \dfrac{1}{2}\sqrt {{{\overrightarrow {AB} }^2}.{{\overrightarrow {AC} }^2} - A{B^2}. A{C^2}.\dfrac{{{{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}}}{{A{B^2}. A{C^2}}}} \)
\(=\dfrac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}.\)