Calculus Planes - General Cartesian Equations

Ulti
Messages
17
Reaction score
0

Homework Statement


Write down the general cartesian equation of a vertical plane (parallel to the z-axis), a
non-vertical plane and a horizontal plane (parallel to the x; y-plane) together with their
normal vectors.
Find the cartesian equations of the line of intersection of the plane
5x + y ¡ 2z = 6 (¤)
and a general horizontal plane.
How is the normal to (¤) related to the horizontal normal to the line of intersection?

Homework Equations


ax + by + cz = d

The Attempt at a Solution


It's been a while since I've been back to the Physics Forums so I'm not sure if this is the right section to post this as it is a homework question but I actually have no idea how to solve it.

I've been given a clue that the general cartesian equation of a vertical plane (parallel to the z-axis) is ax+by-c=0, however I do not understand why - I can only guess at best.

I only need help with getting the general cartesian equations, so if anyone could provide some tips or a link to materials about planes it would be great!

Thank you.

EDIT: Nevermind please close, I kinda understand it now.
 
Last edited:
Physics news on Phys.org
I've been given a clue that the general cartesian equation of a vertical plane (parallel to the z-axis) is ax+by-c=0, however I do not understand why - I can only guess at best.

z doesn't show in the vertical plane expression because so the "height" z is free to move, go wherever you want.
There is no constraint about the z. It can be whatever you want.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top