\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Find the distance of the point(1,1,1) from the plane $x+2y+3z=6$.
Write down the vector equation of the plane.

\medskip

{\bf Answer}

${}$

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Let ${\bf{OP}}$ be ${\bf{i}},\ {\bf{j}},\ {\bf{k}}$ and
${\bf{PQ}}$ be the perpendicular distance from $P$ to the plane.

We require $|{\bf{PQ}}|$.

${\bf{PQ}}$ is in the direction of ${\bf{n}}$ the normal to the
plane, by definition.

Now the equation of the plane is

$$x+2y+3z=6$$

which can be rewritten as

$$(x{\bf{i}}+y{\bf{j}}+z{\bf{k}}) \cdot
({\bf{i}}+2{\bf{j}}+3{\bf{k}})=6$$

${\bf{n}}$ is thus
\begin{eqnarray*} \ds\frac{({\bf{i}}+2{\bf{j}}+3{\bf{k}})}
{|({\bf{i}}+2{\bf{j}}+3{\bf{k}})|} & = &
\ds\frac{({\bf{i}}+2{\bf{j}}+3{\bf{k}})}{\sqrt{1+4+9}}\\ & = &
\ds\frac{1}{\sqrt{14}}{\bf{i}}+\ds\frac{2}{\sqrt{14}}{\bf{j}}+\ds\frac{3}{\sqrt{14}}{\bf{k}}
\end{eqnarray*}

Thus ${\bf{r}} \cdot
\left(\ds\frac{1}{\sqrt{14}}{\bf{i}}+\ds\frac{2}{\sqrt{14}}{\bf{j}}
+\ds\frac{3}{\sqrt{14}}{\bf{k}}\right)=\ds\frac{6}{\sqrt{14}}$

A vector equation of the plane.

This, $\left(\ds\frac{6}{\sqrt{14}}\right)$, is the perpendicular
distance of 0 from the plane $d=|{\bf{OR}}|$.

Clearly $|{\bf{PQ}}|=|{\bf{OR}}|-|{\bf{OP}}| \cos \theta$

What is $|{\bf{OP}}| \cos \theta$?

$$|{\bf{OP}}| \cdot \hat {\bf{n}}=|{\bf{OP}}||\hat {\bf{n}}|\cos
\theta=|{\bf{OP}}|\cos \theta$$

Thus

\begin{eqnarray*} |{\bf{PQ}}| & = & |{\bf{OR}}|-{\bf{OP}} \cdot
\hat{\bf{n}}\\ & = & d-({\bf{OP}} \cdot \hat{\bf{n}})\\ & = &
\ds\frac{6}{\sqrt{14}}-({\bf{i}}+{\bf{j}}+{\bf{k}}) \cdot
\left(\ds\frac{1}{\sqrt{14}}{\bf{i}}+\ds\frac{2}{\sqrt{14}}{\bf{j}}
+\ds\frac{3}{\sqrt{14}}{\bf{k}}\right)\\ & = &
\ds\frac{6}{\sqrt{14}}-\left(\ds\frac{1}{\sqrt{14}}{\bf{i}}+\ds\frac{2}
{\sqrt{14}}{\bf{j}}+\ds\frac{3}{\sqrt{14}}{\bf{k}}\right)\\ & = &
\un{0}!!! \end{eqnarray*}

Thus it looks like (1,1,1) is in the plane.  Ooops it is!

Since if

\begin{eqnarray*} x+2y+3z & = & 6\\ \Rightarrow 1+2+3 & = & 6\\
\Rightarrow 6 & = & 6\end{eqnarray*}

(if $(x,y,z)=(1,1,1)$)

Thus the distance of the point (1,1,1) from the plane is 0, since
it lies in it!!!


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