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?
Best Answer
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.