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

Find a vector in the direction of the line.

Find two points on the line.

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*}

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