After solving a math question, I am left with the following equation. Plotting it on Desmos yields a 3D line, however I am unsure of how I can proceed with turning this equation into a more familiar form.
$13x+3y-4z=5x+12y-5z=11x+3y+8z$
I know that the general Cartesian form of a 3D line is $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$ however I don't know how to approach the algebra for this question. Any help would be appreciated.
We can find two equation from here: $$ 13x + 3y - 4z = 5x + 12y - 5z $$ $$ 5x + 12y - 5z = 11x + 3y + 8z $$
We can simplify each equation:
$$ 13x + 3y - 4z = 5x + 12y - 5z $$
Subtract $ 5x + 12y - 5z $ from both sides: $$ (13x + 3y - 4z) - (5x + 12y - 5z) = 0 $$ $$ 13x - 5x + 3y - 12y - 4z + 5z = 0 $$ $$ 8x - 9y + z = 0 $$
$$ 5x + 12y - 5z = 11x + 3y + 8z $$
Subtract $ 11x + 3y + 8z $ from both sides: $$ (5x + 12y - 5z) - (11x + 3y + 8z) = 0 $$ $$ 5x - 11x + 12y - 3y - 5z - 8z = 0 $$ $$ -6x + 9y - 13z = 0 $$
Now we have two simplified equations:
These equations represent planes in 3D space. To write them in general form of: $$\frac{x-a}{d}=\frac{y-b}{e}=\frac{z-c}{f}$$ Now, we need to find the direction ratios and a point through which the line passes. However, since we have two planes, the intersection of these planes gives us a line in 3D space.
To find the direction ratios ($d$, $e$, $f$) of the line of intersection, we need to take the cross product of the normal vectors of the two planes.
The normal vector of the first plane, ${n_1} $, is $\langle 8, -9, 1 \rangle$. The normal vector of the second plane, ${n_2}$, is $\langle -6, 9, -13 \rangle$.
Calculate the cross product of ${n_1}$ and ${n_2}$:
$$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 8 & -9 & 1 \\ -6 & 9 & -13 \end{vmatrix} $$
$$\mathbf{d} = \mathbf{i} \left((-9)(-13) - (1)(9)\right) - \mathbf{j} \left((8)(-13) - (1)(-6)\right) + \mathbf{k} \left((8)(9) - (-9)(-6)\right) $$
$$\mathbf{d} = \mathbf{i} (117 - 9) - \mathbf{j} (-104 + 6) + \mathbf{k} (72 - 54)$$
$$\mathbf{d} = \mathbf{i} (108) - \mathbf{j} (-98) + \mathbf{k} (18)$$
$$\mathbf{d} = \langle 108, 98, 18 \rangle$$
Thus, the direction ratios are $d = 108$, $e = 98$, and $f = 18$.
To find a point $(a, b, c)$ on the line of intersection, we can solve the system of equations for specific values. Let's set $z = 0$ and solve for $x$ and $y$.
From $8x - 9y + z = 0$: $$8x - 9y = 0 $$ $$8x = 9y $$ $$y = \frac{8}{9}x $$
Substitute $y = \frac{8}{9}x$ into $-6x + 9y - 13z = 0$: $$-6x + 9\left(\frac{8}{9}x\right) = 0 $$ $$ -6x + 8x = 0 $$ $$ 2x = 0 $$ $$ x = 0 $$
So, $y = \frac{8}{9} \cdot 0 = 0$.
Thus, the point $(a, b, c)$ is $(0, 0, 0)$.
Now, we have the point $(a, b, c) = (0, 0, 0)$ and direction ratios $(d, e, f) = (108, 98, 18)$. The symmetric form of the line is:
$$ \frac{x - 0}{108} = \frac{y - 0}{98} = \frac{z - 0}{18} $$
Simplified, it is:
$$ \frac{x}{108} = \frac{y}{98} = \frac{z}{18} $$
$$\begin{align}&\begin{cases}13x+3y-4z&=5x+12y-5z\\5x+12y-5z&=11x+3y+8z\end{cases}\\\iff&\begin{cases}8x-9y+z&=0\\-6x+9y-13z&=0\end{cases} \\\iff&\begin{cases}z&=9y-8x\\0&=(-6+13\cdot8)x+(9-13\cdot9)y\end{cases} \\\iff&\begin{cases}49x&=54y\\z&=\left(9-8\frac{54}{49}\right)y \end{cases} \\\iff&\begin{cases}x&=\frac{54}{49} y\\z&=\frac9{49}y \end{cases}\\\iff&\frac x{54}=\frac y{49}=\frac z9. \end{align}$$
First of all, the line can be expressed in that form only if it is not parallel to any of the coordinate axes. If we assume that it isn't parallel to any of the coordinate axes, then we can take one of the denominators and set it to any non-zero number we want. For instance, we can take $d=1$. Next, we can take $y-b=e(x-a)=ex-ea$, and $z-c=fx-fa$. Looking $13x+3y-4z$, if we substitute in to get everything just in terms of $x$, and then look at just the coefficients of $x$, we get $13(1)+3e-4f$. From the second expression, we get $5+12e-5f$, and from the third we get $11+3e+8f$. Setting these equal, we get
$$13+3e-4f=11+3e+8f$$ $$13-4f=11+8f$$ $$2=12f$$ $$f=\frac 16$$
$$13+3e-4f=5+12e-5f$$ $$9e=f+8=\frac {49}{6}$$ $$e = \frac {49}{54}$$
We got these numbers by taking $d=1$, but we can get integers by taking $d$ to be the least common denominator, which is $54$. So $d=54, e = 49, f=9$ (note that multiplying all three denominators by the same number results in an equivalent set of equations).
We could then take $x,y,$ and $z$ and express two of them in terms of the third (e.g. $y=ex-ea+b, z=fx-fa+c$), substitute back into the equations, and solve for $a,b,c$. If you do this, you'll actually get an infinite number of solutions, and you can just pick one $^{[1]}$. However, $13+3y-4z$ doesn't have any constant terms, which means that it corresponds to a plane that goes through the origin. This is true of the other two expressions as well. This means that $(x, y, z) = (0,0,0)$ must be a solution of the plane equations, and so $(a, b, c)=(0,0,0)$ must be valid values.
[1] Given $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, it would also be valid to have $\dfrac{x-(a+d)}{d}=\dfrac{y-(b+e)}{e}=\dfrac{z-(c+f)}{f}$, since you're just subtracting $1$ from each expression. So there must be an infinite number of values for $(a,b,c)$. These correspond to different "starting points" for the line, and is analogous to the two-dimensional case where if you express a line as $y-y_0=m(x-x_0)$, then $y-(y_0+m)=m(x-(x_0+1))$ describes the same line).
