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Câu hỏi: Trong không gian $Oxyz$, mặt phẳng $\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-1}=1$ có một véc tơ pháp tuyến là
A. $\overrightarrow{n}=\left( \dfrac{1}{2};\dfrac{1}{3};-1 \right)$.
B. $\overrightarrow{n}=\left( -1;\dfrac{1}{3};1 \right)$.
C. $\overrightarrow{n}=\left( 3;2;-1 \right)$.
D. $\overrightarrow{n}=\left( 3;2;3 \right)$.
Mặt phẳng $\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-1}=1\Leftrightarrow \dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-1}-1=0\Rightarrow \overrightarrow{n}=\left( \dfrac{1}{2};\dfrac{1}{3};-1 \right)$.
A. $\overrightarrow{n}=\left( \dfrac{1}{2};\dfrac{1}{3};-1 \right)$.
B. $\overrightarrow{n}=\left( -1;\dfrac{1}{3};1 \right)$.
C. $\overrightarrow{n}=\left( 3;2;-1 \right)$.
D. $\overrightarrow{n}=\left( 3;2;3 \right)$.
Mặt phẳng $\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-1}=1\Leftrightarrow \dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{-1}-1=0\Rightarrow \overrightarrow{n}=\left( \dfrac{1}{2};\dfrac{1}{3};-1 \right)$.
Đáp án A.