Intersection of 2 planes?

Question

The question asks to find the parametric equations of the line of intersection between the planes

$3x+2y-z=28$

$x-4y+2z=0$

I think I know how to do it and I think I got the right answer, but I don't really intuitively understand it:

I multiplied the first equation by $-2$, added the equations together, and the $y's$ and the $z's$ canceled out and was left with $x=8$. Then I substituted $x=8$ into one of the equation, and I think the result ($2y-z=24$) represents the line of intersection. However, I don't REALLY understand what I'm doing. I think the fact that the equations canceled to $x=8$ means that the line of intersection lies on $x=8$, but I don't really know.

Answer 1

Let's write the equations of the planes as

$$z=3x+2y-28 \tag 1$$

and

$$z=-\frac12 x+2y \tag 2$$

If the planes intersect, then obviously the values of $z$ on the line of intersection must be equal. So, setting the right-hand sides of $(1)$ and $(2)$ equal yields

$3x+2y-28=-\frac12 x+2y\implies x=8$

Thus, the two planes intersect at a line on the plane $x=8$. To find the line in the plane $x=8$, we need only look on either plane and see the relationship between $y$ and $z$. To that end, letting $x=8$ in $(1)$ reveals that

$$z=2y-4$$

Therefore, the line of intersection is given parametrically by

$$\vec r(t)=(\hat x8+\hat y2)+(\hat y +\hat 2z)t$$

Answer 2

The line of intersection of the two planes is the solution of the system: $$ \begin{cases} 3x+2y-z=28\\ x-4y+2z=0 \end{cases} $$ solving you find: $x=8$,$y=\frac{t}{2}+2$ and $z=t \quad \forall t \in \mathbb{R}$.

This means that the equation of the line (in vector form) is: $$ \begin{bmatrix} x\\y\\z \end{bmatrix}= \begin{bmatrix} 0\\\frac{1}{2}\\1 \end{bmatrix} t +\begin{bmatrix} 8\\2\\0 \end{bmatrix} $$

Answer 3

I think you may have been confused in this particular example because the equations reduce quickly to $x=8$, and then what...?

A systematic way of finding the line of intersection of two planes is:

1. Write them in Cartesian form
2. Eliminate any one of the variables
3. You now have one linear equation in two variables, for example, $$ ax+by=c$$ Now separate the variables and introduce a parameter, $\lambda$, say, so that you have, for example, $$ax=c-by=\lambda$$
4. Now find each of $x, y$ and $z$ in terms of $\lambda$

Now you have the line of intersection in parametric form.