Cho tam giác ABC. Chứng minh rằng \(\overrightarrow {AB} .\overrightarrow {AC} = \frac{1}{2}\left( {A{B^2} + A{C^2} - B{C^2}} \right)\).

Lời giải

Ta có: \(\frac{1}{2}\left( {A{B^2} + A{C^2} - B{C^2}} \right)\)

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

\( = \frac{1}{2}\left[ {\left( {\overrightarrow {AB} - \overrightarrow {BC} } \right)\left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) + {{\overrightarrow {AC} }^2}} \right]\)

\( = \frac{1}{2}\left[ {\left( {\overrightarrow {AB} - \overrightarrow {BC} } \right)\overrightarrow {AC} + {{\overrightarrow {AC} }^2}} \right]\)

\( = \frac{1}{2}\left[ {\overrightarrow {AB} .\overrightarrow {AC} - \overrightarrow {BC} .\overrightarrow {AC} + {{\overrightarrow {AC} }^2}} \right]\)

\( = \frac{1}{2}\left[ {\overrightarrow {AB} .\overrightarrow {AC} - \left( {\overrightarrow {BA} + \overrightarrow {AC} } \right).\overrightarrow {AC} + {{\overrightarrow {AC} }^2}} \right]\)

\( = \frac{1}{2}\left[ {\overrightarrow {AB} .\overrightarrow {AC} - \overrightarrow {BA} .\overrightarrow {AC} - {{\overrightarrow {AC} }^2} + {{\overrightarrow {AC} }^2}} \right]\)

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