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

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

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

\(= \dfrac{1}{2}\sqrt { A{B^2}.A{C^2}- {{\left(|{\overrightarrow {AB}| .|\overrightarrow {AC}|. \cos BAC} \right)}^2}} \)

\(\begin{array}{l} \Rightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2} - {{\left( {AB.AC.\cos A} \right)}^2}} \\ \Leftrightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2} - A{B^2}.A{C^2}.{{\cos }^2}A }\\ \Leftrightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2}\left( {1 - {{\cos }^2}A} \right)} \end{array}\)

Mà \(1 - {\cos ^2}A = {\sin ^2}A\)

\( \Rightarrow A = \dfrac{1}{2}\sqrt {A{B^2}.A{C^2}.{{\sin }^2}A} \)

\( \Leftrightarrow A = \dfrac{1}{2}.AB.AC.\sin A\) (Vì \({0^o} < \widehat A < {180^o}\) nên \(\sin A > 0\))

Do đó \(A = {S_{ABC}}\) hay \({S_{ABC}} = \dfrac{1}{2}\sqrt {{{\overrightarrow {AB} }^2}.{{\overrightarrow {AC} }^2} - {{\left( {\overrightarrow {AB} .\overrightarrow {AC} } \right)}^2}} .\) (đpcm)

Có vẻ không đúng.

Giả sử \(\overrightarrow{AB}+\overrightarrow{MB}+\overrightarrow{MA}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{MB}+\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{MB}=\overrightarrow{0}\)

\(\Leftrightarrow M\equiv B\) (Vô lí)

Hình vẽ:

a, Chứng minh \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{0}\)

Ta có \(\overrightarrow{AB}+\overrightarrow{BM}+\overrightarrow{MA}=\overrightarrow{BM}+\left(\overrightarrow{MA}+\overrightarrow{AB}\right)=\overrightarrow{BM}+\overrightarrow{MB}=\overrightarrow{0}\)

b, Gọi H là trung điểm \(MC\)

Ta có \(AM=\sqrt{AC^2-MC^2}=\sqrt{4a^2-a^2}=a\sqrt{3}\)

\(AH=\sqrt{AM^2+MH^2}=\sqrt{\left(a\sqrt{3}\right)^2+\left(\dfrac{a}{2}\right)^2}=a.\dfrac{\sqrt{13}}{2}\)

\(\left|\overrightarrow{AM}+\overrightarrow{AC}\right|=\left|2\overrightarrow{AH}\right|=2AH=a\sqrt{13}\)

c, Gọi D là trung điểm AB

\(3\overrightarrow{NA}+3\overrightarrow{NB}+2\overrightarrow{NC}=3\left(\overrightarrow{NA}+\overrightarrow{NB}\right)+2\overrightarrow{NC}=6\overrightarrow{ND}+2\overrightarrow{NC}=\overrightarrow{0}\)

\(\Rightarrow\overrightarrow{NC}=3\overrightarrow{DN}\)

Vậy N thuộc đoạn CD sao cho \(CN=\dfrac{3}{4}CD\)