We’ve been talking about how vectors could be used to come up with the equation of a plane. If you have the normal vector of a plane, and one point containing the plane, then you can find it’s equation. Here’s a problem; find an equation of the plane A with normal vector <3, 2, 5> and containing the point <1, 3, 6>.
Before we learnt if the normal vector n is <a, b, c>, if we know it’s components, then the equation of the plane for that vector is going to be ax plus by plus cz equals d, for some value of d. In this case, our normal vector is <3, 2, 5> so our equation is going to be 3x plus 2y plus 5z equals d. It’s just a matter of figuring out what d is, so that the plane will contain this point.
The way we do that was we plug <1, 3, 6> in for <x, y, z>. 3 times 1, plus 2 times 3, plus 5 times 6 equals d. And this is going to be 3 plus 6, 9, plus 30, 39, that’s our d value. So our plane is 3x plus 2y plus 5z equals 39.
Let’s do another example. This plane has normal vector <2, 0, -3> and contains the point <3, -8, 5>. So we can write the equations 2x plus 0y minus 3z equals d. So we’re not going to have a y term here. And then containing the point <3, -8, 5>, we plug in that point. We get, 2 times 3. I don’t need a y-term, so minus 3 times 5 equals d. This is 6 minus 15, is -9. So I get 2x minus 3z equals -9. You see it’s really easy once you have a normal vector, and the point the plane contains.
This one’s going to be even easier. The normal vector is <0, 0, 3>. So the equation is 0x plus 0y plus 3z equals d. So we plug in the point <-4, 1, 6>. These terms disappear and we just get 3 times the z coordinate, 6 equals d. That’s 18. And so we get 3z equals 18 which is the same as z equals 6. That’s the plane. And it makes sense that if a plane has a vector that’s perpendicular to the xy plane, its only direction is the z direction. That it’s going to have an equation z equals 6. That’s it.
So if you know the normal vector for a plane and one point that the plane contains, you can find the equation of the plane.