The Line of Intersection of Two Planes

[Bassa]

Homework Statement

Find a set of parametric equations for the line of intersection of the planes.

6x-3y+z=5 and -x+y+5z=5[/B]

Homework Equations

The cross product formula
The formula for the parametric equations of a line in three dimensional space:
x=x1+at, y=y1+bt, z=z1+ct
Knowing the fact that the coefficients in front of the variables in the equation of a plane are the components for the normal vector of the plane

The Attempt at a Solution

[/B]
I found the cross product between the normal of the two plane. The resulting vector is going to function as the direction vector for the intersection line.

a= the normal vector of the first plane
b=the normal vector of the second plane

axb=-16i-31j+3k

finding a point on the intersection line:
We could do this by forming a system of equations. We get rid of on of the variables by saying that at some point the value of z in both planes will be the same and at some point it will be zero. This will give us the system of equations:

6x-3y=5
-x+y=5

Would solve for one of the variable by using the method of elimination. We could multiply the second equation by 6. This will give us:

6x-3y=5
-6x+6y=30

We could solve for y:

y=35/3

we could substitute the value of y into the second equation and that will give us a value for x:

-x+(35/3)=5
x=20/3

Now, we have the point ((20/3), (35/3), 0)

From this point it is really easy to plug the numbers into the parametric equations of a line. I referred to the solution of this problem. We both had the same values for the direction vector, but the point on the line was different. This makes the parametric equations of the intersection line have different values for x1, y1, z1. I was wondering if there is a unique answer to this problem? My thought is that there isn't, because we could choose any point on the line and use it to write the parametric equations of the intersection line. Am at right on this?

 

[RUber]

You are correct that any point plus a direction vector defines a unique line. The point itself is in no way unique, and may only be an indicator of which technique was used to find it.
If you are looking for additional or alternative methods, it may be worth trying to understand how the textbook solution was derived.

 

[HallsofIvy]

I can see no reason to worry about the normal vectors, etc. Essentially, a point on the line of intersection, because it lies on both planes, must satisfy both equation. So finding a point on the line of intersection of the two lines is just solving the two equations 6x-3y+z=5 and -x+y+5z=5.
Since that is two equations in three unknowns, you can solve two of the unknowns in terms of the third, then use that third as a parameter.