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\newcommand\ds{\displaystyle}
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QUESTION
\begin{description}

\item[(i)]
Find the vector equation of the line $L$ that passes through the
two points $A(1,-2,1)$ and $B(2,3,1)$.

\item[(ii)]
A plane $P$ passes through the three points $C(2,1,-3),\
D(4,-1,2)$ and $E(3,0,1)$. Obtain two independent vectors that are
parallel to $P$ and hence, or otherwise, show that $P$ has the
vector equation $$\textbf{r}.(1,1,0)=3$$

\item[(iii)]
Find the coordinates of the point of intersection of the line $L$
and the plane $P$.

\item[(iv)]
Derive the vector equation of the plane that contains the line $L$
and is perpendicular to the plane $P$.

\end{description}




ANSWER

\begin{description}

\item[(i)]
$\vec{AB}=(2-1,3-(-2),1-1)=(1,5,0).$ So the equation of $L$ is
$\textbf{r}=(1,-2,1)+s(1,5,0)=(1+s,-2+5s,1)$

\item[(ii)]
$\vec{CD}=(4-2,-1-1,2-(-3))=(2,-2,5)\\
\vec{CE}=(3-2,0-1,1-(-3))=(1,-1,4)\\ \textbf{n}=\vec{CD}\times
\vec{CE}=(-8-(-5),5-8,-2-(-2))=(-3,-3,0)$ \\The equation of the
plane is $\textbf{r.n}=c$ i.e. $\textbf{r}.(-3,-3,0)=c$

$C(2,1,-3)$ lies on the plane,

 therefore
$c=(2,1,-3).(-3,-3,0)=-6-3+0=-9.$

So the equation of the plane is $\textbf{r}.(-3,-3,0)=-9$ or
$\textbf{r}.(1,1,0)=3$.

\item[(iii)]
The line $L$ meets the plane $P$ when $(1+s,-2+5s,1).(1,1,0)=3.$
Therefore $1+s-2+5s+0=6s-1=3,$ with solution $s=\frac{2}{3},$ so
the point of intersection
=$\ds\left(\frac{5}{3},\frac{4}{3},1\right)$

\item[(iv)]
We need to find a second plane parallel to $L$ and \textbf{n}.\\
The normal vector is in the direction
$(1,5,0)\times(1,1,0)=(0,0,-4)$ \\The equation of the plane is
$\textbf{r}.(0,0,-4)=k.$

$A$ is on the plane, so $k=(1,-2,1).(0,0,-4)=-4.$

Hence the equation of the plane is $\textbf{r}.(0,0,-4)=-4$ or
$\textbf{r}.(0,0,1)=1$

\end{description}




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