Câu hỏi: Trong không gian Oxyz, cho 3 điểm A(-1; 2; 2), B(3; -l; -2), C(-4; 0; 3). Gọi I(a; b; c) nằm trên mặt phẳng (Oxz) sao cho biểu thức $\left| \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC} \right|$ đạt giá trị nhỏ nhất. Khi đó a + b + c bằng
A. 2.
B. -2.
C. -17.
D. 17.
A. 2.
B. -2.
C. -17.
D. 17.
Cách 1
Gọi $I\left( a;b;c \right)\in \left( Oxz \right)\Leftrightarrow b=0.$
Ta có $\left\{ \begin{aligned}
& \overrightarrow{IA}=\left( -1-a;2;2-c \right) \\
& \overrightarrow{IB}=\left( 3-a;-1;-2-c \right) \\
& \overrightarrow{IC}=\left( -4-a;0;3-c \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC}=\left( -2a-19;4;-2c+15 \right)$
$P=\left| \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC} \right|=\sqrt{{{\left( 2a+19 \right)}^{2}}+{{\left( 2c-15 \right)}^{2}}+16}\ge 4.$
$MinP=4\Leftrightarrow \left\{ \begin{aligned}
& 2a+19=0 \\
& 2c-15=0 \\
\end{aligned} \right.\Leftrightarrow a=-\dfrac{19}{2};b=0;c=\dfrac{15}{2}\Rightarrow a+b+c=-2$
Cách 2
Gọi M(x; y; z) sao cho $\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}=\overrightarrow{0}\left( 1 \right).$
Ta có $\left\{ \begin{aligned}
& \overrightarrow{MA}=\left( -1-x;2-y;2-z \right) \\
& \overrightarrow{MB}=\left( 3-x;-1-y;-2-z \right) \\
& \overrightarrow{MC}=\left( -4-x;-y;3-z \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}=\left( -2x-19;4-2y;-2z+15 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{aligned}
& -2x-19=0 \\
& 4-2y=0 \\
& -2z+15=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-\dfrac{19}{2} \\
& y=2 \\
& z=\dfrac{15}{2} \\
\end{aligned} \right.\Rightarrow M\left( -\dfrac{19}{2};2;\dfrac{15}{2} \right)$
$\overrightarrow{u}=\overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC}=\left( \overrightarrow{IM}+\overrightarrow{MA} \right)-2\left( \overrightarrow{IM}+\overrightarrow{MB} \right)+3\left( \overrightarrow{IM}+\overrightarrow{MC} \right)$
$=2\overrightarrow{IM}+\left( \underbrace{\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}}_{=\overrightarrow{0}} \right)$
$\Rightarrow \overrightarrow{u}=2\overrightarrow{IM}$
$P=\left| \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC} \right|=\left| \overrightarrow{u} \right|=2\left| \overrightarrow{IM} \right|=2IM$.
$P\min \Leftrightarrow IM\min \Leftrightarrow $ I là hình chiếu của M lên mặt phẳng (Oxz) $\Rightarrow I\left( -\dfrac{19}{2};0;\dfrac{15}{2} \right)$
$\Rightarrow \left\{ \begin{aligned}
& a=-\dfrac{19}{2} \\
& c=\dfrac{15}{2} \\
& b=0 \\
\end{aligned} \right.\Rightarrow a+b+c=-2$
Gọi $I\left( a;b;c \right)\in \left( Oxz \right)\Leftrightarrow b=0.$
Ta có $\left\{ \begin{aligned}
& \overrightarrow{IA}=\left( -1-a;2;2-c \right) \\
& \overrightarrow{IB}=\left( 3-a;-1;-2-c \right) \\
& \overrightarrow{IC}=\left( -4-a;0;3-c \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC}=\left( -2a-19;4;-2c+15 \right)$
$P=\left| \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC} \right|=\sqrt{{{\left( 2a+19 \right)}^{2}}+{{\left( 2c-15 \right)}^{2}}+16}\ge 4.$
$MinP=4\Leftrightarrow \left\{ \begin{aligned}
& 2a+19=0 \\
& 2c-15=0 \\
\end{aligned} \right.\Leftrightarrow a=-\dfrac{19}{2};b=0;c=\dfrac{15}{2}\Rightarrow a+b+c=-2$
Cách 2
Gọi M(x; y; z) sao cho $\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}=\overrightarrow{0}\left( 1 \right).$
Ta có $\left\{ \begin{aligned}
& \overrightarrow{MA}=\left( -1-x;2-y;2-z \right) \\
& \overrightarrow{MB}=\left( 3-x;-1-y;-2-z \right) \\
& \overrightarrow{MC}=\left( -4-x;-y;3-z \right) \\
\end{aligned} \right.\Rightarrow \overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}=\left( -2x-19;4-2y;-2z+15 \right)$
$\left( 1 \right)\Leftrightarrow \left\{ \begin{aligned}
& -2x-19=0 \\
& 4-2y=0 \\
& -2z+15=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x=-\dfrac{19}{2} \\
& y=2 \\
& z=\dfrac{15}{2} \\
\end{aligned} \right.\Rightarrow M\left( -\dfrac{19}{2};2;\dfrac{15}{2} \right)$
$\overrightarrow{u}=\overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC}=\left( \overrightarrow{IM}+\overrightarrow{MA} \right)-2\left( \overrightarrow{IM}+\overrightarrow{MB} \right)+3\left( \overrightarrow{IM}+\overrightarrow{MC} \right)$
$=2\overrightarrow{IM}+\left( \underbrace{\overrightarrow{MA}-2\overrightarrow{MB}+3\overrightarrow{MC}}_{=\overrightarrow{0}} \right)$
$\Rightarrow \overrightarrow{u}=2\overrightarrow{IM}$
$P=\left| \overrightarrow{IA}-2\overrightarrow{IB}+3\overrightarrow{IC} \right|=\left| \overrightarrow{u} \right|=2\left| \overrightarrow{IM} \right|=2IM$.
$P\min \Leftrightarrow IM\min \Leftrightarrow $ I là hình chiếu của M lên mặt phẳng (Oxz) $\Rightarrow I\left( -\dfrac{19}{2};0;\dfrac{15}{2} \right)$
$\Rightarrow \left\{ \begin{aligned}
& a=-\dfrac{19}{2} \\
& c=\dfrac{15}{2} \\
& b=0 \\
\end{aligned} \right.\Rightarrow a+b+c=-2$
Đáp án B.