Question

Cho tam giác ABC có AB = 5cm, BC = 7cm, CA = 8cm

a) Tính \(\overrightarrow{AB}.\overrightarrow{AC}\) rồi suy ra giá trị của góc A ?

b) Tính \(\overrightarrow{CA}.\overrightarrow{CB}\) ?

Answer 1

a) Có
\(\overrightarrow{BC}^2=\left(\overrightarrow{BA}+\overrightarrow{AC}\right)^2=\overrightarrow{BA}^2+\overrightarrow{AC}^2+2\overrightarrow{BA}.\overrightarrow{AC}\)
\(=\overrightarrow{BA}^2+\overrightarrow{AC}^2-2\overrightarrow{AB}.\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=\dfrac{\overrightarrow{BA}^2+\overrightarrow{AC}^2-\overrightarrow{BC^2}}{2}=\dfrac{5^2+8^2-7^2}{2}=20\).
\(cos\widehat{BAC}=\dfrac{\overrightarrow{AB}.\overrightarrow{AC}}{\left|\overrightarrow{AB}\right|.\left|\overrightarrow{AC}\right|}=\dfrac{20}{5.8}=\dfrac{1}{2}\).
Vì vậy \(\widehat{BAC}=60^o\).
b) Tương tự:
\(\overrightarrow{CA}.\overrightarrow{CB}=\dfrac{CA^2+CB^2-AB^2}{2}=\dfrac{7^2+8^2-5^2}{2}=44\).