# Examples on Equations of Three Dimensional Lines Set 1

Go back to  'Three Dimensional Geometry'

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.

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