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Câu hỏi: Trong không gian Oxyz, tìm tọa độ của véc tơ $\overrightarrow{u}=-6\overrightarrow{i}+4\overrightarrow{k}+8\overrightarrow{j}.$
A. $\overrightarrow{u}=\left( -3;2;4 \right)$
B. $\overrightarrow{u}=\left( -3;4;2 \right)$
C. $\overrightarrow{u}=\left( -6;4;8 \right)$
D. $\overrightarrow{u}=\left( -6;8;4 \right)$
A. $\overrightarrow{u}=\left( -3;2;4 \right)$
B. $\overrightarrow{u}=\left( -3;4;2 \right)$
C. $\overrightarrow{u}=\left( -6;4;8 \right)$
D. $\overrightarrow{u}=\left( -6;8;4 \right)$
Phương pháp:
Cho vecto $\overrightarrow{u}=a\overrightarrow{i}+b.\overrightarrow{j}+c\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( a;b;c \right).$
Cách giải:
Ta có: $\overrightarrow{u}=-6\overrightarrow{i}+4\overrightarrow{k}+8\overrightarrow{j}=-6\overrightarrow{i}+8\overrightarrow{j}+4\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( -6;8;4 \right).$
Cho vecto $\overrightarrow{u}=a\overrightarrow{i}+b.\overrightarrow{j}+c\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( a;b;c \right).$
Cách giải:
Ta có: $\overrightarrow{u}=-6\overrightarrow{i}+4\overrightarrow{k}+8\overrightarrow{j}=-6\overrightarrow{i}+8\overrightarrow{j}+4\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( -6;8;4 \right).$
Đáp án D.