Find equation of all points equidistant

[MarkSA]

Hello,

Q: Find equation of the set of all points equidistant from the point (0,0,4) and the xy-plane. This is a quadric surface.

I think I have the general idea of this one. I need the distance from a point (x,y,z) in my equation and (0,0,4) and the xy-plane to be the same.
I can do the distance equation on (x,y,z) and (0,0,4)...
d = sqrt[(x-0)^2 + (y - 0)^2 + (z - 4)^2]

And this needs to equal the distance from (x,y,z) to the xy-plane. I have a formula.. the distance from a plane to a point using the normal vector of the plane. Could this be used? Would the normal vector of the xy-plane just be <1,1,0>? But the equation requires a 'd' from the equation of a plane, which I don't seem to have.

 

[Deleted member 4993]

Hello,

Q: Find equation of the set of all points equidistant from the point (0,0,4) and the xy-plane. This is a quadric surface.

I think I have the general idea of this one. I need the distance from a point (x,y,z) in my equation and (0,0,4) and the xy-plane to be the same.
I can do the distance equation on (x,y,z) and (0,0,4)...
d = sqrt[(x-0)^2 + (y - 0)^2 + (z - 4)^2] <<< This is an equation of the surface of a sphere.

And this needs to equal the distance from (x,y,z) to the xy-plane. I have a formula.. the distance from a plane to a point using the normal vector of the plane. Could this be used? Would the normal vector of the xy-plane just be <1,1,0>?

No the normal vector to xy plane is z-axis - which is <0,0,1>

But the equation requires a 'd' from the equation of a plane, which I don't seem to have.

The plane passes through the point <0,0,d>

You got the equation of all the points at a distance 'd' from (0,0,4)

Now think of a plane at a distance of 'd' from x-y plane.

The intersection of that plane - with the surface of the sphere will give you the locii of the points (it would be a circle0.

 

[MarkSA]

I'm still not seeing it.

This is what I have:
|0(x) + 0(y) + 1(z) + d?|/1 is distance from a point to a plane equation
sqrt[x^2 + y^2 + (z-4)^2] is distance from a point to (0,0,4) equation
My point is (x,y,z). So these two should be equal.

sqrt[x^2 + y^2 + (z-4)^2] = |0(x) + 0(y) + 1(z) + d|
sqrt[x^2 + y^2 + (z-4)^2] = |z + d|

If I knew d, I would have my equation for for the set of all points equidistant from (0,0,4) and xy-plane I believe.

I'm not sure how to get d though. I'm not real sure what it represents in the equation of a plane either. I know <a,b,c> is the normal vector. But d?

 

[dts5044]

what does the xy plane mean? It is a plane involving both the x axis and the y axis...so it cannot contain the z axis (z=0)

so the distance from any point P(x,y,z) to the xy plane is |z|

sqrt(x^2 + y^2 + (z-4)^2) = |z|

both sides are positive so you can square and not introduce or remove a solution

...

 

[galactus]

If we are in the xy-plane, then z=0

 

[pka]

The surface is a paraboloid up with focus (0,0,4).