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

{\bf Question}

Two geometrical objects are defined by the cartesian equations

$$2x+7y+5z=3$$

$$\ds\frac{x+3}{2}=\ds\frac{y-5}{-1}=\ds\frac{z-2}{-3}$$

One is a line, and the other is a plane. State which is which.

Convert the cartesian equations to vector equations and hence find
the intersection between the objects.

Find the angle between the line and a normal to the plane.


\medskip

{\bf Answer}

$2x+7y+5z=3$ is a plane

$\ds\frac{x+3}{2}=\ds\frac{y-5}{-1}=\ds\frac{z-2}{-3}$ is a line

Plane can be rewritten as

$$(2,7,6) \cdot (x,y,z)=3$$

for general ${\bf{r}}=(x,y,z)$

i.e.,

$$\un{{\bf{r}} \cdot (2{\bf{i}}+7{\bf{j}}+5{\bf{k}})=3}$$

For the line:

If, $\ds\frac{x+3}{2}=\lambda,\ \ds\frac{y-5}{-1}=\lambda$ and
$\ds\frac{z-2}{-3}=\lambda$ simultaneously

Therefore $\left\{ \begin{array} {rcl} x & = & 2\lambda-3\\ y & =
& -\lambda+5\\ z & = & -3\lambda+2 \end{array} \right.$

Therefore if ${\bf{r}}=x{\bf{i}}+y{\bf{j}}+z{\bf{k}}$,

${\bf{r}}=(2\lambda-3){\bf{i}}+(5-\lambda){\bf{j}}+(2-3\lambda){\bf{k}}$

Therefore
\un{${\bf{r}}=-3{\bf{i}}+5{\bf{j}}+2{\bf{k}}+\lambda(2{\bf{i}}-{\bf{j}}-3{\bf{k}})$}

\newpage
Intersection is when

$[(-3{\bf{i}}+5{\bf{j}}+2{\bf{k}})+\lambda(2{\bf{i}}-{\bf{j}}-3{\bf{k}})]
\cdot (2{\bf{BA}}+7{\bf{BA}}+5{\bf{BA}})=3$

$\Rightarrow (2\lambda-3) \times 2+(5-\lambda) \times
7+(2-3\lambda) \times 5=3$

$\Rightarrow 4\lambda-6+35-7\lambda+10-15\lambda=3$

$\Rightarrow -18\lambda+39=3$

$\Rightarrow -18\lambda=-36$

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

Therefore intersection at

${\bf{r}}=(2\times
2-3){\bf{i}}+(5-2){\bf{j}}+(2-3\times2){\bf{k}}$

$$\un{{\bf{r}}={\bf{i}}+3{\bf{j}}-4{\bf{k}}}$$

Normal to plane is given by a vector

$$2{\bf{i}}+7{\bf{j}}+5{\bf{k}}\ \ \rm{or}\ \
\ds\frac{2{\bf{i}}+7{\bf{j}}+5{\bf{k}}}{\sqrt{4+49+25}}=\hat{\bf{n}}$$

Direction of line is given by vector

$$2{\bf{i}}-{\bf{j}}-3{\bf{k}}\ \ \rm{or}\ \
\ds\frac{2{\bf{i}}-{\bf{j}}-3{\bf{k}}}{\sqrt{4+1+9}}=\hat{\bf{d}}$$

\setlength{\unitlength}{.5in}

\begin{picture}(4,4)

\put(1,0){\vector(0,1){3}}

\put(0,0){\vector(3,2){3}}

\put(1,3){\makebox(0,0)[b]{$\hat{\bf{n}}$}}

\put(3,2){\makebox(0,0)[l]{$\hat{\bf{d}}$}}

\put(1.2,1){\makebox(0,0){$\theta$}}
\end{picture}

So by scalar product

\begin{eqnarray*} {\bf{n}} \cdot {\bf{d}} & = &
|{\bf{n}}||{\bf{d}}|\cos\theta\\ \rm{or}\ \hat{\bf{n}} \cdot
\hat{\bf{n}} & = & \cos\theta \end{eqnarray*}

\newpage
$\Rightarrow$

\begin{eqnarray*} \cos\theta & = & \ds\frac{(2,7,5) \cdot
(2,-1,-3)}{\sqrt{4+49+25}\sqrt{4+1+9}}\\ & = &
\ds\frac{-18}{\sqrt{78}\sqrt{14}}\\ & = & -0.544705
\end{eqnarray*}

$\Rightarrow \un{\theta=123.004 ^{\circ}}$ (or
$180-123.004=56.9955^{\circ}$)

\end{document}