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?
Best Answer
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.