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Câu hỏi: Cho $\overrightarrow{u}=\left( 1;1;1 \right)$ và $\overrightarrow{v}=\left( 0;1;m \right)$. Để góc giữa hai vectơ $\overrightarrow{u},\overrightarrow{v}$ có số đo bằng ${{45}^{0}}$ thì $m$ bằng
A. $\pm \sqrt{3}$
B. $2+\pm \sqrt{3}$
C. $\sqrt{3}.$
D. $1\pm \sqrt{3}$
A. $\pm \sqrt{3}$
B. $2+\pm \sqrt{3}$
C. $\sqrt{3}.$
D. $1\pm \sqrt{3}$
Vì $\left( \overrightarrow{u},\overrightarrow{v} \right)={{45}^{0}}$ nên $\cos \left( \overrightarrow{u},\overrightarrow{v} \right)=\cos {{45}^{0}}=\dfrac{\sqrt{2}}{2}$
$\Leftrightarrow \dfrac{\left| 1.0+1.1+1.m \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}.\sqrt{{{0}^{2}}+{{1}^{2}}+{{m}^{2}}}}=\dfrac{\sqrt{2}}{2}$
$\Leftrightarrow 2\left| m+1 \right|=\sqrt{6\left( {{m}^{2}}+1 \right)}\Leftrightarrow 2{{m}^{2}}-8m+2=0\Leftrightarrow m=2\pm \sqrt{3}.$
$\Leftrightarrow \dfrac{\left| 1.0+1.1+1.m \right|}{\sqrt{{{1}^{2}}+{{1}^{2}}+{{1}^{2}}}.\sqrt{{{0}^{2}}+{{1}^{2}}+{{m}^{2}}}}=\dfrac{\sqrt{2}}{2}$
$\Leftrightarrow 2\left| m+1 \right|=\sqrt{6\left( {{m}^{2}}+1 \right)}\Leftrightarrow 2{{m}^{2}}-8m+2=0\Leftrightarrow m=2\pm \sqrt{3}.$
Đáp án B.