How to quickly convert to the parametric form of equation of line?

I have a straight line given by: $$\begin{cases} x - y - 4z + 12 = 0 \\ 2x + y -2z + 3 = 0 \end{cases}$$

I'm looking to convert it to the parametric form of straight line like this:

$$\vec{r} = p_1 + t\vec{v}$$

The way I currently do it is:

1. Eliminate one variable - in this case $-3x-3y+6=0 \implies x+y=2$
2. Let one variable be $t$ - in this case $y=t$
3. Write $t$ in terms of other variables - in this case it's $t=2-x$ and $t=\frac{14-4z}{2}$
4. Write it all in one line: $\frac{x-2}{-1} = y = \frac{z-\frac{14}{4}}{- \frac{1}{2}}$
5. Write the parametric form: $\vec{r} = (2, 0, \frac{14}{4}) + t (-1, 1, -\frac{1}{2})$

First, is my method correct? Secondly, this is a rather long method and produces ugly parametric form, can I do this quicker and better?

• Place in row-reduced eschalon form and it is simple. Jan 16, 2018 at 2:13

The equations can be written as $$\begin{bmatrix} 1&-1\\ 2&1 \end{bmatrix} \begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} 4z-12\\ 2z-3 \end{bmatrix}$$ Invert the matrix to get \begin{align} \begin{bmatrix} x\\ y \end{bmatrix} &=\frac13\begin{bmatrix} 1&1\\ -2&1 \end{bmatrix} \begin{bmatrix} 4z-12\\ 2z-3 \end{bmatrix}\\ &=\begin{bmatrix} 2z-5\\ -2z+7 \end{bmatrix} \end{align} Thus, a parametric form is $$\begin{bmatrix} x\\ y\\ z \end{bmatrix} =\begin{bmatrix} 2\\ -2\\ 1 \end{bmatrix}t +\begin{bmatrix} -5\\ 7\\ 0 \end{bmatrix}$$

• nice and general. Where do i learn stuff like this ? Nov 17, 2022 at 10:20
• I would go to a text on Linear Algebra
– robjohn
Nov 17, 2022 at 11:39

You do not need step 4. you can directly settle for

$\begin{cases} x=2-t\\ y=t\\ z=\dfrac{7-t}2 \end{cases}$

$M=(2,0,\frac 72)+t(-1,1,-\frac 12)$

Anyway, you can also select two values of $t$ that give "nice" points $A$ and $B$ if you are not satisfied with the fractionnal coefficients.

For instance $t=1$ gives $A=(1,1,3)$ and $t=-1$ gives $B=(3,-1,4)$

Calculate the direction vector $\vec u=\vec{AB}=(2,-2,1)$

And you get your line : $M=A+t\vec u=(1+2t,1-2t,3+t)$

In general form, the way you have expressed the two planes, the normal to each plane is given by the variable coefficients. If you just take the cross product of those two norms, you get the direction vector for the line. So, $$(1,-1,-4)\otimes (2,1,-2)=(6,-6,3)$$ Now you need a point that is on the line. To get that, pick one of the variables and set it to zero, then solve for the other two. That's pretty much what you did. If I let z=0, then $A=(-5,7,0)$

So the line is $$(x,y,z)=(-5,7,0)+t(6,-6,3)$$