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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, cho ba điểm $A\left( a;0;0 \right)$, $B\left( 0;b;0 \right)$, $C\left( 0;0;c \right)$ với $a,b,c$ là các số thực dương thay đổi tùy ý sao cho ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3$. Khoảng cách từ O đến mặt phẳng $\left( ABC \right)$ lớn nhất bằng
A. $\dfrac{1}{3}$.
B. 3.
C. $\dfrac{1}{\sqrt{3}}$.
D. 1.
A. $\dfrac{1}{3}$.
B. 3.
C. $\dfrac{1}{\sqrt{3}}$.
D. 1.
Phương trình mặt phẳng $\left( ABC \right):\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{c}{z}=1\left( abc\ne 1 \right)$.
Khi đó: $d\left( O;\left( ABC \right) \right)=\dfrac{\left| \dfrac{0}{a}+\dfrac{0}{b}+\dfrac{0}{c}-1 \right|}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}$
Ta có: $\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}\ge \dfrac{9}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\dfrac{9}{3}=3\Rightarrow \dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}\le \dfrac{1}{\sqrt{3}}$
Hay $d\left( O;\left( ABC \right) \right)\le \dfrac{1}{\sqrt{3}}$.
Dấu "=" xảy ra khi $\left\{ \begin{aligned}
& a=b=c>0 \\
& {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3 \\
\end{aligned} \right.\Leftrightarrow a=b=c=1$.
Vậy $d{{\left( O;\left( ABC \right) \right)}_{\max }}=\dfrac{1}{\sqrt{3}}$ khi $a=b=c=1$.
Khi đó: $d\left( O;\left( ABC \right) \right)=\dfrac{\left| \dfrac{0}{a}+\dfrac{0}{b}+\dfrac{0}{c}-1 \right|}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}$
Ta có: $\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}\ge \dfrac{9}{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}=\dfrac{9}{3}=3\Rightarrow \dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}\le \dfrac{1}{\sqrt{3}}$
Hay $d\left( O;\left( ABC \right) \right)\le \dfrac{1}{\sqrt{3}}$.
Dấu "=" xảy ra khi $\left\{ \begin{aligned}
& a=b=c>0 \\
& {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=3 \\
\end{aligned} \right.\Leftrightarrow a=b=c=1$.
Vậy $d{{\left( O;\left( ABC \right) \right)}_{\max }}=\dfrac{1}{\sqrt{3}}$ khi $a=b=c=1$.
Đáp án D.