Cho tam giác ABC có AB = 2cm, BC = 3cm, CA = 5cm. Tính $\overrightarrow{CA}$. $\overrightarrow{CB}$
2 câu trả lời
$\begin{array}{l}
\overrightarrow {CA} .\overrightarrow {CB} = CA.CB.\cos \left( {\overrightarrow {CA} ,\overrightarrow {CB} } \right)\\
= CA.CB.\cos \widehat {ACB}\\
= CA.CB.\dfrac{{C{A^2} + C{B^2} - A{B^2}}}{{2CA.CB}}\\
= \dfrac{{C{A^2} + C{B^2} - A{B^2}}}{2} = \dfrac{{{5^2} + {3^2} - {2^2}}}{2}\\
= 15
\end{array}$
