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\newcommand\ds{\displaystyle}

\begin{document}

\parindent=0pt

QUESTION

A line $L_1$ passes through the two points $A(2,-4,1)$ and
$B(5,2,-2)$ and a second line $L_2$ passes through the two points
$C(2,-2,5)$ and $D(2,-1,7)$.

\begin{description}

\item[(i)]
Obtain the vector equations of the lines $L_1$ and $L_2$.

\item[(ii)]
Find the point of intersection of $L_1$ and $L_2$.

\item[(iii)]
Show that the lines $L_1$ and $L_2$ are perpendicular.

\item[(iv)]
Find the area of the triangle $ABD$.

\item[(v)]
Find the acute angle between the line $L_2$ and the line joining
the points $B$ and $D$.

\end{description}

\bigskip

ANSWER

\setlength{\unitlength}{.3in}
\begin{center}
\begin{picture}(10,7)

\put(0,1){\line(5,2){10}}

\put(7,.5){\line(-1,2){3}}

\put(1,1){$A(2,-4,1)$}

\put(1,1.3){$\bullet$}

\put(9,4.2){$B(5,2,-2)$}

\put(8.8,4.4){$\bullet$}

\put(6.7,1.25){$C(2,-2,5)$}

\put(6.4,1.2){$\bullet$}

\put(4.5,5.5){$D(2,-1,7)$}

\put(4.2,5.7){$\bullet$}

 \put(0,0.5){$L_1$}

\put(7.1,0.5){$L_2$}
\end{picture}
\end{center}

\begin{description}

\item[(i)]
$\vec{AB}=(5-2,2-(-4),-2-2)=(3,6,-3)$

Therefore the equation of $L_1$ is $\textbf{r}=(2,-4,1)+s(3,6,-3)$

$\vec{CD}=(2-2,-1-(-2),7-5)=(0,1,2)$

Therefore the equation of $L_2$ is $\textbf{r}=(2,-2,5)+t(0,1,2)$

\item[(ii)]
At the point of intersection
\begin{eqnarray*}
2+3s&amp;=&amp;2,\ \Rightarrow s=0\\ -4+6s&amp;=&amp;-2+t,\Rightarrow t=-2\\
1-3s&amp;=&amp;5+2t\\
\end{eqnarray*}
The third equation is satisfied by $s=0,t=-2$ therefore the point
of intersection is $(2,-4,1)+0(3,6,-3)=(2,-4,1),$ i.e. point $A$.

\item[(iii)]
$L_1$ is parallel to $(3,6,-3),$\ $L_2$ is parallel to $(0,1,2).$

$(3,6,-3).(0,1,2)=0+6-6=0$ therefore the lines are perpendicular.

\item[(iv)]
Since $L_1$ and $L_2$ intersect at $A$, area
$ABD=\frac{1}{2}(AB)(AD)$\\ But\\
$AB=\{(5-2)^2+(2-(-4))^2+(-2-1)^2\}^\frac{1}{2}=\{9+36+9\}^\frac{1}{2}=\sqrt{54}\\
AB=\{(2-2)^2+(-1-(-4))^2+(7-1)^2\}^\frac{1}{2}=\{0+9+36\}^\frac{1}{2}=\sqrt{45}$\\
Therefore the area is
$\frac{1}{2}\sqrt{54}\sqrt{45}=\frac{1}{2}3\sqrt{6}\
3\sqrt{5}=\frac{9}{2}\sqrt{30}\sim24.65$

\item[(v)]
$\vec{BD}=(2-5,-1-2,7-(-2))=(-3,-3,9)=3(-1,-1,3)\\ L_2$ is
parallel to $(0,1,2)$ therefore
$$\cos\theta=\frac{(-1,-1,3).(0,1,2)}{\sqrt{((-1)^2+(-1)^2+3^2)}\sqrt{(1^2+3^2)}}=\frac{0-1+6}{\sqrt{11}\sqrt{5}}=\frac{5}{\sqrt{55}}$$
$\theta=47.6^{\circ}(=0.831 $ radians)

\end{description}




\end{document}
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