The line of intersection of two planes

Question

Let's assume we want to find the line of intersection of two planes.

$x + y - z = 7$ and $2x + 3y - 4z = 2$

The first method is the method I've learnt from the textbook:

Find the normal vector of the two normal vectors of the planes:

$(1, 1, -1) \times (2, 3, -4) = (-1, 2, 1)$

then set $x = 0 $in both equations to find a point of intersection

This gives us $y - z = 7$ and $3y - 4z = 2$

Which gives $y = 26$ and $z = 19$

So a point of intersection is $(0, 26, 9)$

And so the line of intersection is $(0, 26, 19) + s(-1, 2, 1)$

However, my tutor who is an undergraduate said this method would only be valid if the planes do not meet on a line parallel to the $x = 0$ plane.

Is he correct? Is the method invalid in this case?

Answer 1

Yes, your tutor is correct.

The reason is that if the planes meet in a line parallel to the plane $x=0$, then their intersection contains no points of the form $(0,y,z)$, i.e. points where the $x$-coordinate is $0$.

Points where the $x$-coordinate is $0$ are exactly the ones in the plane $x=0$.

So in case the line of intersection is parallel to the plane $x=0$, you can choose instead to put $y=0$ in both equations and continue in the same way.

Answer 2

We can write the first plane as $x=-y+z+7$ and substitute in the second $2(-y+z+7)+3y-4z=2 \Leftrightarrow -2y+2z+14+3y-4z=2 \Leftrightarrow y-2z=-12 \Leftrightarrow y=2z-12 $. Then if we set $z=s$ we get $y=2s-12$ and $x=-2s+12+s+7=-s+19$
Then your line is given by $(x,y,z)=(19,-12,0)+s(-1,2,1)$

Answer 3

I like the following way.

Let $z=t$.

Thus, $$x+y=7+t$$ and $$2x+3y=2+4t,$$ which gives $$x=19-t,$$ $$y=-12+2t$$ and with $$z=t$$ it gives the answer: $$(19,-12,0)+t(-1,2,1).$$