# Plane point distance

Let’s start by representing the point $A$ and the plane $ax + by + cz = d$: $P_{1}$ is a point in the given plane. From the plane we have the point $P_{0}$ and also a vector $\vec{b}$ that defines the segment $\overset{\longrightarrow}{P_{0}P_{1}}$:

$$\vec{b} = <x_{1} – x_{0},y_{1} – y_{0},z_{1} – z_{0} >$$

The distance $D$ of $P_{1}$ to the plane is equal to the absolute value of the scalar projection of $\vec{b}$ on the normal vector $\vec{n}$:

$$D = \left| comp_{\vec{n}}\vec{b}\right| = \left| \cfrac{\vec{n}\cdot \vec{b}}{|\vec{n}|} \right|$$

The above equation is read as the absolute value of the dot product of the normal vector $n$ multiplied by the vector $b$ divided by the absolute value of the normal vector $n$. Remember that the normal vector $n$ is represented with the coefficients of the unknowns of the plane equation respectively, that means that if the equation of the plane is $ax + by + cz = d$, then the normal vector is $\vec{n} = <a, b, c>$. Here is the equation obtained from the operations of the previous equation:

$$D =\cfrac{a(x_{1} – x_{0}) + b(y_{1} – y_{0}) + c(z_{1} – z_{0})}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

Multiplying the parentheses is as follows:

$$D =\cfrac{ax_{1} – ax_{0} + by_{1} – by_{0} + cz_{1} – cz_{0}}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

We sort all the unknowns with sub-index $1$ together and the unknowns with sub-index $0$ together:

$$D =\cfrac{ax_{1} + by_{1} + cz_{1} – ax_{0} – by_{0} – cz_{0}}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

We factor the negative and it looks like this:

$$D =\cfrac{ax_{1} + by_{1} + cz_{1} – (ax_{0} + by_{0} + cz_{0})}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

But as the general equation of the plane is:

$$a(x – x_{0}) + b(y – y_{0}) + c(z – z_{0}) = 0$$

$$ax – ax_{0} + by – by_{0} + cz – cz_{0} = 0$$

$$ax + by + cz = ax_{0} + by_{0} + cz_{0}$$

Note that $d = ax_{0} + by_{0} + cz_{0}$, then the equation of the distance from the point to the plane can be written finally as follows:

$$D = \cfrac{ax_{1} + by_{1} + cz_{1} – d}{\sqrt{a^{2} + b^{2} + c^{2}}}$$

## Exercise of distance between a point and a plane

Find the distance from point $(3,-2,7)$ to the plane $4x-6y+z=5$

It is not necessary to graph the point and the plane, but we are going to do it: The exercise is solved using a simple formula, we have $P_{1}$ which is:

$$P_{1} = (3,-2,7)$$

The normal vector is the one that will give us the values of $a$, $b$ and $c$; but the normal vector are the coefficients of the unknowns of the plane equation respectively:

$$\begin{array}{c c c c c c} \vec{n} & = < & 4, & -6, & 1 & >\\ & & a & b & c & \end{array}$$

With our point $P_{1}$, our normal vector and $d$, we simply substitute in the formula:

$$D = \cfrac{(4)(3) + (-6)(-2) + (1)(7) – 5}{\sqrt{(4)^{2} + (-6)^{2} + (1)^{2}}}$$

And we grab a calculator to make the operations easier to finally have our distance between the point $P_ {1}$ and the plane:

$$D = \cfrac{26\sqrt{53}}{53} \ \text{u}$$

The $\text{u}$ means “units”.

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