# Perpendicular planes, solve for parameter a

1. Mar 25, 2014

### concon

1. The problem statement, all variables and given/known data
Determine for what value/s of the parameter a the following planes in R3 are perpendicular:
ax + 0y - 5z = 3
ax + ay + 5z = -26
Write answer in form {a,b} or {a}

2. Relevant equations
I know that in R2 two vectors are perpendicular if
u*v = 0
What what do I use for R3 with planes?

3. The attempt at a solution
To solve this problem I need to know what it means for two planes to be perpendicular. I know in R2 that means there dot product is zero, what does that mean in R3? Also dot product?

2. Mar 25, 2014

### Staff: Mentor

Two planes are perpendicular if their normals are perpendicular. That can be determined by getting the dot product of the two normals.

3. Mar 25, 2014

### maajdl

You need to know:

- what a vector perpendicular to a plane is,
- how to find a vector perpendicular to a plane (and why)
- that two planes are perpendicular when vectors perpendicular (normal) to them are perpendicular (and why)

then you can proceed easily (knowing what you do).

Try to think more precisely!
In R2 there is no plane, there are lines.

Do you know how to check if two line in R2 are perpendicular?
For example: when are these lines perpendicular?

ax - 5 y= 1458966.1973
ax + 5 y= -125896.5586

dot product of what? That's the point!

4. Mar 25, 2014

### concon

Okay I got the solution: {-5,5}

Normal of the first equation n = (a,0,-5)
Normal of the second equation m = (a,a,5)

thus n*m = a^2 -25 = 0
thus a = -5 or 5

Thanks to both of y'all for the replies. It helped a lot and I appreciate it!