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Câu hỏi: . Trong không gian với hệ tọa độ Oxyz, cho hai vectơ $\vec{a}=\left( -2;-3;1 \right),\vec{b}=\left( 1;0;1 \right).$ Tính $\cos \left( \vec{a},\vec{b} \right).$
A. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{-1}{2\sqrt{7}}.$
B. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{1}{2\sqrt{7}}.$
C. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{-3}{2\sqrt{7}}.$
D. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{3}{2\sqrt{7}}.$
A. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{-1}{2\sqrt{7}}.$
B. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{1}{2\sqrt{7}}.$
C. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{-3}{2\sqrt{7}}.$
D. $\cos \left( \vec{a},\vec{b} \right)=\dfrac{3}{2\sqrt{7}}.$
Ta có: $\cos \left( \overrightarrow{a},\overrightarrow{b} \right)=\dfrac{\overrightarrow{a}.\overrightarrow{b}}{\left| \overrightarrow{a} \right|.\left| \overrightarrow{b} \right|}=\dfrac{-2.1+\left( -3 \right).0+1.1}{\sqrt{{{\left( -2 \right)}^{2}}+{{\left( -3 \right)}^{2}}+{{1}^{2}}}.\sqrt{{{1}^{2}}+{{0}^{2}}+{{1}^{2}}}}=\dfrac{-1}{2\sqrt{7}}$.
Đáp án A.