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\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
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\begin{document}

{\bf Question}
\begin{description}
\item[(i)]
Convert the cartesian equation of the plane

$$x-2y+2z=5$$

into a vector equation of the form ${\bf{r}} \cdot
\hat{{\bf{n}}}={\bf{d}}$, where $\hat{{\bf{n}}}$ is a unit vector.

What is the geometric significance of $\hat{{\bf{n}}}$?

State the perpendicular distance of the plane from the origin.

\item[(ii)]
Find the intersections (if any) of the above plane with the
following lines

\begin{description}
\item[(a)]
${\bf{r}}=-5{\bf{i}}+3{\bf{j}}-{\bf{k}}+\lambda({\bf{i}}-2{\bf{j}}+2{\bf{k}})$

\item[(b)]
${\bf{r}}=6{\bf{i}}+{\bf{j}}+{\bf{k}}+\mu(2{\bf{i}}+{\bf{j}})$
\end{description}

Comment on result of (b).
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
$x-2y+2z=5$

Set ${\bf{r}}=x{\bf{i}}+y{\bf{j}}+z{\bf{k}}$ and we have

$${\bf{r}} \cdot (\underbrace{{\bf{i}}-2{\bf{j}}+2{\bf{k}}})=5$$

                       This is not a unit vector

so \begin{eqnarray*} \hat{\bf{n}} & = &
\ds\frac{{\bf{i}}-2{\bf{j}}+2{\bf{k}}}{|{\bf{i}}-2{\bf{j}}+2{\bf{k}}|}\\
& = & \ds\frac{{\bf{i}}-2{\bf{j}}+2{\bf{k}}}{\sqrt{1+4+4}}\\ & = &
\ds\frac{1}{3}{\bf{i}}-\ds\frac{2}{3}{\bf{j}}+\ds\frac{2}{3}{\bf{k}}
\end{eqnarray*}

so divide both sides by 3 and get

$${\bf{r}} \cdot
\left(\underbrace{\ds\frac{1}{3}{\bf{i}}-\ds\frac{2}{3}{\bf{j}}
+\ds\frac{2}{3}{\bf{k}}}\right)=\underbrace{\ds\frac{5}{3}}$$

              $\hat{\bf{r}}$           $d$

$\hat{\bf{n}}$ is \un{a unit normal to the plane}

$d$ is the perpendicular distance of the plane from the origin.

So here $d=\ds\frac{5}{3}$

\item[(ii)]
\begin{description}
\item[(a)]
We have

$[-{\bf{i}}+3{\bf{j}}-{\bf{k}}+\lambda({\bf{i}}-2{\bf{j}}+2{\bf{k}})]
\cdot \left[\ds\frac{1}{3}{\bf{i}}-\ds\frac{2}{3}{\bf{j}}
+\ds\frac{2}{3}{\bf{k}}\right]=\ds\frac{5}{3}$

$\Rightarrow (-5+\lambda,3-2\lambda,-1+2\lambda) \cdot
\left(\ds\frac{1}{3},-\ds\frac{2}{3},\ds\frac{2}{3}\right)=\ds\frac{5}{3}$

$\Rightarrow
\ds\frac{-5+\lambda}{3}-\ds\frac{2}{3}(3-2\lambda)+\ds\frac{2}{3}(-1+2\lambda)=\ds\frac{5}{3}$

$\Rightarrow -5+\lambda-6+4\lambda-2+4\lambda=5$

$\Rightarrow 9\lambda=18$

$\Rightarrow \un{\lambda=2}$

So there exists a single intersection point of line and plane at

\begin{eqnarray*}
{\bf{r}} & = &
-5{\bf{i}}+3{\bf{j}}-{\bf{k}}+2({\bf{i}}-2{\bf{j}}+2{\bf{k}})\\ &
= & \un{-3{\bf{i}}-{\bf{j}}+3{\bf{k}}} \end{eqnarray*}

\item[(b)]
We have

$[(6+2\mu){\bf{i}}+(1+\mu){\bf{j}}+{\bf{k}}] \cdot
\left[\ds\frac{1}{3}{\bf{i}}-\ds\frac{2}{3}{\bf{j}}+
\ds\frac{2}{3}{\bf{k}}\right]=\ds\frac{5}{3}$

Therefore $6+2\mu-2(1+\mu)+2=5$

Therefore $6+2\mu-2-2\mu+2=5$

$\Rightarrow 6=5$!!

Therefore no value of $\mu$ can satisfy this equation so no
intersection. Line must be parallel to plane.

\end{description}
\end{description}
\end{document}