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QUESTION

A line $L_1$ passes through the two points $A(1,2,-1)$     and
$B(2,-3,1)$. A second line $L_2$, which is parallel to the vector
$(0,-2,1),$ passes through the point $C(2,1,-1)$.

\begin{description}

\item[(i)]
Obtain the vector equations of the lines $L_1$ and $L_2$ and show
that they intersect at the point $B$.

\item[(ii)]
Find a vector $\mathbf{n}$ which is perpendicular to $L_1$ and
$L_2$, and hence obtain the vector equation of the plane $P_1$
which contains the lines $L_1$ and $L_2$ and passes through $B$.

\item[(iii)]
Obtain the vector equation of a second plane $P_2$ which is
parallel to $P_1$ and passes through $D(-2,1,-1)$.

\item[(iv)]
Find the distance between the planes $P_1$ and $P_2$.

\end{description}




\bigskip

ANSWER

$A(1,2,-1),\ B(2,-3,1),\ C(2,1,-1)$

\begin{description}

\item[(i)]
$L_1:\ \vec{AB}=(2-1,-3-2,1-(-1))=(1,-5,2)$ so the equation of the
line is\\ $\mathbf{r}=(1,2,-1)+s(1,-5,2)=(1+s,2-5s,-1+2s)\\ L_2:\
\mathbf{r}=(2,1,-1)+t(0,-2,1)=(2,1-2t,-1+t)$\\ The lines intersect
when $(1+s,2-5s,-1+2s)=(2,1-2t,-1+t)$\\ i.e. $1+s=2\Rightarrow
s=1\\2-5s=1-2t\Rightarrow 2t=5s-1=5(1)-1=4\Rightarrow t=2$\\
{[}Check :\ $-1+2s=-1+2=1;\ -1+t=-1+2=1${]}

Therefore the point of intersection is $(1+1,2-5,-1+2)=(2,-3,1)$

\item[(ii)]
A vector parallel to $L_1$ is $(1,-5,2)$\\ a vector parallel to
$L_2$ is $(0,-2,1)$\\ Therefore
\begin{eqnarray*}
\textbf{n}&=&(1,-5,2)\times(0,-2,1)\\&=&(-5(1)-2(-2),2(0)-1(1),1(-2)-0(-5))\\
&=&(-5+4,0-1,-2-0)=(-1,-1,-1)
\end{eqnarray*}
The equation of $P_1$ is $\mathbf{r\cdot n}=C_1.$\\ $A$ lies on
the plane so
\begin{eqnarray*}
C_1&=&{\bf a}\cdot{\bf
n}=(1,2,-1).(-1,-1,-2)\\&=&1(-1)+2(-1)+(-1)(-2)=-1-2+2=-1
\end{eqnarray*}
Therefore $\mathbf{r}\cdot(-1,-1,-2)=-1$ or
$\mathbf{r}\cdot(1,1,2)=1$.

\item[(iii)]
Parallel planes have the same normal so the equation of the plane
is

$\mathbf{r}\cdot(-1,-1,-2)=k$\\ $D$ is on the plane so
\begin{eqnarray*}k&=&(-2,1,-1)\cdot(-1,-1,-2)\\&=&-2(-1)+1(-1)-1(-2)=2-1+2=3
\end{eqnarray*}
i.e. $\mathbf{r}\cdot(-1,-1,-2)=3$ or $\mathbf{r}\cdot(1,1,2)=-3$

\item[(iv)]
There are many ways to obtain this answer, for example

\setlength{\unitlength}{.5in}
\begin{picture}(7,4)
\put(1,0){\line(1,0){3}} \put(1,0){\line(1,1){1}}
\put(2,1){\line(1,0){3}} \put(4,0){\line(1,1){1}}
\put(0.5,0.5){$P_2$} \put(2.5,0.5){$D$}

\put(1,2){\line(1,0){3}} \put(1,2){\line(1,1){1}}
\put(2,3){\line(1,0){3}} \put(4,2){\line(1,1){1}}
\put(0.5,2.5){$P_1$} \put(2,2.2){$A$} \put(3,2.7){$B$}
\put(3.7,2.5){$C$} \put(4.5,2.7){\vector(0,1){1}}
\put(4.6,3.6){\bf n}

\end{picture}



\begin{eqnarray*}
\textrm{distance}&=&|\vec{DB}\cdot\hat{\mathbf{n}}|\\
&=&\left|(2-(-2),-3-1,1-(-1))\cdot\frac{(1,1,2)}{\sqrt{1^2+1^2+2^2}}\right|\\
&=&|(4,-4,2)\cdot(1,1,2)|\frac{1}{\sqrt{6}}\\
&=&\frac{4-4+4}{\sqrt{6}}=\frac{4}{\sqrt{6}}
\end{eqnarray*}

\end{description}



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