Please note I just need pointing out where my line of reasoning goes wrong.
Given a plane defined by a fixed point $A$ with coordinates $(x_a, y_a, z_a)$ position vector $\boldsymbol{a}$ and a normal vector $\boldsymbol{n}=\begin{pmatrix}a \\ b \\ c\end{pmatrix}$
We have $\boldsymbol{r}.\boldsymbol{n} = \boldsymbol{a}.\boldsymbol{n}$, where $\boldsymbol{r}$ is position vector of a general point in that plane. And the equation of a plane is thus:
$\boldsymbol{r}.\boldsymbol{n} = d$ in vector form
or
$ax+by+cz=d$ in Cartesian form.
Further, I am happy that the perpendicular distance from some point $(x_0, y_0, z_0)$ is given by
$$\frac{|ax_0 + by_0 + cz_0 – d|}{\sqrt{a^2+b^2+c^2}}$$
I also know from various sources (which state it without proof) that if a unit normal vector $\boldsymbol{\hat{n}}=\frac{\boldsymbol{n}}{|\boldsymbol{n}|} = \begin{pmatrix}\frac{a}{\sqrt{a^2+b^2+c^2}} \\ \frac{b}{\sqrt{a^2+b^2+c^2}} \\ \frac{c}{\sqrt{a^2+b^2+c^2}}\end{pmatrix}$ is used in the vector equation of a plane
$$\boldsymbol{r}.\boldsymbol{\hat{n}} = d$$
then $d$ is the perpendicular distance from the origin to the plane.
I am struggling to derive this result for myself.
(1) Using the formula for perpendicular distance of any general point:
$\frac{|a\times 0 + b\times 0 + c\times 0 – d|}{\sqrt{a^2+b^2+c^2}} = \frac{|- d|}{\sqrt{a^2+b^2+c^2}} = \frac{|-(ax_a + bx_a + cz_a)|}{\sqrt{a^2+b^2+c^2}} = \pm\frac{ax_a + bx_a + cz_a}{\sqrt{a^2+b^2+c^2}}$
(2) Using the vector plane equation:
$\boldsymbol{a}.\boldsymbol{\hat{n}} = \frac{ax_a + bx_a + cz_a}{\sqrt{a^2+b^2+c^2}}$
The only difference is the $\pm$ sign which comes from dropping the modulus. But the modulus is in the formula for a good reason — to account for cases when the general point $(x_0, y_0, z_0)$ is on either side of the plane.
Is it more correct then to state that in the plane equation $\boldsymbol{r}.\boldsymbol{\hat{n}} = d$, the distance from the origin is $|d|$, rather than $d$?
Best Answer
Your reasoning goes wrong in this line:
It should be:
Also you can't drop the modules that way: $$|something|=\pm something$$