Distance from a point to a plane

[GwtBc]

Homework Statement

What is the distance from the point P to the plane S?

Homework Equations

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

The Attempt at a Solution

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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?

 

[Orodruin]

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.

 

[GwtBc]

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.

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

 

[Orodruin]

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

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!)

 

[GwtBc]

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!)

Got it. Thanks for the help.

 

[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.

 

[Orodruin]

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 $\vec V \times \hat n$.

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.

 

[LCKurtz]

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.

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