3

If d=0, then we obviously see that the equation is the Euclidean inner product with (a,b,c) and (x,y,z) that equals zero - and so (a,b,c) is the normal vector to the plane.

What if $d \ne 0$?

Then I don't have the same intuitive way of finding the normal - the "inner product" wouldn't be 0, and we couldn't conclude the orthogonality of the vectors.

Thanks,

User001
  • 1

2 Answers2

7

Since the gradient $\nabla f(x,y,z)$ is perpendicular to the level sets of $f(x,y,z)$, we can use the gradient $\nabla(ax+by+cz-d)=(a,b,c)$ which is perpendicular to the plane.

robjohn
  • 345,667
  • Oh this is so freakin cool, @robjohn :-) I always really enjoy your work on MSE. Thanks so much! – User001 Dec 31 '15 at 19:41
5

If $(x_0,y_0z_0)$ and $(x_1,y_1,z_1)$ are any two points in the plane $ax+by+cz=d$, we have $ax_0+by_0+cz_0=ax_1+by_1+cz_1$, whence $$a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)=0.$$

Bernard
  • 175,478
  • Ok got it - thanks so much @bernard :-) – User001 Dec 31 '15 at 19:26
  • It's an especially nice reminder to sometimes switch from thinking of vectors as points in space and vice versa ... especially in some of the 4-dimensional plane equation problems :-) – User001 Dec 31 '15 at 19:38
  • Excellent, thank you. It is worth reading the above, with this answer if anyone has any confusion: https://math.stackexchange.com/a/4400640/379107 – BenKoshy Feb 03 '23 at 01:18
  • Perhaps it might be worth adding another line separating the constants on the other side of the equation> – BenKoshy Feb 03 '23 at 01:21