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Câu hỏi: Trong không gian Oxyz, cho mặt phẳng $\left( P \right):x+y+z-2=0$ và đường thẳng $d:\dfrac{x}{1}=\dfrac{y-1}{-1}=\dfrac{z-1}{1}$. Mặt phẳng $\left( Q \right):ax+by+cz=3$ chứa d và tạo với $\left( P \right)$ một góc nhỏ nhất. Tính $a+b+c$.
A. 2.
B. 3.
C. 4.
D. 5.
A. 2.
B. 3.
C. 4.
D. 5.
Ta có $A\left( 0;1;1 \right)\in d,B\left( 1;0;2 \right)\in d$.
Mà $d\subset \left( Q \right)\Rightarrow \left\{ \begin{aligned}
& A\in \left( Q \right) \\
& B\in \left( Q \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& b+c-3=0 \\
& a+2c-3=0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& b=3-c \\
& a=3-2c \\
\end{aligned} \right.$.
Như vậy $\begin{aligned}
& \left\{ \begin{aligned}
& \overrightarrow{{{n}_{Q}}}=\left( 3-2c;3-c;c \right) \\
& \overrightarrow{{{n}_{P}}}=\left( 1;1;1 \right) \\
\end{aligned} \right.\Rightarrow \cos \left( \left( Q \right);\left( P \right) \right)=\dfrac{\left| \left( 3-2c \right)+\left( 3-c \right)+c \right|}{\sqrt{{{\left( 3-2c \right)}^{2}}+{{\left( 3-c \right)}^{2}}+{{c}^{2}}}.\sqrt{3}} \\
& =\dfrac{\left| 6-2c \right|}{\sqrt{6{{c}^{2}}-18c+18}.\sqrt{3}}=\dfrac{2}{\sqrt{18}}.\dfrac{\left| c-3 \right|}{\sqrt{{{c}^{2}}-3c+3}} \\
\end{aligned}$
Góc giữa $\left( Q \right)$ và $\left( P \right)$ nhỏ nhất khi $\cos \left( \left( Q \right);\left( P \right) \right)$ lớn nhất.
Xét $f\left( c \right)=\dfrac{{{\left( c-3 \right)}^{2}}}{{{c}^{2}}-3c+3}\Rightarrow f'\left( c \right)=\dfrac{2\left( c-3 \right)\left( {{c}^{2}}-3c+3 \right)-{{\left( c-3 \right)}^{2}}\left( 2c-3 \right)}{{{\left( {{c}^{2}}-3c+3 \right)}^{2}}}=0$.
$\begin{aligned}
& \Rightarrow 2\left( {{c}^{2}}-3c+3 \right)=\left( c-3 \right)\left( 2c-3 \right)=2{{c}^{2}}-9c+9\Rightarrow 3c=3\Rightarrow c=1 \\
& f\left( c \right)\le f\left( 1 \right)=4\Rightarrow \cos \left( \left( Q \right);\left( P \right) \right)\le \dfrac{2}{\sqrt{18}}.\sqrt{4}=\dfrac{2\sqrt{2}}{3} \\
\end{aligned}$
Dấu "=" xảy ra $\Leftrightarrow c=1\Rightarrow \overrightarrow{{{n}_{Q}}}=\left( 1;2;1 \right)$.
$\Rightarrow \left( Q \right):x+2\left( y-1 \right)+\left( z-1 \right)=0\Leftrightarrow x+2y+z-3=0$.
Mà $d\subset \left( Q \right)\Rightarrow \left\{ \begin{aligned}
& A\in \left( Q \right) \\
& B\in \left( Q \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& b+c-3=0 \\
& a+2c-3=0 \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& b=3-c \\
& a=3-2c \\
\end{aligned} \right.$.
Như vậy $\begin{aligned}
& \left\{ \begin{aligned}
& \overrightarrow{{{n}_{Q}}}=\left( 3-2c;3-c;c \right) \\
& \overrightarrow{{{n}_{P}}}=\left( 1;1;1 \right) \\
\end{aligned} \right.\Rightarrow \cos \left( \left( Q \right);\left( P \right) \right)=\dfrac{\left| \left( 3-2c \right)+\left( 3-c \right)+c \right|}{\sqrt{{{\left( 3-2c \right)}^{2}}+{{\left( 3-c \right)}^{2}}+{{c}^{2}}}.\sqrt{3}} \\
& =\dfrac{\left| 6-2c \right|}{\sqrt{6{{c}^{2}}-18c+18}.\sqrt{3}}=\dfrac{2}{\sqrt{18}}.\dfrac{\left| c-3 \right|}{\sqrt{{{c}^{2}}-3c+3}} \\
\end{aligned}$
Góc giữa $\left( Q \right)$ và $\left( P \right)$ nhỏ nhất khi $\cos \left( \left( Q \right);\left( P \right) \right)$ lớn nhất.
Xét $f\left( c \right)=\dfrac{{{\left( c-3 \right)}^{2}}}{{{c}^{2}}-3c+3}\Rightarrow f'\left( c \right)=\dfrac{2\left( c-3 \right)\left( {{c}^{2}}-3c+3 \right)-{{\left( c-3 \right)}^{2}}\left( 2c-3 \right)}{{{\left( {{c}^{2}}-3c+3 \right)}^{2}}}=0$.
$\begin{aligned}
& \Rightarrow 2\left( {{c}^{2}}-3c+3 \right)=\left( c-3 \right)\left( 2c-3 \right)=2{{c}^{2}}-9c+9\Rightarrow 3c=3\Rightarrow c=1 \\
& f\left( c \right)\le f\left( 1 \right)=4\Rightarrow \cos \left( \left( Q \right);\left( P \right) \right)\le \dfrac{2}{\sqrt{18}}.\sqrt{4}=\dfrac{2\sqrt{2}}{3} \\
\end{aligned}$
Dấu "=" xảy ra $\Leftrightarrow c=1\Rightarrow \overrightarrow{{{n}_{Q}}}=\left( 1;2;1 \right)$.
$\Rightarrow \left( Q \right):x+2\left( y-1 \right)+\left( z-1 \right)=0\Leftrightarrow x+2y+z-3=0$.
Đáp án C.