Question

Prove that, In $R^{3}$ the distance $D$ between the point $P_{0}\left(x_{0}, y_{0}, z_{0}\right)$ and the plane $a x+b y+c z+d=0$ is $$ D=\frac{\left|a x_{0}+b y_{0}+c z_{0}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}} $$

Answer 1

Proof 

Let $Q\left(x_{1}, y_{1}, z_{1}\right)$ be any point in the plane, and let $\mathbf{n}=(a, b, c)$ be a normal vector to the plane that is positioned with its initial point at $Q .$ It is now evident that the distance $D$ between $P_{0}$ and the plane is simply the length (or norm) of the orthogonal projection of the vector $\overrightarrow{Q P}_{0}$ on $\mathbf{n}$,  is
$$
D=\left\|\operatorname{proj}_{\mathbf{n}} \overrightarrow{Q P_{0}}\right\|=\frac{\left|\overrightarrow{Q P}_{0} \cdot \mathbf{n}\right|}{\|\mathbf{n}\|}
$$
But
$$
\begin{aligned}
&\overrightarrow{Q P_{0}}=\left(x_{0}-x_{1}, y_{0}-y_{1}, z_{0}-z_{1}\right) \\
&\overrightarrow{Q P_{0}} \cdot \mathbf{n}=a\left(x_{0}-x_{1}\right)+b\left(y_{0}-y_{1}\right)+c\left(z_{0}-z_{1}\right) \\
&\|\mathbf{n}\|=\sqrt{a^{2}+b^{2}+c^{2}}
\end{aligned}
$$
Thus
$$
D=\frac{\left|a\left(x_{0}-x_{1}\right)+b\left(y_{0}-y_{1}\right)+c\left(z_{0}-z_{1}\right)\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}
$$
Since the point $Q\left(x_{1}, y_{1}, z_{1}\right)$ lies in the given plane, its coordinates satisfy the equation of that plane; thus
$$
a x_{1}+b y_{1}+c z_{1}+d=0
$$
or
$$
d=-a x_{1}-b y_{1}-c z_{1}
$$