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Question

Lập ( Q ) qua M ( -4;-9;12) cắt ox, oy, oz tại A , B, C thỏa mãn
$\left\{ \begin{array}{l} OC = OA + OB\\ \frac{ 4}{OC} = \frac{1}{OA} + \frac{1}{OB}\end{array} \right. $

score 2 · Answer 1

Bình chọn tăng 2 Bình chọn giảm

Giả sử $(Q)$ là mặt phẳng thỏa mãn bài toán. Khi đó:
$\frac{4}{OC}=\frac{1}{OA}+\frac{1}{OB}\geq \frac{4}{OA+OB}=\frac{4}{OC}$.
Do đó $OA=OB=\frac{OC}{2}$.
Giả sử $A(a,0,0),B(0,a,0),C(0,0,-2a)$ thì phương trình của $(Q)$ có dạng $\frac{x}{a}+\frac{y}{a}+\frac{z}{-2a}=1.$
Vì $M(-4,-9,12)\in (Q)$ nên $\frac{-4}{a}+\frac{-9}{a}+\frac{12}{-2a}=1$, suy ra $a=-19$.
Vậy $(Q):2x+2y-z+38=0$

score 6 · Answer 2

Giả sử $A(a,0,0),B(0,b,0),C(0,0,c)$.
Phương trình theo đoạn chắn của $(ABC)$ là: $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Vì $\left\{ \begin{array}{l} M(-4;-9;12)\in(ABC)\\OC=OA+OB\\\frac{4}{OC}=\frac{1}{OA}+\frac{1}{OB} \end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} \frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\\|c|=|a|+|b| \\\frac{4}{|c|}=\frac{1}{|a|}+\frac{1}{|b|}\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} \frac{4}{|a|+|b|}=\frac{|a|+|b|}{|ab|}\\|c|=|a|+|b| \\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} (|a|+|b|)^2=4|ab|\\|c|=|a|+|b| \\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.$
$\Leftrightarrow \left\{ \begin{array}{l} |a|=|b|\\|c|=|a|+|b| \\\frac{-4}{a}+\frac{-9}{b}+\frac{12}{c}=1\end{array} \right.\Leftrightarrow \left[ \begin{array}{l} a=-7;b=-7;c=-14\\a=-19;b=-19;c=38\\a=-1;b=1;c=2\\a=11;b=-11;c=22 \end{array} \right.$
Khi đó, phương trình của $(ABC)$ là: $\left[ \begin{array}{l} \frac{x}{-7}+\frac{y}{-7}+\frac{z}{-14}=1\Leftrightarrow 2x+2y+z+14=0\\\frac{x}{-19}+\frac{y}{-19}+\frac{z}{38}=1\Leftrightarrow 2x+2y-z+38=0\\\frac{x}{-1}+\frac{y}{1}+\frac{z}{2}=1\Leftrightarrow 2x-2y-z+2=0\\\frac{x}{11}+\frac{y}{-11}+\frac{z}{22}=1\Leftrightarrow 2x-2y+z-22=0 \end{array} \right.$