The equation of plane:

$ Ax + By + Cz = d$

Where $ A \hat i + B \hat j + C \hat k$ is the normal vector $ \vec n$ perpendicular to the plane.

The position vector $ \vec r = x \hat i + y \hat j + z \hat k$


$ \vec r \cdot \vec n = d$

This last equation says that a position vector to any point on the plane has the same projection along $ \vec n$ because $d$ is a constant.

But I can't see that graphically. Different position vectors to the plane have different slopes, while the normal vector of the plane is always perpendicular and doesn't change, so how does every position vector to any point on the plane have the same projection along $ \vec n$? Thank you.

  • $\begingroup$ Yes, they have. It is the position vector of the orthogonal projection of the origin onto the plane. $\endgroup$ – Bernard Feb 21 '19 at 16:19
  • $\begingroup$ Have a look at goo.gl/images/HjdXrF. $\endgroup$ – amd Feb 21 '19 at 23:37

I like to think of it in terms of orthogonality. Let $(x_0,y_0,z_0)$ be a point on the plane, and consider the position vector $\vec r_0 = x_0 \hat\imath + y_0 \hat\jmath + z_0 \hat k$. Since the point is on the plane $\vec r_0 \cdot \vec n = d$ too. So the plane equation becomes $$ \vec r \cdot \vec n = \vec r_0 \cdot \vec n \implies (\vec r-\vec r_0)\cdot \vec n = 0 $$ This says that the vector drawn from $(x_0,y_0,z_0)$ to $(x,y,z)$ is perpendicular to $\vec n$.

  • $\begingroup$ Yes that makes sense, but what I'm trying to understand is that some points on the plane are assigned by different position vectors with different slopes, so it's not very intuitive to think about it graphically. $\endgroup$ – khaled014z Feb 21 '19 at 16:28
  • $\begingroup$ I agree. That's why I don't think about points on the plane in terms of position vectors. $\endgroup$ – Matthew Leingang Feb 21 '19 at 16:32
  • $\begingroup$ I'm glad I'm not the only one, thank you. $\endgroup$ – khaled014z Feb 21 '19 at 16:33
  • 1
    $\begingroup$ To elaborate, axes and an origin are devices we attach to space, but they're not really necessary to describe it geometrically. Vectors can be combined without reference to any specific origin, and more often it's better to do it that way. $\endgroup$ – Matthew Leingang Feb 21 '19 at 16:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.