Linear 3D Geometry

[NaN-Gram]

Hello all, I have a couple of problems I need help with on planar geometry.

I'm given a plane P, with coords 3x-4y+z=7.

How can I find a parallel plane, if it goes through the point (1,2,-5)?

Additionally, how can I prove whether or not my given plane is perpendicular to the xy plane?

 

[Eduleo]

A parallel plane is one with the same gradient (same coeficients, or scaled ones), but with different contstant value. Pluggin the point coordinates in the LHS of your plane equation would give the constant value to a parallel plane at that point, since both planes has the same gradient.

A perpendicular vector to the plane xy is, for example, unit vector k. So, if your plane is perpendicular to the xy plane, then its gradient has to be perpendicular to vector k.

 

[pka]

Hello all, I have a couple of problems I need help with on planar geometry.
I'm given a plane P, with coords 3x-4y+z=7.
How can I find a parallel plane, if it goes through the point (1,2,-5)?
Additionally, how can I prove whether or not my given plane is perpendicular to the xy plane?

Two planes are parallel if and only it they have the same normal.
If \(N=\left<n_1,n_2,n_3\right>\) then the planes \(\Pi_1=n_1x=+n_2y+n_3z=d_1~\&~\Pi_2- n_1x=+n_2y+n_3z=d_2\) are parallel.

 

[NaN-Gram]

A parallel plane is one with the same gradient (same coeficients, or scaled ones), but with different contstant value. Pluggin the point coordinates in the LHS of your plane equation would give the constant value to a parallel plane at that point, since both planes has the same gradient.

A perpendicular vector to the plane xy is, for example, unit vector k. So, if your plane is perpendicular to the xy plane, then its gradient has to be perpendicular to vector k.

I actually have a few options when it comes to proving that my given vector is perpendicular to the xy plane.

Of these options, which is the best, and why?

 

[Eduleo]

A general plane [math]ax+by+cz=d[/math] is perpendicular to the plane xy if its gradient is perpendicular to the unit vector k. That is, the dot product of the plane gradient and unit vector k is zero.

So [math]<a,b,c> · <0,0,1> = 0[/math]so [math]\boxed{c=0}[/math] (your plane has [math]c\ne 0[/math], so it is not perpendicular to the xy plane.)
so, any general plane [math]ax+by=d[/math] is perpendicular to the plane xy.

 

[pka]

I'm given a plane P, with coords 3x-4y+z=7.
How can I find a parallel plane, if it goes through the point (1,2,-5)?

The reason My first reply is in the form it is, I had hoped that you could that the answer is simply:
\(\large 3x-4y+z=d\) to find \(d\) use the point \((x=1,y=2,z=-5)\)