# Examples on Equations of Three Dimensional Lines Set 1

**Example – 17**

Find the direction cosines of the line \(6x-2=3y+1=2z-2.\)

**Solution:** We have ,

\[\begin{align}&\qquad6\left( x-\frac{1}{3} \right)=3\left( y+\frac{1}{3} \right)=2\left( z-1 \right) \\\\ & \Rightarrow\quad \frac{x-\frac{1}{3}}{1}=\frac{y+\frac{1}{3}}{2}=\frac{z-1}{3} \\ \end{align}\]

Comparing this with the symmetrical form of the equation of a line, we can say that the direction ratios of this line are proportional to 1, 2, 3. Thus, the direction cosines are

\[l=\frac{1}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{3}^{2}}}}=\frac{1}{\sqrt{14}},\,\,\,\,\,m=\frac{2}{\sqrt{14}},\,\,\,\,\,n=\frac{3}{\sqrt{14}}\]

\(\Rightarrow \) The direction cosines are \(\begin{align}\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}.\end{align}\)

**Example – 18**

Find the distance of the point *A *(1, –2, 3) from the plane \(x-y+z=5\) measured parallel to the line\(\begin{align}\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}.\end{align}\)

**Solution:** The direction cosines of the line parallel to whom we wish to measure the distance, can be evaluated to be

\[\frac{2}{7},\frac{3}{7},-\frac{6}{7}\]

Thus, any point on the line through *A* with these direction cosines, at a distance *r* from *A*, will have the coordinates

\[\left( 1+\frac{2r}{7},\,\,-2+\frac{3r}{7},\,\,3-\frac{6r}{7} \right)\]

If this point lies on the given plane, we have

\[\begin{align} &\qquad \left( 1+\frac{2r}{7} \right)-\left( -2+\frac{3r}{7} \right)+\left( 3-\frac{6r}{7} \right)=5 \\\\ & \Rightarrow\quad r=1 \\ \end{align}\]

Thus, the required distance is 1 unit.