Distance from a point in a plan in Cartesian space

In the Euclidean space, the points can be defined using their coordinated known as Cartesian.

That is to say the $P$ plan and the point $A$ in Cartesian space. One calls $(x_A, y_A, z_A)$ the coordinates of the point $A$ and $Ax+By+Cz+D=0$ the equation representative of the $P$ plan: then the distance from the point $A$ in the plan $P$ , $d$ is worth:

$d = \ frac {\ left| Ax_A + By_A + Cz_A + D \ right|} {\ sqrt {A^2 + B^2 + C^2}}$

Demonstration:

That is to say $H: = (X, there, Z)$ the projected orthogonal one of $A$ on $P$ Is $N: = (has, B, C)$ a normal vector with $P$ .

There is $\ overrightarrow {AH} = \ lambda. \ vec N$ and $H \ in P$ thus one can write (in terms of components):

$(x-x_A; there there _A; z-z_A) = \ lambda (has; B; C)$

This amounts solving the following system: $\begin{Bmatrix} x=\lambda A+x_A \\ y=\lambda B+y_A \\ z=\lambda C+z_A \\ Ax+By+Cz+D=0
\end{Bmatrix}$

The substitution of X, there and Z in the 4^ème equation by their values obtained in the 3 first makes it possible to write:

$A (\ lambda A+x_A) +B (\ lambda B+y_A) +C (\ lambda C+z_A) +D=0$ .

Or:

$Ax_A+By_A+Cz_A+D + \ lambda (A^2+B^2+C^2) =0$ .

P being a plan, has, B, C are not all null: one has

$\ lambda = - \ frac {Ax_A+By_A+Cz_A+D} {A^2+B^2+C^2}$

However, the distance from $A$ with $P$ , is not other than the length of the vector $\ overrightarrow {AH}$ ; thus:

$d: =AH= \ left| \lambda \right| \ begin {Vmatrix} \ vec NR \ end {Vmatrix}$

$d= \ left| \ frac {- (Ax_A+By_A+cz_A+D)} {A^2+B^2+C^2} \ right| \sqrt{A^2+B^2+C^2}$

$d = \ frac {\ left| Ax_A + By_A + Cz_A + D \ right|} {\ sqrt {A^2 + B^2 + C^2}}$ This finishes the proof.

See too

concept of distance in Mathematical
the orthogonal Projection

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