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{\bf Question}

The equations

$$\ds\frac{3x+3}{2}=\ds\frac{-2y+1}{7}=\ds\frac{2z+6}{3}$$

determines a straight line.  Express this in its equivalent
parametric form and hence as a vector equation.
\medskip

{\bf Answer}

If $\ds\frac{3x+3}{2}=\ds\frac{-2y+1}{7}=\ds\frac{2z+6}{3}$ we
form the parametric equation by reversing the steps of Q5:

so set

$\ds\frac{3x+3}{2}=\lambda,\ \ds\frac{-2y+1}{7}=\lambda,\
\ds\frac{2z+6}{3}=\lambda.$

$\Rightarrow \left\{x=\ds\frac{2\lambda-3}{3};\
y=\ds\frac{1-7\lambda}{2};\ z=\ds\frac{3\lambda-6}{2}\right\}$
\un{parametric equations}

Vector equation is again got by reversing the steps of Q5:

If $x=\ds\frac{2\lambda-3}{3}$

$\Rightarrow {\bf{i}}$ component of vector equation is
$\ds\frac{2\lambda}{3}-\ds\frac{3}{3}=\ds\frac{2\lambda}{3}-1$

If $y=\ds\frac{1-7\lambda}{2}$

$\Rightarrow {\bf{j}}$ component of vector equation is
$\ds\frac{1}{2}-\ds\frac{7}{2}\lambda$

If $z=\ds\frac{3\lambda-6}{2}$

$\Rightarrow {\bf{k}}$ component of vector equation is
$\ds\frac{3\lambda}{2}-3$

Thus
${\bf{r}}=\left(\ds\frac{2\lambda}{3}-1\right){\bf{i}}+\left(\ds\frac{1}{2}-
\ds\frac{7}{2}\lambda\right){\bf{j}}+\left(\ds\frac{3\lambda}{2}-3\right){\bf{k}}$

or
$\un{{\bf{r}}=-{\bf{i}}+\ds\frac{1}{2}{\bf{j}}-3{\bf{k}}+\lambda\left(\ds\frac{2}{3}{\bf{i}}
-\ds\frac{7}{2}{\bf{j}}+\ds\frac{3}{2}{\bf{k}}\right)}$

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