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Câu hỏi: Trong không gian với hệ tọa độ $Oxyz$, hãy tính góc giữa hai vectơ $\overrightarrow{a}\left( 1;2;-2 \right)$ và $\overrightarrow{b}\left( -1;-1;0 \right)$.
A. $\left( \overrightarrow{a},\overrightarrow{b} \right)=120{}^\circ .$
B. $\left( \overrightarrow{a},\overrightarrow{b} \right)=45{}^\circ .$
C. $\left( \overrightarrow{a},\overrightarrow{b} \right)=60{}^\circ .$
D. $\left( \overrightarrow{a},\overrightarrow{b} \right)=135{}^\circ .$
A. $\left( \overrightarrow{a},\overrightarrow{b} \right)=120{}^\circ .$
B. $\left( \overrightarrow{a},\overrightarrow{b} \right)=45{}^\circ .$
C. $\left( \overrightarrow{a},\overrightarrow{b} \right)=60{}^\circ .$
D. $\left( \overrightarrow{a},\overrightarrow{b} \right)=135{}^\circ .$
Gọi $\alpha $ là góc giữa hai vectơ.
Ta có: $\cos \alpha =\dfrac{1.\left( -1 \right)+2\left( -1 \right)+\left( -2 \right).0}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -2 \right)}^{2}}}.\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{0}^{2}}}}=\dfrac{-1}{\sqrt{2}}\Rightarrow \alpha =-135{}^\circ .$
Ta có: $\cos \alpha =\dfrac{1.\left( -1 \right)+2\left( -1 \right)+\left( -2 \right).0}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{\left( -2 \right)}^{2}}}.\sqrt{{{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}+{{0}^{2}}}}=\dfrac{-1}{\sqrt{2}}\Rightarrow \alpha =-135{}^\circ .$
Đáp án D.