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Câu hỏi: Trong không gian với hệ tọa độ $\left( O;\overrightarrow{i};\overrightarrow{j};\overrightarrow{k} \right)$, cho $\overrightarrow{u}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$. Tính $\left| \overrightarrow{u} \right|$.
A. $\left| \overrightarrow{u} \right|=\sqrt{6}.$
B. $\left| \overrightarrow{u} \right|=2.$
C. $\left| \overrightarrow{u} \right|=4.$
D. $\left| \overrightarrow{u} \right|=\sqrt{5}.$
A. $\left| \overrightarrow{u} \right|=\sqrt{6}.$
B. $\left| \overrightarrow{u} \right|=2.$
C. $\left| \overrightarrow{u} \right|=4.$
D. $\left| \overrightarrow{u} \right|=\sqrt{5}.$
Ta có: $\overrightarrow{u}=2\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( 2;-1;1 \right)\Rightarrow \left| \overrightarrow{u} \right|=\sqrt{{{2}^{2}}+{{\left( -1 \right)}^{2}}+{{1}^{2}}}=\sqrt{6}.$
Lưu ý: $\overrightarrow{u}=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( a;b;c \right)\Rightarrow \left| \overrightarrow{u} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
Lưu ý: $\overrightarrow{u}=a\overrightarrow{i}+b\overrightarrow{j}+c\overrightarrow{k}\Rightarrow \overrightarrow{u}=\left( a;b;c \right)\Rightarrow \left| \overrightarrow{u} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$.
Đáp án A.