# Find the angle between two planes

1. Mar 9, 2013

### hamsterB

1. The problem statement, all variables and given/known data
Find the angle between the plane 3x+5y+7z = 1 and the plane z = 0.

2. Relevant equations
a.b=|a||b|cosθ

3. The attempt at a solution
Hi, I know that I need to have both these planes in the form (x,y,z) and then find the dot product to find the angle between them. The problem I am having is with putting them in that form, at first I assumed plane 1 would just be (3,5,7) and plane 2 would be (0,0,1), but I have also read that to find the angle between to planes I need the normal vector to each plane, and this has confused me. Using these vectors I came up with the answer 30.8°, but I don't know if what I did was right! Any help would be appreciated!

2. Mar 9, 2013

### Dick

(3,5,7) IS the normal vector to the plane, not the plane itself. Same for (0,0,1). It sounds like you are doing it correctly. I don't get the answer you got though. Can you show how your numbers worked?

3. Mar 9, 2013

### fwang6

I think the answer should be 39.8

4. Mar 9, 2013

### fwang6

can anyone help me with this question:
Find the angle between
(a) the line L1 given by the equations y = 2z, x = 0, and
(b) the line L2 given by the equations x = 3z, y = 0.

5. Mar 10, 2013

### hamsterB

Sorry, working it out again I got 39.79;
a.b=(3x0+5x0+7x1) = 7
|a|=√83
|b|=√1
∴θ=cos-1(a.b/|a||b|)= 39.79

So for any equation ax+by+cz=d, will the normal vector always be (a,b,c)?