# Examples on Equations of Three Dimensional Lines Set 2

Go back to  'Three Dimensional Geometry'

Example – 19

Find a set of direction ratios of the line

\begin{align} & {{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0\,\,;{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0 \\\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{a}_{1}}:{{b}_{1}}:{{c}_{1}}\ne {{a}_{2}}:{{b}_{2}}:{{c}_{2}} \\ \end{align}

Solution: The equation of the line has been specified in unsymmetric form, i.e., as the intersection of two non-parallel planes.

Visualise in your mind that when two planes intersect, the line of intersection will be perpendicular to normals to both the planes. Normal vectors to the two planes can be taken to be

\begin{align} & {{{\vec{n}}}_{1}}={{a}_{1}}\hat{i}+{{b}_{1}}\hat{j}+{{c}_{1}}\hat{k} \\\\ & {{{\vec{n}}}_{2}}={{a}_{2}}\hat{i}+{{b}_{2}}\hat{j}+{{c}_{2}}\hat{k} \\ \end{align}

Thus, the line of intersection will be parallel to i.e. to

$\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\\end{matrix} \right|=\hat{i}\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)+\hat{j}\left( {{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}} \right)+\hat{k}\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)$

A set of direction ratios of the line of intersection can be taken to be

$\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right),\left( {{c}_{1}}{{a}_{2}}-{{a}_{1}}{{c}_{2}} \right),\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)$

Example – 20

Find the equation of the plane passing through the line

$2x+y-z-3=0=5x-3y+4z+9$

and parallel to the line  \begin{align}\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-5}{5}\end{align}

Solution: In terms of a parameter $$\lambda$$ the equation of the plane that we require can be written as

\begin{align} &\qquad\;\;\; \left( 2x+y-z-3 \right)+\lambda \left( 5x-3y+4z+9 \right)=0 \\\\ & \Rightarrow\quad \left( 2+5\lambda \right)x+\left( 1-3\lambda \right)y+\left( 4\lambda -1 \right)z+\left( 9\lambda -3 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ldots \left( 1 \right) \\ \end{align}

For this plane to be parallel to the given line, its normal must be perpendicular to the given line. Using the condition for perpendicularity, we thus have

\begin{align} &\qquad\;\; 2\left( 2+5\lambda \right)+4\left( 1-3\lambda \right)+5\left( 4\lambda -1 \right)=0 \\\\ & \Rightarrow\quad 3+18\lambda =0 \\\\ & \Rightarrow\quad \lambda =-\frac{1}{6} \\ \end{align}

Using this value of  $$\lambda$$ in (1), we get the required equation of the plane as  $$7x+9y-10z=27.$$

Learn from the best math teachers and top your exams

• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school