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Câu hỏi: Trong không gian với hệ trục tọa độ Oxyz, cho hai vectơ $\overrightarrow{a}=\left( 2;1;0 \right)$ và $\overrightarrow{b}=\left( -1;0;-2 \right)$. Khi đó $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)$ bằng
A. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=-\dfrac{2}{25}.$
B. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=-\dfrac{2}{5}.$
C. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{2}{25}.$
D. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{2}{5}.$
A. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=-\dfrac{2}{25}.$
B. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=-\dfrac{2}{5}.$
C. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{2}{25}.$
D. $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{2}{5}.$
Ta có: $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|}=\dfrac{-2}{\sqrt{5}.\sqrt{5}}=-\dfrac{2}{5}.$
Đáp án B.