What is the proof for the distance formula from a point to a plane?

[Reshma]

The distance 'D' of a point (x₀,y₀,z₀) from a plane ax + by +cz + d = 0 is given by the formula:

$$D = \left|\frac{ax_0 + by_0 + cz_0 + d}{a^2 + b^2 + c^2}\right|$$

Could someone give me some explanation\links on the proof\derivation of this formula? Thanks in advance.

 

[arildno]

First, do NOT use "d" in two distinct meanings! :grumpy:

I'll call the distance D hereafter.

Now, what should we mean by D?

Please answer the following question:
What is a meaningful definition of the distance from a plane to a point?

 

[HallsofIvy]

Arildno, as often as I have complained about people using capital letters ("D") and small letters ("d") to mean the same thing, I have no problem with using "D" and "d" to mean different things!

Reshma, there are two ways to do that:

1. For any point (x,y,z) in the plane, the square of the distance from that point to (x₀, y₀, z₀) is
(x- x₀)²+ (y- y₀)²+ (z- z₀)²). You can minimize that subject to the condition that ax+ by+ cz+ d= 0 by using Lagrange multipliers or by replacing z in the distance formula by z= (-ax-by-d)/c and then setting partial derivatives to 0.

2. (Simpler) Argue that, geometrically, the line from (x₀,y₀,z₀) to the nearest point on the plane is perpendicular to the plane (any other line would be the hypotenuse of a right triangle and so longer than a leg). Find the equation of the line through (x₀, y₀, z₀) in the direction of a normal vector to the plane (which is ai+ bj+ ck). Determine where that intersects the plane.

 

[arildno]

If you look at Reshma's edit, that was made while I wrote my post!
I agree, Reshma did correct his usage of small "d" to mean two different things (for distance AND for the constant term).

 

[Reshma]

Thanks HallsofIvy and arildno. I will follow the technique that you have suggested and post my proof soon.

If you look at Reshma's edit, that was made while I wrote my post!
I agree, Reshma did correct his usage of small "d" to mean two different things (for distance AND for the constant term).

It is "her". Never mind, not your fault. :tongue2:

 

[Reshma]

The vector normal to the plane ax + by + cz + d = 0 is:
$$\vec n = a\hat i + b\hat j + c\hat k$$
The line from any point (x, y, z) on the plane to a point P(x₀,y₀,z₀) lying outside the plane is:
$$\vec m = (x - x_0)\hat i + (y - y_0)\hat j + (z - z_0)\hat k$$

So distance 'D' of P(x₀,y₀,z₀) from the plane is equal to the projection of $\vec m$ on $\vec n$.

$$D = \frac{\left |\vec m \cdot \vec n\right|}{|\vec n|}$$
$$D = \frac{a(x - x_0) + b(y - y_0) + c(z - z_0)}{\sqrt{a^2 + b^2 + c^2}}$$

$$D = \left|\frac{ax_0 + by_0 + cz_0 + d}{\sqrt{a^2 + b^2 + c^2}\right|}$$

considering the absolute values...