Cho các vectơ \(\overrightarrow{a}=\left(1;2;1\right);\overrightarrow{b}=\left(2;1;2\right);\overrightarrow{c}=\left(3;2;-1\right)\) . Tính tọa độ vectơ \(\overrightarrow{a}^2\overrightarrow{b}+\overrightarrow{b}^2\overrightarrow{c}+\overrightarrow{c}^2\overrightarrow{a}\).
\(\left(53;-52;-17\right)\).\(\left(53;52;-17\right)\).\(\left(53;52;17\right)\).\(\left(53;-52;17\right)\).Hướng dẫn giải:\(\overrightarrow{a}^2=\overrightarrow{a}.\overrightarrow{a}=1^2+2^2+1^2=6;\) \(\overrightarrow{b}^2=2^2+1^2+2^2=9;\)\(\overrightarrow{c}^2=3^2+2^2+\left(-1\right)^2=14.\)
\(\overrightarrow{a}^2\overrightarrow{b}=6\left(2;1;2\right)=\left(12;6;12\right);\) \(\overrightarrow{b}^2\overrightarrow{c}=9\left(3;2;-1\right)=\left(27;18;-9\right);\) \(\overrightarrow{c}^2\overrightarrow{a}=14\left(1;2;1\right)=\left(14;28;14\right).\)
\(\overrightarrow{a}^2\overrightarrow{b}+\overrightarrow{b}^2\overrightarrow{c}+\overrightarrow{c}^2\overrightarrow{a}=\left(53;52;17\right)\)
