## Proof of Formula For the Least Distance Between Plane and Origin

The normal
$\vec{n}$
to the plane
$ax+by+cz=d$
is
$\begin{pmatrix}a\\b\\c\end{pmatrix}$

. The line
$r(t)=\begin{pmatrix}a\\b\\c\end{pmatrix} t$
is perpendicular to the plane and passes through the origin passes through the origin. For this line, the
$x, \; y, \; z$
coordinates are
$at, \; bt, \; ct$
respectively. Substitute these into the equation of the plane.
$a(at)+b(bt)+c)ct)=d \rightarrow t=\frac{d}{a^2+b^2+c^2}$
.
Substitute this value of
$t$
into the equation of the line to get the coordinate of the intersection of the line with the plane.
$\begin{pmatrix}a\\b\\c\end{pmatrix}^T \frac{d}{a^2+b^2+c^2}= (\frac{a}{a^2+b^2+c^2}, \frac{b}{a^2+b^2+c^2}, \frac{c}{a^2+b^2+c^2} )$
.
The distance between this point and the origin
$(0,0,0)$
is
\begin{aligned} d &=\sqrt{(\frac{a}{a^2+b^2+c^2} -0)^2+(\frac{b}{a^2+b^2+c^2} -0)^2+(\frac{c}{a^2+b^2+c^2} -0)^2} \\ &= \sqrt{\frac{d^2(a^2+b^2+c^2)}{(a^2+b^2+c^2)^2}} \\ &= \frac{d}{\sqrt{a^2+b^2+c^2}}\end{aligned}

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