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{\bf Question}

Find the vector and Cartesian equation of the plane containing the
point $3{\bf{i}}-{\bf{k}}$ and also containing the vectors
${\bf{a}},\ {\bf{b}}$ where

${\bf{a}}={\bf{i}}+2{\bf{j}}+{\bf{k}}$,

${\bf{b}}=-{\bf{i}}+2{\bf{j}}+2{\bf{k}}$.

\medskip

{\bf Answer}

${}$

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${\bf{a}}={\bf{i}}+2{\bf{j}}+{\bf{k}}$

${\bf{b}}=-{\bf{i}}+2{\bf{j}}+2{\bf{k}}$

Vector equation is

${\bf{r}}={\bf{c}}+\lambda{\bf{a}}+\mu {\bf{b}}$

$\Rightarrow \un{{\bf{r}}=3{\bf{i}}-{\bf{k}}+\lambda({\bf{i}}
+2{\bf{j}}+{\bf{k}})+\mu(-{\bf{j}}+2{\bf{j}}+2{\bf{k}})}$

Cartesian equation is given by the set of points
${\bf{r}}=(x,y,z)$ satisfying $(x,y,z) \cdot \hat{\bf{n}}=d$ where
$\hat {\bf{n}}$ is a unit vector to the plane and $d$ is the
perpendicular distance from the origin(|$OP$|). First we need
$\hat {\bf{n}}$. This is given by $\ds\frac{{\bf{a}} \times
{\bf{b}}}{|{\bf{a}} \times {\bf{b}}|}$, since ${\bf{a}},\
{\bf{b}}$ lie in the plane.

\begin{eqnarray*} & & {\bf{a}}\times{\bf{b}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 1 & 2 & 1\\ -1 & 2 & 2\end{array}\right|\\ &
= & {\bf{i}}(2 \times 2)-{\bf{i}}(2 \times 1)-{\bf{j}}(1 \times
2)\\ & & +{\bf{k}}(1 \times 2)+{\bf{j}}(-1 \times 1)-{\bf{k}}(-1
\times 2)\\ & = &
4{\bf{i}}-2{\bf{i}}-2{\bf{j}}+2{\bf{k}}-{\bf{j}}+2{\bf{k}}\\ & = &
2{\bf{i}}-3{\bf{j}}+4{\bf{k}}
\end{eqnarray*}

$\ds\frac{{\bf{a}} \times {\bf{b}}}{|{\bf{a}} \times
{\bf{b}}|}=\ds\frac{2{\bf{i}}-3{\bf{j}}+4{\bf{k}}}{\sqrt{2^2+3^2+4^2}}
=\ds\frac{2}{\sqrt{29}}{\bf{i}}-\ds\frac{3}{\sqrt{29}}{\bf{j}}+\ds\frac{4}{\sqrt{29}}{\bf{k}}$

What is $d$? It's given by ${\bf{c}} \cdot \hat {\bf{n}}$ (see
diagram above) $\Rightarrow$ \begin{eqnarray*} d
 & = & (3{\bf{i}}-{\bf{k}}) \cdot
\left(\ds\frac{2}{\sqrt{29}}{\bf{i}}-\ds\frac{3}{\sqrt{29}}{\bf{j}}
+\ds\frac{4}{\sqrt{29}}{\bf{k}}\right)\\ & = &
\ds\frac{6}{\sqrt{29}}-\ds\frac{4}{\sqrt{29}}=\ds\frac{2}{\sqrt{29}}
\end{eqnarray*}

Thus we have

$\ (x,y,z) \cdot
\left(\ds\frac{2}{\sqrt{29}},-\ds\frac{3}{\sqrt{29}},
\ds\frac{4}{\sqrt{29}}\right)=\ds\frac{2}{\sqrt{29}}$

$\Rightarrow
\ds\frac{2x}{\sqrt{29}}-\ds\frac{2y}{\sqrt{29}}+\ds\frac{4z}{\sqrt{29}}=\ds\frac{2}{\sqrt{29}}$

$\Rightarrow \un{2x-3y+4z=2} \ (\star)$

Check: from vector equation above taking ${\bf{i}},\ {\bf{j}},\
{\bf{k}}$.

Components: \begin{eqnarray*} x=(3+\lambda-\mu)\\
y=(2\lambda+2\mu)\\ z=(-1+\lambda+2\mu) \end{eqnarray*}

Substitute into LHS of $(\star)$:

\begin{eqnarray*} & &
2(3+\lambda-\mu)-3(2\lambda+2\mu)+4(-1+\lambda+2\mu)\\ & = &
6+2\lambda-2\mu-6\lambda-6\mu-4+4\lambda+8\mu\\ & = & 2\\ & = &
RHS\ (\star)\surd \end{eqnarray*}

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