Cho các véctơ \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\) thỏa mãn \(\left|\overrightarrow{a}\right|=x,\left|\overrightarrow{b}\right|=y,\left|\overrightarrow{c}\right|=z\) và \(\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}\) Tính \(A=\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}\)
Ta có:
\(\overrightarrow{a}+\overrightarrow{b}+3\overrightarrow{c}=\overrightarrow{0}\Leftrightarrow\overrightarrow{a}+\overrightarrow{b}=-3\overrightarrow{c}\Leftrightarrow\left(\overrightarrow{a}+\overrightarrow{b}\right)^2=9\overrightarrow{c}^2\)
<=> \(\overrightarrow{a}^2+\overrightarrow{b}^2+2\overrightarrow{a}\overrightarrow{b}=9\overrightarrow{c}^2\)
<=> \(\overrightarrow{a}\overrightarrow{b}=\dfrac{9z^2-x^2-y^2}{2}\)
Tương tự ta có: \(\overrightarrow{b}+3\overrightarrow{c}=-\overrightarrow{a}\) <=> \(\left(\overrightarrow{b}+3\overrightarrow{c}\right)^2=\overrightarrow{a}^2\)
<=> \(\overrightarrow{b}.\overrightarrow{c}=\dfrac{x^2-y^2-9z^2}{2}\)
Và lại có : \(\overrightarrow{a}\overrightarrow{c}=\dfrac{y^2-x^2-9z^2}{2}\)
Suy ra: A=\(\dfrac{9z^2-x^2-y^2}{2}+\dfrac{x^2-y^2-9z^2}{2}+\dfrac{y^2-x^2-9z^2}{2}=\dfrac{3z^2-z^2-y^2}{2}\)