Equation of a plane through a line

artie ziff

i'm hoping that someone here can help. i'm sure the answer is glaringly obvious but there's conflicting info in my notes and textbook.....

if given the equations of 2 planes which intersect in a line AB, how do i find the general equation of a plane through AB?

any help would be greatly appreciated

Michael Collins

Well you can certaintly find the equation of the line where the planes A & B intersect but there are an infinite number of planes that will contain that line. You need some more information - such as a point on the 3rd plane (which isn't on the line AB) or the slope of the plane etc...

artie ziff

well the two planes given are 2x-7y+5z+1=0 and x+4y-3z=0

i tried to make three points by getting the normal of the equations above and treating them as i,j,k vectors but this is either wrong or more complicated than it needs to be?

Michael Collins

The problem is though you don't have enough information. You've given the formula of two planes there in your last post, that will at best get you a line in 3D space. There are any number of planes going through that line. If you're after any old plane through that line, then just pick a point - that isn't on that line, say p1 - and now pick two that are, say p2 & p3.

Now we can get two vectors parallel to the plane:

v1 = p2 - p1, v2 = p3 - p1. (Can you see that these must be parallel to the plane?)

Now take the cross product of these two vectors:

n = v1 x v2

This must be a normal to the plane. This normal consists of three components n = [n1, n2, n3] (the i, j & k components if you like).

The plane you're after will then simply be P: n1.x+n2.y+n3.z + D = 0 (This follows from simple algebra)

You can find D by subbing in any point in the plane...

I hope this is what you're after. If not, maybe if you explained why you're trying to do this I be able to help you.

MathsManiac

I note the original post refers to "the general equation" of the plane containing AB.

Possibly what is required, therefore, is a parameterised family of planes.

Just as the "general" equation of a line through the point of intersection of two lines with equations L1=0 and L2=0 is given by L1+kL2=0, the equivalent is true of planes.

If this is the intention of the question, then the "general" equation of the plane required is (2x-7y+5z+1) + k(x+4y-3z) =0. Depending on what you wanted to do with this, you might rearrange it as: (2+k)x + (-7+4k)y + (5-3k)z + 1 = 0.

As already pointed out, there are infinitely many planes through the line concerned and if you want a unique one, you need a further piece of information. The advantage of writing the general version as above is that, if that further piece of information comes along, it might be very easy to use it to find k. For example, if you're told the plane has to pass through a given point, you just substitute that point in and solve for k.

Of course, you mightn't be looking for a unique plane. Suppose the question was: determine the set of planes that contain AB and that intersect the negative z-axis. In that case, the condition on k is given by setting x and y equal to 0 and solving z<0 for k.

(Pedantic footnote: there is one plane through AB that does not correspond to any value of k. It is the second of the two given planes.)

Michael Collins

That's a nice method there Mathsmaniac. Pretty much follows from the Leaving Cert course for doing the same with two lines intersecting.