Examples on Direction Cosines and Ratios of Lines Set 2

Go back to  'Three Dimensional Geometry'

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 \(\begin{align}{\cos ^{ - 1}}\left( {\frac{1}{6}} \right).\end{align}\)