# What is the locus (unsure if this is the right word) defined by these equations?

1. Sep 9, 2011

### tamtam402

1. The problem statement, all variables and given/known data

Using the Ax+By+Cz+D= 0 equation for planes, where D = -(Axo, Byo, Czo)

Find the locus given by the equation if A and B = 0

note: I'm not 100% sure if locus is the right word, english isn't my primary language.

2. Relevant equations

Given above.

3. The attempt at a solution

Well, if A and B = 0, the equations can be written as Cz+D = 0.

Which means z = -D/C = (0x + 0y Czo)/C = zo

Since we're in a 3 dimensions space, the way I see it both x and y are unrestricted and can take any values. Does that mean the locus defines a plane parallel to the X and Y axis, at a Z value of Zo?

2. Sep 9, 2011

### tamtam402

One more thing. The book gives an example, stating that if A=0 then the equation defines a plane parallel to the X axis. Shouldn't it say perpendicular, instead of parallel?

3. Sep 9, 2011

### Dick

Your answer, that's it's a plane parallel to the X and Y axes, is correct. But why are you writing D = -(Axo, Byo, Czo)? That looks like a vector to me. That doesn't make sense, a form like Ax+By+Cz+D=0. D should be a scalar, i.e. a simple number.

Last edited: Sep 9, 2011
4. Sep 9, 2011

### Dick

The plane z=1 is parallel to x axis, isn't it? That has A=0.

5. Sep 9, 2011

### tamtam402

Yes sorry, I meant D = - (Axo + Byo + Czo),

A,B and C being the components of the normal vector N=(A,B,C)
and xo, yo, zo being the coordinates of a point on the plane Po:(xo,yo,zo).

6. Sep 9, 2011

### tamtam402

Yeah I noticed my mistake after posting that. I somehow misunderstood A=0 for x=0. Thanks for the help!