Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My professor did this question (Finding the shortest distance from a point to a plane) in class, but before doing the example he showed us a formula for calculating it, but didn't really explain how he got the formula. I was hoping someone could tell me how or why the formula works.

The question was as follows: Find the formula for the shortest distance from a point $P_0(x_0,y_0,z_0)$ to a plane $Ax+By+Cz+D=0$.

So the shortest distance would be a straight line from the point to the plane, which means that straight line would have to be in the direction of a normal vector to the plane. Let me denote the normal by $\vec{N}=[A,B ,C]^T$.

Next the professor said that $||\vec{P_{1}P_{0}}||\cos \theta=\frac{||\vec{P_{1}P{0}||||\vec{N}||\cos \theta}}{||\vec{N}||}=\frac{(\vec{P_1}P_{0}\cdot\vec{N})}{||\vec{N}||}=\frac{Ax_{0}+By_{0}+Cz_{0}+D.}{\sqrt{A^2+B^2+C^2}}$

I understood how he manipulated it to go to the last formula, however I guess I am having trouble understanding how/where the $\cos \theta$ in this formula: $||\vec{P_{1}P_{0}}||\cos \theta$ came from.

If it is for the simple reason that $\text{adjacent}=\text{hypotenuse} \cos \theta$, then I don't understand how the $||\vec{P_{1}P_{0}}|$ is the hypotenuse.

Can anyone shed any light on any of this?

share|cite|improve this question
up vote 3 down vote accepted

It is a projection.

Make a vector with the starting point as the given point, and the ending point as any point one the plane. We here call is $v$. The distance is exactly the projection of $v$ on an unit vector $n$ perpendicular to the plane. And $n$ can be easily calculated according to the plane.

And this is the result.

share|cite|improve this answer
No, $cos$ means projection. – eccstartup Jun 19 '13 at 2:41
Oh, I never noticed that. – Sujaan Kunalan Jun 19 '13 at 2:54
Thank you. Very helpful explanation! – Sujaan Kunalan Jun 19 '13 at 4:24

He took the dot product $\vec{P_1P_0}\cdot \vec{N}$, which is the same as $|\vec{P_1P_0}||\vec{N}|\cos\theta$ (where theta is the angle between the vectors).

He is finding the magnitude of the component of $\vec{P_1P_0}$ in the direction of the unit normal vector (hence dividing by the magnitude of $\vec{N}$) to the plane. If you draw a picture (as in a right triangle, with $\vec{P_1P_0}$ an arbitrary vector from the point of interest $P_0$ to some point $P_1$ on the plane$ you will find is indeed the perpendicular distance.

share|cite|improve this answer

See P800 Example 8, 12.5, Calculus. 6th Ed, by James Stewart. Here's a modified picture to avail:

enter image description here Espy that $\mathbb{n} \neq D$ necessarily.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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