A line is defined by the symmetric equations $\dfrac{x-6}{3}=\dfrac{y+2}{1}=\dfrac{z+3}{-2}$.

Solution (a) From the denominators of the symmetric equations, we find $a=3$, $b=1$, and $c=-2$. So, ${\bf D}= 3\mathbf{i}+\mathbf{j} -2\mathbf{k}$ is a vector in the direction of the line.

(b) Since the line is defined by the symmetric equations $\dfrac{x-6}{3}=\dfrac{y+2}{1}=\dfrac{z+3}{-2},$ one point on the line is $( 6,-2,-3)$. To find a second point, we assign a value to $x$ say, $x=0.$ Then $$\begin{eqnarray*} \frac{0-6}{3} &=&\frac{y+2}{1}=\frac{z+3}{-2} \\[4pt] -2 &=&\frac{y+2}{1}=\frac{z+3}{-2} \end{eqnarray*}$$

736

Now we solve for $y$ and $z,$ and find $y=-4$ and $z=1$. So, another point on the line is $(0,-4,1)$.