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Câu hỏi: Trong không gian với hệ tọa độ $Oxyz,$ cho ba vectơ $\overrightarrow{a}=\left( 2;-1;0 \right)$, $\overrightarrow{b}=\left( 1;2;3 \right),$ $\overrightarrow{c}=\left( 4;2;-1 \right)$. Khẳng định nào sau đây đúng?
A. $\overrightarrow{a}.\overrightarrow{c}=5$.
B. $\overrightarrow{c}=2\overrightarrow{a}$.
C. $\left| \overrightarrow{b} \right|=\sqrt{6}$.
D. $\overrightarrow{a}\bot \overrightarrow{b}$.
A. $\overrightarrow{a}.\overrightarrow{c}=5$.
B. $\overrightarrow{c}=2\overrightarrow{a}$.
C. $\left| \overrightarrow{b} \right|=\sqrt{6}$.
D. $\overrightarrow{a}\bot \overrightarrow{b}$.
Ta có: $\overrightarrow{a}.\overrightarrow{c}=2.4+\left( -1 \right).2+0.\left( -1 \right)=6\ne 5.$
$\overrightarrow{c}\ne 2\overrightarrow{a}.$
$\overrightarrow{b}=\left( 1;2;3 \right)\Rightarrow \left| \overrightarrow{b} \right|=\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}=\sqrt{14}\ne \sqrt{6}.$
$\overrightarrow{a}.\overrightarrow{b}=2.1+\left( -1 \right).2+0.3=0\Rightarrow \overrightarrow{a}\bot \overrightarrow{b}.$.
$\overrightarrow{c}\ne 2\overrightarrow{a}.$
$\overrightarrow{b}=\left( 1;2;3 \right)\Rightarrow \left| \overrightarrow{b} \right|=\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}=\sqrt{14}\ne \sqrt{6}.$
$\overrightarrow{a}.\overrightarrow{b}=2.1+\left( -1 \right).2+0.3=0\Rightarrow \overrightarrow{a}\bot \overrightarrow{b}.$.
Đáp án D.