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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, cho hai đường thẳng ${{d}_{1}}:\dfrac{x-1}{1}=\dfrac{y+2}{2}=\dfrac{z}{-1}$ và ${{d}_{2}}:\dfrac{x+2}{2}=\dfrac{y-1}{-1}=\dfrac{z}{2}$. Phương trình mặt phẳng $\left( P \right)$ chứa ${{d}_{1}}$ sao cho góc giữa mặt phẳng $\left( P \right)$ và đường thẳng ${{d}_{2}}$ là lớn nhất là: $ax-y+cz+d=0$. Giá trị của $T=a+c+d$ bằng
A. $T=0$
B. $T=3$
C. $T=-\dfrac{13}{4}$
D. $T=-6$
A. $T=0$
B. $T=3$
C. $T=-\dfrac{13}{4}$
D. $T=-6$
Ta có: ${{d}_{1}}=\dfrac{x-1}{1}=\dfrac{y+2}{2}=\dfrac{z}{-1}\Rightarrow \left\{ \begin{aligned}
& 2\text{x}-y-4=0\left( \alpha \right) \\
& y+2\text{z}+2=0\left( \beta \right) \\
\end{aligned} \right.$
Khi đó ${{d}_{1}}\subset \left( P \right)\Rightarrow \left( P \right):m\left( 2\text{x}-y-4 \right)+n\left( y+2\text{z}+2 \right)=0,{{\text{m}}^{2}}+{{n}^{2}}>0$
$\Rightarrow \overrightarrow{{{n}_{P}}}=\left( 2m;-m+n;2n \right)$ là VTPT của $\left( P \right)$.
Mặt khác, ${{d}_{2}}$ có VTCP là $\overrightarrow{{{u}_{2}}}=\left( 2;-1;2 \right)$.
Xét $\sin \measuredangle \left( {{d}_{2}};(P) \right)=\left| \cos \left( \overrightarrow{{{u}_{{{d}_{2}}}}};\overrightarrow{{{n}_{P}}} \right) \right|=\dfrac{\left| 4m+m-n+4n \right|}{3\sqrt{4{{m}^{2}}+{{\left( n-m \right)}^{2}}+4{{n}^{2}}}}=\dfrac{\left| 5m+3n \right|}{3\sqrt{5{{m}^{2}}-2mn+5{{n}^{2}}}}$
$\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{1}{9}.\dfrac{25{{m}^{2}}+30mn+9{{n}^{2}}}{5{{m}^{2}}-2mn+5{{n}^{2}}}$.
TH1: $n=0\Rightarrow m\ne 0$, ta chọn $m=1\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{5}{9}\Rightarrow \sin \measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{\sqrt{5}}{3}$.
TH2: $n\ne 0$, ta chọn $n=1\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{1}{9}.\dfrac{25{{m}^{2}}+30m+9}{5{{m}^{2}}-2m+5}=f\left( m \right)$.
${f}'\left( m \right)=\dfrac{1}{9}.\dfrac{-200{{m}^{2}}+160m+168}{{{\left( 5{{m}^{2}}-4m+8 \right)}^{2}}}\xrightarrow{f=0}\left[ \begin{aligned}
& m=-\dfrac{3}{5}\Rightarrow f\left( -\dfrac{3}{5} \right)=0 \\
& m=\dfrac{7}{5}\Rightarrow f\left( \dfrac{7}{5} \right)=\dfrac{25}{27} \\
\end{aligned} \right.$.
Lập bảng biến và nhận xét: $\measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Leftrightarrow \sin \measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Leftrightarrow {{\sin }^{2}}\measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Rightarrow m=\dfrac{7}{5}$.
Khi đó $\dfrac{7}{5}\left( 2\text{x}-y-4 \right)+\left( y+2\text{z}+2 \right)=0\Rightarrow 7\text{x}-y+5\text{z}-9=0\Rightarrow a=7,c=5,d=-9\Rightarrow T=a+c+d=3$.
& 2\text{x}-y-4=0\left( \alpha \right) \\
& y+2\text{z}+2=0\left( \beta \right) \\
\end{aligned} \right.$
Khi đó ${{d}_{1}}\subset \left( P \right)\Rightarrow \left( P \right):m\left( 2\text{x}-y-4 \right)+n\left( y+2\text{z}+2 \right)=0,{{\text{m}}^{2}}+{{n}^{2}}>0$
$\Rightarrow \overrightarrow{{{n}_{P}}}=\left( 2m;-m+n;2n \right)$ là VTPT của $\left( P \right)$.
Mặt khác, ${{d}_{2}}$ có VTCP là $\overrightarrow{{{u}_{2}}}=\left( 2;-1;2 \right)$.
Xét $\sin \measuredangle \left( {{d}_{2}};(P) \right)=\left| \cos \left( \overrightarrow{{{u}_{{{d}_{2}}}}};\overrightarrow{{{n}_{P}}} \right) \right|=\dfrac{\left| 4m+m-n+4n \right|}{3\sqrt{4{{m}^{2}}+{{\left( n-m \right)}^{2}}+4{{n}^{2}}}}=\dfrac{\left| 5m+3n \right|}{3\sqrt{5{{m}^{2}}-2mn+5{{n}^{2}}}}$
$\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{1}{9}.\dfrac{25{{m}^{2}}+30mn+9{{n}^{2}}}{5{{m}^{2}}-2mn+5{{n}^{2}}}$.
TH1: $n=0\Rightarrow m\ne 0$, ta chọn $m=1\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{5}{9}\Rightarrow \sin \measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{\sqrt{5}}{3}$.
TH2: $n\ne 0$, ta chọn $n=1\Rightarrow {{\sin }^{2}}\measuredangle \left( {{d}_{2}};(P) \right)=\dfrac{1}{9}.\dfrac{25{{m}^{2}}+30m+9}{5{{m}^{2}}-2m+5}=f\left( m \right)$.
${f}'\left( m \right)=\dfrac{1}{9}.\dfrac{-200{{m}^{2}}+160m+168}{{{\left( 5{{m}^{2}}-4m+8 \right)}^{2}}}\xrightarrow{f=0}\left[ \begin{aligned}
& m=-\dfrac{3}{5}\Rightarrow f\left( -\dfrac{3}{5} \right)=0 \\
& m=\dfrac{7}{5}\Rightarrow f\left( \dfrac{7}{5} \right)=\dfrac{25}{27} \\
\end{aligned} \right.$.
Lập bảng biến và nhận xét: $\measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Leftrightarrow \sin \measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Leftrightarrow {{\sin }^{2}}\measuredangle {{\left( {{d}_{2}};(P) \right)}_{\max }}\Rightarrow m=\dfrac{7}{5}$.
Khi đó $\dfrac{7}{5}\left( 2\text{x}-y-4 \right)+\left( y+2\text{z}+2 \right)=0\Rightarrow 7\text{x}-y+5\text{z}-9=0\Rightarrow a=7,c=5,d=-9\Rightarrow T=a+c+d=3$.
Đáp án B.