Find the vector equation for the line of intersection of the two planes.

[marek]

Hello everyone, I would like to check my work on this problem.

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

x+2y+3z=1 (V1)
x-y+z=1 (V2)

I took the cross product V1 X V2 and got <5, 2, -3>.

I know I need to put this into the equation r = r(sub zero) + t(V1XV2), which is the equation for a line in three dimensions.

To get the r(sub zero) portion, I set y = 0 on the two plane equations.

Solving the system of equations of two variables, I got z = 0, and x = 1. So a point on this line would be (1, 0, 0).


So the vector equation of this line should be:

r = <1, 0 , 0> + t<5, 2, -3>

Please correct me if I made any mistakes, thank you.

 

[HallsofIvy]

It's easy to check- does every point on this line lie in both planes? Any point on this line is of the form (x, y, z)= (5t+ 1, 2t, -3t). The equation of the first plane is x+ 2y+ 3z= (5t+ 1)+ 2(2t)+ 3(-3t)= 5t+ 1+ 4t- 9t= 5t+ 1- 5t= 1. Yes, the entire line lies in the first plane. The equation of the second plane is x- y+ z= (5t+ 1)- (2t)+ (-3t)= 5t+ 1- 5t= 1. Yes, the entire line lies in the second plane. Therefore this is the unique line of intersection.

But rather than taking the cross product of the normal vectors, I would just solve the two equations. Any point on the intersection of the two planes must satisfy both equations. Adding twice the second equation to the first eliminates y and gives 3x+ 5z= 3 so that x= 1- (5/3)z.
Putting that into the second equation, x- y+ z= 1- (5/3)z- y+ z= 1, y= z- (5/3)z= (-2/3)z.

Taking t= z/3, x= 1- 5t, y= -2t, z= 3t are parametric equations for the line of intersection. That is equivalent to the vector equation <x, y, z>= <1, 0, 0>+ t<-5, -2, 3>. That is the same equation as yours except that your "t" is the negative of my "t".