Now let's work out the equation of the plane in component form. Suppose that, in some coordinates system, the normal vector is n = ai + bj + ck and the initial point has position vector r0 = x0i + y0j + z0k. For a generic point on the plane with position vector
r = xi + yj + zk,

r - r0 = (x - x0)i + (y - y0)j + (z - z0)k

so the point-normal equation n•(r - r0) = 0 transforms into

a(x - x0) + b(y - y0) + c(z - z0) = 0.

This equation is known as the standard form of the equation of the plane. If you expand it and simplify, you get

ax + by + cz = d

where d = n•r0 = ax0 + by0 + cz0.

If you work with the unexpanded form of the standard equation, it's easy to see where the initial point and normal vector occur: the coordinates of the point are subtracted from x, y and z inside the brackets and the components of the normal vector multiply the brackets. Thus the plane through the point (1, 0, -2) with normal vector [3, -5, 1] has standard equation

3(x - 1) + (-5)(y - 0) + 1(z - (-2)) = 0.

which can be simplified to 3x -5y + z = 1.

Any equation in the form ax + by + cz = d represents a plane with normal vector n = ai + bj + ck, so, for example, the equation
2x - 5y + 7z =11 represents a plane with normal vector n = 2i - 5j + 7k.

Suppose you are given a plane with standard equation 3x - 4y + z = 5. How would you describe the plane 6x - 8y + 2z = 10 relative to the original plane? the plane 3x - 4y + z = 6?  the plane
3x - 4y + z = 0?

Standard equations for some special planes.

• Planes through the origin must have standard equations of the form
  ax + by + cz = 0, since the values x = 0, y = 0, z = 0 must satisfy the equation.
  
  
• The x-y-plane passes through the origin and has normal vector k, so its standard equation is
                     0(x - 0) + 0(y - 0) + 1(z - 0) = 0 ,   i.e.  z = 0.
  Similarly, the y-z-plane has standard equation x = 0 and the x-z-plane has standard equation y = 0.
  
  
• A plane parallel to the x-y-plane must have a standard equation z = d for some d, since it has normal vector k. A plane parallel to the y-z-plane has equation x = d, and one parallel to the x-z-plane has equation y = d.
  
  
• The normal of a plane parallel to the z-axis must be perpendicular to k, so the k-component of the normal vector is 0. The plane thus has the form ax + by = d. Similarly, a plane parallel to the y-axis has equation ax + cz = d, and one parallel to the x-axis has equation by + cz = d.