How do I convert the equations $$\left\{\begin{array}{} 3a +2b+ c & = & 0\\ a+4b+4c & = & 0 \end{array}\right.$$ to the form $$\frac{a}{4}=\frac{b}{-11}=\frac{c}{10}=k.$$ Thank you
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$\begingroup$One way of doing this is to relate $\frac a4$ to $b$ and $c$, separately.
First, multiplying the first equation by $2$ gives
$2\times (3a+2b+c)=2\times0 \Leftrightarrow 6a+4b+2c=0$.
Subtracting the second equation from this result gives
$5a-2c=0$
$5a=2c$
$a=\frac{2}{5}c$
$\frac{a}{4}=\frac{c}{10}$.
Similarly, to relate $\frac{a}{4}$ to $b$, you can start by multiplying the second equation by $3$, then subtracting it from the first equation to get
$(3a+2b+c)-(3a+12b+12c)=0$
$-10b-11c=0$
$-11c=10b$
$c=\frac{10b}{-11}$.
Dividing everything by $10$ gives
$\frac{c}{10}=\frac{10b}{-110}=\frac{b}{-11}$.
But we know, $\frac{a}{4}=\frac{c}{10}$, so $\frac{a}{4}=\frac{-b}{11}$.
This simplifies to
$\frac{a}{4}=\frac{b}{-11}=\frac{c}{10}$,
as required.
You can google "linear simultaneous equations" for more details on how to solve this kind of questions.
$\endgroup$ 3 $\begingroup$Solve for $a%$ and $b$ in terms of $c$
Then
$$3a+2b=-c$$ $$a+4b = -4c$$
Multiplying the first by $-2$ and adding,
$$-5a = -2c \implies \frac{a}{2}=\frac{c}{5}$$
and $$-10b = 11c \implies \frac{b}{-11} = \frac{c}{10} $$
$$\implies \frac{a}{4}=\frac{b}{-11}=\frac{c}{10}$$
$\endgroup$ 1 $\begingroup$Each individual equation in this system describes a plane and the system describes the line which is their intersection. This line is orthogonal to the normals of both planes, so we can find its direction by computing the cross product $$(3,2,1)\times(1,4,4)=(4,-11,10),$$ i.e., the line consists of all points $(a,b,c)=k\,(4,-11,10)=(4k,-11k,10k)$. The symmetric form is obtained by isolating $k$: $$\frac a4=\frac b{-11}=\frac c{10}=k.$$
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