<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}


{\bf Question}

The cartesian co-ordinates of the points $A,B,C$ are $(-1,1,0),\,
(1,4,6),$

$(3,5,7)$ respectively.  Find
\begin{description}
\item[(i)] The components of $\vec{AB}$ and $\vec{AC}$.
\item[(ii)] The direction cosines of line $BC$.
\item[(iii)] The parametric form of the equation $BC$ and give its
cartesian form.
\item[(iv)] The parametric form of the equation of the plane
$\pi$ containing $A,B,C$.
\item[(v)] The sines of the angle $BAC$.
\item[(vi)] The components of the unit vector $\hat{{\bf n}}$
perpendicular to the plane $\pi$ such that $\vec{AB}, \vec{AC},
\hat{{\bf n}}$ form a right-handed system.
\item[(vii)] The \lq normal \rq form the equation of the plane
$\pi$ and check it agrees with part (iv)
\item[(viii)] What is the shortest distance from $O$ to the plane
$\pi$?
\item[(xi)] The shortest distance between the lines $BC$ and $OA$.
\item[(x)] The equation of the line perpendicular to both $BC$ and
$OA$.
\end{description}

\vspace{.25in}

{\bf Answer}

$A = (-1,1,0),\, B = (1,4,6), \, C = (3,5,7)$

\begin{description}
\item[(i)] $\vec{AB} = (2,3,6)$

$\vec{AC} = (4,4,7)$

${}$
\item[(ii)] $\ds\vec{BC} = (2,1,1) \hspace{.2in} {\rm So\ }
\hat{BC} = \left( \frac{2}{\sqrt 6}, \frac{1}{\sqrt 6},
\frac{1}{\sqrt 6}\right)$

${}$
\item[(iii)] $\ds {\bf r} = (1,4,6) + t(2,1,1)$

$\ds \frac{x-1}{2} = \frac{y-4}{1} = \frac{z-6}{1}$

${}$
\item[(iv)] $r = \vec{OA} + u\vec{AB} + u\vec{AC} =
(-1,1,0)+u(2,3,6) + v(4,4,7)$

$x = -1 + 2 + 4v \hspace{.2in} y = 1 + 3u + 4v \hspace{.2in} z =
6u + 7v$

${}$
\item[(v)] $\ds \cos BAC = \frac{AB \cdot AC}{|AB||AC|} =
\frac{62}{63}$

${}$
\item[(vi)] $\ds \vec{AB} \times\vec{AC} = (-3,10,-4) \hspace{.2in}
\hat{n} = \left(-\frac{3}{\sqrt{125}}, \frac{10}{\sqrt{125}},
-\frac{4}{\sqrt{125}} \right)$

${}$
\item[(vii)] The equation of $\pi$ is $-3x + 10y-4z=k$

$(-1,1,0) \epsilon \pi$ So $k = 13$

Check with (iv) $-3(1 +2u +4v) + 10(1 + 3u + 4v) - 4(6u + 7v) = 13
\surd$

${}$
\item[(viii)] Shortest distance from ${\bf p}$ to ${\bf r \cdot a}
= k$ is $\ds\left| \frac{a \cdot p - k}{|a|} \right|$ so when $p=0
\, \,$

$\ds d = \frac{|k|}{|a|}  =\frac{13}{|(-3,10,-4)|} =
\frac{13}{\sqrt{125}}$

${}$
\item[(xi) \&amp; (x)]${}$


The line $BC$ has parametric equation ${\bf r}=(1,4,6) + t(2,1,1)
\hspace{.2in} -L$

The line $OA$ has parametric equation ${\bf r}=(0,0,0) + t(-1,1,0)
\hspace{.2in} -M$

Let $P,Q$ be points on $L,M$

$\vec{QP} = (1+2k+l, \, 4+k-l, \, 6+k)$

We want $\vec{QP} \cdot(2,1,1)=0$ and $\vec{QP} \cdot (-1,1,0)=0$

$\begin{array}{rcc}{\rm So\ } 2+4k+2;+4+k-l+6+k=0 &amp; &amp; 6k+l=-12 \\
-1-2k-l+4+k-l=0 &amp; &amp; -k-2l=-3 \end{array}$

So $4k = -\frac{27}{11} \hspace{.2in} l = \frac{30}{11}
\hspace{.2in} P=\left(-\frac{43}{11},\frac{17}{11}, \frac{39}{11}
\right) \, Q = \left(-\frac{30}{11}, \frac{30}{11},0\right)$

$\vec{QP} = \frac{13}{11}(-1,-1,3)$ so $|QP| = \frac{13}{11} \cdot
\sqrt{11}$

The equation of $QP$ is ${\bf r} = \left(-\frac{43}{11},
\frac{17}{11}, \frac{39}{11} \right) + t \left(-\frac{30}{11},
\frac{30}{11},0\right)$

$\ds \frac{x+\frac{30}{11}}{-\frac{13}{11}} =
\frac{y-\frac{30}{11}}{-\frac{13}{11}} = \frac{z}{\frac{39}{11}}
{\rm \ \ or\ \ } 11x + 30 = 11y -  30 = -\frac{11}{3} z$
\end{description}

\end{document}
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