Đáp án:
b. \(\overrightarrow {AB} .\overrightarrow {AC} = \frac{{ - 15}}{2}\)
\( = {60^o}\)
c. \(\overrightarrow {BC} .\overrightarrow {CA} = \frac{{ - 65}}{2}\)
\(\overrightarrow {CB} .\overrightarrow {CA} = \frac{{65}}{2}\)
Giải thích các bước giải:
a. \(\begin{array}{l}
A{B^2} - 2.\overrightarrow {AB} .\overrightarrow {AC} + A{C^2}\\
= {(\overrightarrow {AB} - \overrightarrow {AC} )^2}\\
= {(\overrightarrow {CB} )^2} = B{C^2}
\end{array}\) (đpcm)
b. \(\begin{array}{l}
B{C^2} = A{B^2} - 2.\overrightarrow {AB} .\overrightarrow {AC} + A{C^2}\\
\to \overrightarrow {AB} .\overrightarrow {AC} = \frac{{A{B^2} + A{C^2} - B{C^2}}}{2} = \frac{{{3^2} + {5^2} - {7^2}}}{2} = \frac{{ - 15}}{2}
\end{array}\)
\(\cos = \frac{{\left| {\overrightarrow {AB} .\overrightarrow {AC} } \right|}}{{|\overrightarrow {AB} |.|\overrightarrow {AC} |}} = \frac{{\left| {\frac{{ - 15}}{2}} \right|}}{{3.5}} = \frac{1}{2} \to = {60^o}\)
c. \(\begin{array}{l}
B{C^2} - 2\overrightarrow {BC} .\overrightarrow {AC} + A{C^2} = {(\overrightarrow {BC} - \overrightarrow {AC} )^2} = A{B^2}\\
\to \overrightarrow {BC} .\overrightarrow {CA} = \frac{{A{B^2} - B{C^2} - A{C^2}}}{2} = \frac{{ - 65}}{2}
\end{array}\)
\(\overrightarrow {CB} .\overrightarrow {CA} = \frac{{65}}{2}\)
