Giải Câu 7 Bài 2: Hai đường thẳng vuông góc

Ta có: $\overrightarrow{AB}.\overrightarrow{AC}=AB.AC. cos\widehat{A}=>cos\widehat{A}=\frac{\overrightarrow{AB}.\overrightarrow{AC}}{AB.AC}$

Lại có:

\(S_{ABC}=\frac{1}{2}AB.AC.sin\widehat{A} =\)\(\frac{1}{2}AB.AC.\sqrt{1-cos^{2}\widehat{A}}\)

\(=\frac{1}{2}AB.AC.\sqrt{1-\left ( \frac{\overrightarrow{AB}.\overrightarrow{AC}}{AB.AC} \right )^{2}}\)

\(=\frac{1}{2}.AB.AC.\frac{\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}}{AB.AC}\)

\(=\frac{1}{2}\sqrt{\overrightarrow{AB}^{2}.\overrightarrow{AC}^{2}-(\overrightarrow{AB}.\overrightarrow{AC})^{2}}.\)