\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}


{\bf Question}

The three vectors ${\bf a, b, c}$ are linearly independent.
\begin{description}
\item[(a)] What does the equation $${\bf r} = (1+2u -v){\bf a} +
(3v - 6u){\bf b + c}$$ $u,v \epsilon {\bf R}$ represent?
\item[(b)] When ${\bf p} = {\bf a} + 2{\bf c}, \, {\bf q} = 2{\bf a}
-{\bf b}+ 3{\bf c}, \, {\bf s} = 2{\bf b} -3{\bf a}- 4{\bf c}$ are
${\bf p, q, s}$ linearly independent?  If ${\bf p, q, s}$ are
position vectors of the points $P,Q,S$ are $P,Q,S$ colinear?
\item[(c)] Find the point(s) of intersection of the line $${\bf r} = (1+t){\bf a}
+ (t-2){\bf b}+ (2-t){\bf c} \hspace{.5in} t\epsilon {\bf R}$$
with
\begin{description}
\item[(i)]
The lines

$\begin{array}{rlllr}  (a) & {\bf r} & = & (2+u){\bf a} +
(u-1){\bf b}+ (2u+1){\bf c}  & \hspace{.5in} u\epsilon {\bf R}
\\ (b) &  {\bf r} & = &  (1+u)(2{\bf a} + {\bf b}) + (3+u){\bf c}
& u\epsilon {\bf R} \\ (c) & {\bf r} & = &  u{\bf a} + (u-3){\bf
b}+ (3-u){\bf c} & u\epsilon {\bf R}
\end{array}$
\item[(ii)]

The planes

$\begin{array}{rlllr}  (a) &  {\bf r} & = & {\bf a} + u{\bf b}+
v{\bf c} & \hspace{.5in} u,v\epsilon {\bf R}
\\ (b) & {\bf r} & = &  (1+u){\bf a} + (u-2){\bf b} + v{\bf c} &
u,v\epsilon {\bf R} \\ (c) & {\bf r} & = &  (1+v){\bf a} +
(u-2){\bf b}+ (3-v){\bf c} & \hspace{.5in} u,v\epsilon {\bf R}
\end{array}$
\end{description}
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
\begin{eqnarray*} {bf r} & = & (1 + 2u - v) {\bf a} + (3v -
6v){\bf b} + {\bf c} \\ & = & {\bf a+ c} + u(2{\bf a} - 6{\bf b})
+ v(3{\bf b-a}) \\ & = &{\bf a + c} + (2u - v)({\bf a } - 3{\bf
b})
\end{eqnarray*}  This is a line through ${\bf a + c}$ with
direction vector ${\bf a} - 3{\bf b}$

\item[(b)]
$$ {\bf p} = {\bf a} + 2{\bf c} \hspace{.3in} {\bf q} = 2{\bf a -
b} + 3{\bf c} \hspace{.3in} {\bf s} = 2{\bf b} - 3{\bf a} - 4{\bf
c}$$

$$\alpha{\bf p} + \beta{\bf q} + \gamma{\bf s} = {\bf
a}(\alpha+2\beta-3\gamma) + {\bf b}(-\beta+2\gamma) + {\bf
c}(2\alpha+3\beta-4\gamma) = {\bf 0}$$ if and only if


$\left.\begin{array}{rcl}\alpha+2\beta-3\gamma & = & 0 \\
-\beta+2\gamma & = & 0 \\ 2\alpha+3\beta-4\gamma &  = & 0
\end{array}\right\}$ Solution $(-\gamma,2\gamma,\gamma)$

So ${\bf p, \, q, \, s,}$ are not linearly independent.

$\alpha+\beta+\gamma=2\gamma=0$ iff $\alpha=\beta=\gamma=0.$  So
$PQR$ are not collinear.


\item[(c)]

${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

(in standard form ${\bf r} = {\bf a} - 2{\bf b} + 2{\bf c} +
t({\bf a+b-c})$)

\begin{description}
\item[(i)]
\begin{description}
\item[(a)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r}=(2+u){\bf a} + (u-1){\bf b}+ (2u+1){\bf c}$

These lines meet where

$\begin{array}{rclrclrcl} 1 + t & = & 2+u & t-2 & = & u-1 & 2-t &
= & 2u+1 \\ t & = & 1+u & t & = & 1+u & t&  = & 1-2u \end{array}$

So we require $1+u = 1+2u$ $v=0$ so $t=1$

Thus the lines meet at ${\bf r} = 2{\bf a} - {\bf b} + {\bf c}$

\item[(b)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r} = (1+u)(2{\bf a} + {\bf b}) + (3+u){\bf c}$

These meet where $$(1+t) = (2+2u) \hspace{.2in} \underbrace{(t-2)
= (1+u) \hspace{.2in}(2-t) = 3+u}_{\ds 1 + u = -3-3u \Rightarrow u
= -2, \,  t=1} $$ \\ which doesn't fit 1st equation.

Therefore the lines do not meet.


\item[(c)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r}  =   u{\bf a} + (u-3){\bf b}+ (3-u){\bf c}$

$u = 1+t$ so the lines are identical.
\end{description}
\item[(ii)]
\begin{description}
\item[(a)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r}  =  {\bf a} + u{\bf b}+ v{\bf c}$

$\begin{array}{rclrclrcl} 1+t & = & 1, & t - 2 & = & u, & 2-t & =
& v \\ t & = & 0 , u & = & -2, & v & = & 2 \end{array}$

So ${\bf r} = {\bf a} - 2{\bf 2} + 2{\bf c}$ is the point of
intersection.

\item[(b)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r} = (1+u){\bf a} + (u-2){\bf b} + v{\bf c}$

$t=u, \, v = -t+2$ So the line lies in the plane.

\item[(c)] ${\bf r} = (1+t){\bf a} + (t-2){\bf b}+ (2-t){\bf c}$

${\bf r}  =   (1+v){\bf a} + (u-2){\bf b}+ (3-v){\bf c}$

$t = v , \, t - u \, t = v-1$

So the line does not meet the plane.  It is parallel to the plane.
\end{description}
\end{description}

\end{description}
\end{document}