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Câu hỏi: Trong không gian $Oxyz$, cho hai vectơ $\overrightarrow{a}=\left( 2;1;0 \right)$ và $\overrightarrow{b}=\left( -1;0;-2 \right)$. Tính $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)$.
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: $\left| {\vec{a}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}}=\sqrt{5};\ \left| {\vec{b}} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{5}$.
Vậy $\cos \left( \vec{a};\vec{b} \right)=\dfrac{\vec{a}.\vec{b}}{\left| {\vec{a}} \right|.\left| {\vec{b}} \right|}=-\dfrac{2}{5}$
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: $\left| {\vec{a}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}}=\sqrt{5};\ \left| {\vec{b}} \right|=\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{5}$.
Vậy $\cos \left( \vec{a};\vec{b} \right)=\dfrac{\vec{a}.\vec{b}}{\left| {\vec{a}} \right|.\left| {\vec{b}} \right|}=-\dfrac{2}{5}$
Đáp án B.