# Math Help - Vector parallel to both planes?

1. ## Vector parallel to both planes?

The two planes are
x - y + z = 4 and 3x + y - z = 5

I know that a vector cannot be parallel to the planes themselves but to the intersections of the two planes. Uhh, supposedly Im supposed to use a matrix involving
|1, -1, 1|
|3, 1, -1|
|i, j, k |
I'm not sure why we do this step though. Can someone explain how to solve this and why a matrix needs to be used.

2. You're right, a vector can't be parallel to the planes - so what you actually are finding is a vector in the same direction as the intersection of the planes.

I'm sure you learned about cross products, which is what you have to use to find this line. That matrix, when you take the determinant of it, is the cross product between the two planes' normal vectors.

A normal vector is the vector that is perpendicular to the plane. Taking a cross product between two vectors gives you a vector perpendicular to those two vectors. Sooooo when you take the cross product between the two planes' normal vectors, you get a vector that is perpendicular to both those normal vectors. So that means this vector you get after computing the determinant of the matrix will be "parallel" to both planes.

The "i" "j" and "k" are normal vectors of length 1 in the x, y, and z direction. If you have issues actually taking the cross product/determinant of that matrix, this is a good site: Pauls Online Notes : Calculus II - Cross Product

3. Originally Posted by hellojellojw
The two planes are
x - y + z = 4 and 3x + y - z = 5

I know that a vector cannot be parallel to the planes themselves but to the intersections of the two planes. Uhh, supposedly Im supposed to use a matrix involving
|1, -1, 1|
|3, 1, -1|
|i, j, k |
I'm not sure why we do this step though. Can someone explain how to solve this and why a matrix needs to be used.
you are not actually using the matrix, you are using the determinant. That determinant, with "i", "j", and "k" the unit vectors in the x, y, and z directions, respectively, is the cross product of the vectors i- j+ k and 3i+ j- k, the normal vectors to the planes. Their cross product is then normal to both of them and so in the same direction as the line of intersection.

If that is too "abstract" for you, here is a more elementary way to solve this problem. Add the two equations together to get 4x= 9. That immediately gives x= 9/4 for all points on the line of intersection. Setting x= 9/4 in the first equation, 9/4- y+ z= 4, gives z= y+ 7/4. Setting x= 9/4 in the second equation, 27/4+ y- z= 5, gives y= z+ 7/4 also. Of course, that had to happen.

In any case, we can take z itself as parameter and write the equation of the line of intersection as x= 9/4, y= t+ 7/4, z= t and write down a vector parallel to that line immediately.