Vector Line Equation of Plane Intersection

[Tarmac27]

Hi

I have discovered something that I need clarification on whilst working on this plane intersection problem.

Question:

Find vector equation of line of the intersection of the two planes:

[math]\Pi_1: 2x+3y+3z=10[/math] [math]\Pi_2: x+3y+2z=4[/math]
Method 1:
I used Gaussian elimination to form the new system of equations;

[math]3y+z=-2[/math][math]2x+3y+3z=10[/math]
Then by letting [math]z=\lambda[/math] and therefore [math]y=\frac{-2-\lambda}{3}[/math], [math]x=6-\lambda[/math]
Giving the vector equation:

[math]r(\lambda)=\begin{pmatrix}6-\lambda\\ \frac{-2-\lambda}{3}\\ \lambda\end{pmatrix}[/math]
Method 2:

I found the vector line equation using the cross product of the two normals to find the direction vector as [math]\vec{v} = \begin{pmatrix}-3\\ -1\\ 3\end{pmatrix}[/math]By just letting z=0 in the plane equations I found a point for the position vector which was [math]\vec{u}=\begin{pmatrix}6\\ \frac{-2}{3}\\ 0\end{pmatrix}[/math]
Hence,

[math]r(t)=\begin{pmatrix} 6\\ \frac{-2}{3}\\ 0\end{pmatrix}+t\begin{pmatrix}-3\\ -1\\ 3\end{pmatrix}[/math]
By scaling down the direction vector of the method 2 line by 1/3 and combining vectors it gives the same equation as method 1. So does this mean you can scale the direction vector as large or as small as desired it just consequently means that different t values will be needed to get to the same point on the line?

 

[Dr.Peterson]

By scaling down the direction vector of the method 2 line by 1/3 and combining vectors it gives the same equation as method 1. So does this mean you can scale the direction vector as large or as small as desired it just consequently means that different t values will be needed to get to the same point on the line?

Yes.

The direction vector is needed only to give the direction of the line; any scalar multiple of it has the same direction and will result in the same line, though with a different parameter.

 

[pka]

I have discovered something that I need clarification on whilst working on this plane intersection problem.
Question:
Find vector equation of line of the intersection of the two planes:
[math]\Pi_1: 2x+3y+3z=10[/math] [math] x+3y+2z=4[/math]
By scaling down the direction vector of the method 2 line by 1/3 and combining vectors it gives the same equation as method 1. So does this mean you can scale the direction vector as large or as small as desired it just consequently means that different t values will be needed to get to the same point on the line?

I don't fully understand your point. You found the direction vector [imath]\left<-3,-1,3\right>[/imath]
Any multiple of that is also a direction vector vector so I would use [imath]\left<3,1,-3\right>[/imath]
Now I would find a point on the intersection. Let [imath]z=0[/imath] then solve the system
[imath]2x+3y =10 \\ x+3y=4 [/imath]
[imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath][imath][/imath]

 

[HallsofIvy]

Your two planes are given by
2x+ 3y+ 3z= 10 and
x+ 3y+ 2z=4.

An obvious thing to do is to eliminate y by subtracting the second equatioin from the first: any point on the line of intersection satisfies
x+ z= 6. And taking z= 6- x the first equation becomes
2x+ 3y+ 3(6- x)= -x+ 3y+ 18= 10 so x= 3y+ 8 and then z= 6- x= -2- 3y.

(x, y, z)= (3y+ 8, y, -2- 3y). We can write the vector equation as <8, 0, -2>+ <3, 1, -2>t,