Cho vecto u = ( x1; y1), vecto v = ( x2; y2). Tính tích vô hướng của vecto u, vecto v

Hướng dẫn giải:

Vì \(\overrightarrow u = \left( {{x_1};\,\,{y_1}} \right),\,\,\overrightarrow v = \left( {{x_2};\,\,{y_2}} \right)\).

Nên ta có: \(\overrightarrow u = {x_1}\overrightarrow i + {y_1}\overrightarrow j \,;\,\,\overrightarrow v = {x_2}\overrightarrow i + {y_2}\overrightarrow j \,\).

Do đó \(\overrightarrow u \,.\,\overrightarrow v = \left( {{x_1}\overrightarrow i + {y_1}\overrightarrow j } \right).\left( {\overrightarrow u = {x_2}\overrightarrow i + {y_2}\overrightarrow j } \right)\)\( = {x_1}{x_2}.{\overrightarrow i ^2} + {x_1}{y_2}.\left( {\overrightarrow i \,\,.\,\overrightarrow j } \right) + {y_1}{x_2}.\,\left( {\overrightarrow j \,\,.\,\overrightarrow i } \right) + {y_1}{y_2}.\,{\overrightarrow j ^2}\)

\( = {x_1}{x_2} + {y_1}{y_2}\) (do \({\overrightarrow i ^2} = {\left| {\overrightarrow i } \right|^2} = 1;\,\,{\overrightarrow j ^2} = {\left| {\overrightarrow j } \right|^2} = 1\); \(\overrightarrow i \,\,.\,\,\overrightarrow j = \overrightarrow j \,\,.\,\,\overrightarrow i = 0\))

Vậy \(\overrightarrow u \,\,.\,\,\overrightarrow v \)\( = {x_1}{x_2} + {y_1}{y_2}\).