# Finding intersection of three planes

1. Sep 25, 2014

### Tarrius

Hi!

I'm having trouble with this question, any help would be much appreciated! :)

Q1: Given the three vectors:

n1 = (1, 2, 3)
n2 = (3, 2, 1)
n3 = (1, −2, −5)

Find the intersection of the three planes ni*x = 0. What happens if n3 = (1, −2, −4)? Why is this different?

2. Sep 25, 2014

### da_nang

Are you familiar with vector spaces and matrix rank?

3. Sep 25, 2014

### mathman

Your original n3=(1,-2,-5) = n2-2n1 so the three planes don't intersect. The pairwise intersections (of the planes) are 3 parallel lines.

The other n3 would work since it is not a linear combination of the others.

4. Sep 26, 2014

### da_nang

Actually they do intersect, just not in a single point as can be shown by the Rouché-Capelli theorem.

5. Sep 26, 2014

### HallsofIvy

You don't really need to know linear algebra- just the basics of systems of equations.
The planes defined by the first three vectors are
x+ 2y+ 3z= 0
3x+ 2y+ z= 0
x- 2y- 5z= 0.

Find the general solution to that system (there is NOT a unique solution because the determinant of coefficients is 0). What does that define, geometrically. The second set of equations do NOT have 0 determinant so have a unique solution. What that solution is should be obvious.

6. Sep 26, 2014

### mathman

I forgot that the three parallel lines could be coincident.