# Distance from a point to a plane

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1. Aug 28, 2016

### GwtBc

1. The problem statement, all variables and given/known data
What is the distance from the point P to the plane S?

2. Relevant equations
$S = \left \{ r_{0} + s(u_{1},u_{2},u_{3})+t(v_{1},v_{2},v_{3}) | s,t \in \mathbb{R} \right \}$

3. The attempt at a solution

I found an expression for the general distance between point P and a point on S, then found an expression for the distance and took the partial derivatives , $\frac{\partial r}{\partial s}$ and $\frac{\partial r}{\partial t}$ (both of the argument of the square root function) and set both to zero, then solved the resulting linear equations. Is this the right method?

2. Aug 28, 2016

### Orodruin

Staff Emeritus
It will work, but there is an easier way to do it by finding a normal vector of the plane and projecting the difference vector between the point and any point in the plane on it.

3. Aug 28, 2016

### GwtBc

Wouldn't that just give me another vector in terms of x, y and z that I'd have to minimize?

4. Aug 28, 2016

### Orodruin

Staff Emeritus
No. It gives you the difference vector between the point and the closest point in the plane. You get the distance by computing its magnitude. (Note the projection part of the procedure!)

5. Aug 28, 2016

### GwtBc

Got it. Thanks for the help.

6. Aug 28, 2016

### LCKurtz

Equivalently, if $\vec V$ is a vector from P to any point Q on the plane, and $\hat n$ is a unit normal to the plane, then$$|\vec V \times \hat n| = |\vec V|\cdot 1 \cdot \sin\theta = d$$so all you have to remember is to take the magnitude of $\vec V \times \hat n$.
Edit, added: Please ignore this post. This method works for finding the distance from a point to a line, where $\hat n$ is a unit direction vector along the line.

Last edited: Aug 29, 2016
7. Aug 29, 2016

### Orodruin

Staff Emeritus
This is not equivalent. You want $\vec V\cdot \hat n$. The cross product you quote is also dependent on which point in the plane you chose.

8. Aug 29, 2016

### LCKurtz

You're correct of course. I must have posted that before I was fully awake.