# Examples on Direction Cosines and Ratios of Lines Set 2

Go back to  'Three Dimensional Geometry'

## Examples on Direction Cosines and Ratios of Lines

Example – 6

Two lines L1 and L2 have direction cosines  $$\left\{ {{l_1},\,{m_1},\,{n_1}} \right\}$$ and  $$\left\{ {{l_2},\,{m_2},\,{n_2}} \right\}$$  respectively. Find the angle at which L1 and L2 are inclined to each other respectively.

Solution: The unit vectors $${\hat u_1}$$ and $${\hat u_2}$$ along L1 and L2 respectively can be written as

${{\hat{u}}_{1}}={{l}_{1}}\hat{i}+{{m}_{1}}\hat{j}+{{n}_{1}}\hat{k},{{\hat{u}}_{2}}={{l}_{2}}\hat{i}+{{m}_{2}}\hat{j}+{{n}_{2}}\hat{k}$

The angle between  $${\hat u_1}$$ and  $${\hat u_2}$$ (and hence L1 and L2 ) is given by

\begin{align} &\cos \theta = {{\hat u}_1} \cdot {{\hat u}_2} = {l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} \\\\ &\Rightarrow \quad \theta = {\cos ^{ - 1}}\left( {{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}} \right)\\ \end{align}

We can dedude the following conditions on the direction cosines of L1 and L2.

If L1 and L2  are parallel  :\begin{align}&\qquad\;\;{{\vec u}_1} = \lambda {{\vec u}_2} \\ \\ &\Rightarrow \quad \boxed{\frac{{{l_1}}}{{{l_2}}} = \frac{{{m_1}}}{{{m_2}}} = \frac{{{n_1}}}{{{n_2}}}} \\ \end{align}

:

If  L1 and L2   are  perpendicular :  \begin{align}&\qquad\;\;{{\vec u}_1} \cdot {{\vec u}_2} = 0 \\ \\ &\Rightarrow \quad \boxed{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} = 0} \\ \end{align}

What will be the corresponding conditions had a set of direction ratios been specified instead of the direction cosines?

Example – 7

For the lines L1 and L2 of the previous example, find the direction cosines of the line L3 perpendicular to both L1 and L2 .

Solution: Let the unit vector along L3 be  $${\vec u_3}.$$ We have,

\begin{align}&{{\vec u}_3} = {{\vec u}_1} \times {{\vec u}_2} \\\\ \,\,\,\,\,\, &\quad= \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ {{l_1}}&{{m_1}}&{{n_1}} \\ {{l_2}}&{{m_2}}&{{n_2}} \end{array}} \right| \\\\ \,\,\,\,\, &\;\;\;\;= \hat i\left( {{m_1}{n_2} - {m_2}{n_1}} \right) + \hat j\left( {{n_1}{l_2} - {n_2}{l_1}} \right) + \hat k\left( {{l_1}{m_2} - {l_2}{m_1}} \right) & \\ \end{align}

Since  $${\hat u_3}$$  is a unit vector itself, the direction cosines of L3 are simply

$\left( {{m_1}{n_2} - {m_2}{n_1}} \right),\left( {{n_1}{l_2} - {n_2}{l_1}} \right),\left( {{l_1}{m_2} - {l_2}{m_1}} \right)$

Example – 8

Find the angle between the lines whose direction cosines are given by the equations

$3l + m + 5n = 0,\,\,\,\,\,6mn - 2nl + 5lm = 0$

Solution: Using the value of m from the first equation in the second, we have

\begin{align} & - 6\left( {3l + 5n} \right)n - 2nl - 5l\left( {3l + 5n} \right) = 0 \\ \\ &\Rightarrow\quad 45ln + 30{n^2} + 15{l^2} = 0 \\ \\ &\Rightarrow \quad 2{n^2} + 3ln + {l^2} = 0 \\\\ &\Rightarrow\quad \left( {2n + l} \right)\left( {n + l} \right) = 0 \\\\ &\Rightarrow\quad 2n = - l\,\,\,\,{\text{or}}\,\,\,\,n = - l \\ \end{align}

For  $$l = - 2n,$$  we obtain m = n. A set of direction ratios of one line is therefore $$\left\{ { - 2n,\,\,n,\,\,n} \right\}.$$

For  $$l = - n,$$  we obtain  $$m = - 2n.$$  A set of direction ratios of the other line is therefore $$\left\{ { - n,\, - 2n,\,n} \right\}.$$

Using the result of example 6 (the one that you were asked to prove at the end of the question), the angle between the two lines can now be evaluated to be \begin{align}{\cos ^{ - 1}}\left( {\frac{1}{6}} \right).\end{align}

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

0