Question

Cho tam giác ABC . CMR :

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

 

Answer 1

Ta có : \(\dfrac{1}{2}\sqrt{\overrightarrow{AB}^2\overrightarrow{AC}^2-\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2}\)

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

\(=\dfrac{1}{2}.\sqrt{AB^2AC^2-AB^2.AC^2.Cos^2BAC}\)

\(=\dfrac{1}{2}\sqrt{AB^2AC^2\left(1-Cos^2BAC\right)}\)

Thấy : \(Sin^2a+Cos^2a=1\)

\(\Rightarrow Sin^2a=1-Cos^2a\)

\(\Rightarrow\dfrac{1}{2}\sqrt{AB^2AC^2Sin^2BAC}=\dfrac{1}{2}\left|AB.AC.SinBAC\right|=\dfrac{1}{2}AB.AC.SinBAC=S\)

=> ĐPCM

 

Answer 2

Sao đề là lạ đoạn kia là \(\left(\overrightarrow{AB}.\overrightarrow{AC}\right)^2\)à