Finding the point on the line of intersection between planes

[Bishamonten]

Homework Statement

Given two planes, P1, P2. Find parametric equations for the line of intersection between the two planes.

Homework Equations

P1 = 2x -3y + 4z = 3
P2 = x + 4y - 2z = 7

The Attempt at a Solution

Let N1 be the normal vector to P1, and N2 be the normal vector to P2. Then,

N1 = (2, -3, 4) ; N2 = (1, 4, -2)

Then, the direction numbers for our desired line will come from the components of the cross product between the normal vectors, because it will be parallel to our desired line.

Then, N1xN2 = (-10, 8, 11).

Then, Eq. of line: x = x? -10t ; y = y? +8t ; z = z? + 11t

I currently need to find a point that would be on our desired line, so that I can complete the equation for this line. The book we're using unsatisfactorily just says in it's example version of this type of problem to "let z = 0" in both the given equations of the two planes in it's example, and then solve the system that emerges from that.

My problem is, why? How would I know ahead of time to let one of these three unknowns be 0?(In this case, the back of the book suggests to let y = 0 for my problem here, but I want to know what line of reasoning would let me come to that conclusion).

 

[andrewkirk]

You can solve the problem by setting any one of the coordinates to zero, provided the line is not parallel to but not in the plane where that coordinate is zero. The line will be parallel to such a plane if the corresponding component of the line vector is zero. So you can pick any component for which the corresponding component of the line vector is nonzero. In this case that's any of the three coordinates, because none of the components of the line vector are zero.

What you are doing by setting a coordinate equal to zero is obtaining two equations for the other two coordinates of the point where the line intersects the plane where that coordinate is zero. Those two equations will be linear in the other two coordinates, and hence can be solved to find the values of those two coordinates at the intersection point. Substituting both those values into the equation of either of the planes will allow you to calculate the value of the third coordinate at that point.

 

[Bishamonten]

You can solve the problem by setting any one of the coordinates to zero, provided the line is not parallel to but not in the plane where that coordinate is zero. The line will be parallel to such a plane if the corresponding component of the line vector is zero. So you can pick any component for which the corresponding component of the line vector is nonzero. In this case that's any of the three coordinates, because none of the components of the line vector are zero.

What you are doing by setting a coordinate equal to zero is obtaining two equations for the other two coordinates point where the line intersects the plane where that coordinate is zero. Those two equations will be linear in the other two coordinates, and hence can be solved to find the values of those two coordinates at the intersection point. Substituting both those values into the equation of either of the planes will allow you to calculate the value of the third coordinate at that point.

So I can set any of them to be 0? In this case, when I set z = 0, I get a system in which when solved yields x = 3, y = 1. Substituting these two back into one of the plane equations just tells me that z = 0.

These three values do not match the parametric equation found in the back of the book for this problem, which says that x = 17/4 - 10t ; y = 8t ; z = -11/8 + 11t, so there must be something I am not understanding here.

 

[andrewkirk]

there must be something I am not understanding here

The bit you are not understanding is that parametrisations are not unique. You can pick any point on the line as the 'zero point' of the parametrised line that corresponds to where the parameter value is zero, so there are as many parametrisations as there are points on the line.

Both your parametrisation and the one in the book are correct. To see that, note that the intersection point you found is when $t=0$ in your parametrisation and $t=1/8$ in the book parametrisation.

But your parametrisation is nicer, because it has no fractions in it.

 

[Bishamonten]

Well that's really convenient of parameters. Thank you for clarifying that point to me!

 

[Ray Vickson]

So I can set any of them to be 0? In this case, when I set z = 0, I get a system in which when solved yields x = 3, y = 1. Substituting these two back into one of the plane equations just tells me that z = 0.

These three values do not match the parametric equation found in the back of the book for this problem, which says that x = 17/4 - 10t ; y = 8t ; z = -11/8 + 11t, so there must be something I am not understanding here.

You can use the equations to solve for any two of the variables in terms of the third. For example, you can write the equations as
$$\begin{array}{lcr}2x - 3y &=&3-4z\\
x + 4y &=& 7 + 2z
\end{array}
$$
This is a simple system of 2 linear equations in 2 unknowns, and is easily solved using elementary algebra, to get
$$x = 3 - \frac{10}{11} z, \; y = 1 + \frac{8}{11} z $$

So, one possible parametrization of the line would be $x = 3 -(10/11)t, y = 1 + ( 8/11) t, z = t$. Of course there are infinitely many points on the intersection line, but one convenient point would be obtained by putting $t = 11$, to eliminate the fractions: $x = -7, y = 9, z = 11$.