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Saturday, 9 August 2014

CHAPTER 4-Worked Out Examples 2

     Example: 6    

Two lines {L_1} and {L_2} 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 {L_1} and {L_2} are inclined to each other respectively.
Solution: 6

The unit vectors {\hat u_1} and {\hat u_2} along {L_1}and {L_2} 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 {\rm{\theta }} between {\hat u_1} and {\hat u_2} (and hence {L_1} and {L_2} ) is given by
\cos \theta  = {\hat u_1} \cdot {\hat u_2} = {l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}
 \Rightarrow  \,\,\,\, \theta  = {\cos ^{ - 1}}\left( {{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}} \right)
We can dedude the following conditions on the direction cosines of {L_1} and {L_2}.
If {L_1} and {L_2} are parallel
{\vec u_1} = \lambda {\vec u_2}
\Rightarrow\,\,\,\,{\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{{n_1}}}{{{n_2}}}}
If {L_1} and {L_2} are perpendicular
{\vec u_1} \cdot {\vec u_2} = 0
\Rightarrow\,\,\,\,\,{{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} = 0}
What will be the corresponding conditions had a set of direction ratios been specified instead of the direction cosines?
     Example: 7     

For the lines {L_1} and {L_2} of the previous example, find the direction cosines of the line {L_3} perpendicular to both {L_1} and {L_2} .
Solution: 7

Let the unit vector along {L_3} be {\vec u_3}. We have,
{\vec u_3} = {\vec u_1} \times {\vec u_2}
 = \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)
Since {\hat u_3} is a unit vector itself, the direction cosines of {L_3} 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: 8

Using the value of m from the first equation in the second, we have
 - 6\left( {3l + 5n} \right)n - 2nl - 5l\left( {3l + 5n} \right) = 0
 \Rightarrow  \,\,\,\, 45ln + 30{n^2} + 15{l^2} = 0
 \Rightarrow  \,\,\,\, 2{n^2} + 3ln + {l^2} = 0
 \Rightarrow  \,\,\,\, \left( {2n + l} \right)\left( {n + l} \right) = 0
 \Rightarrow  \,\,\,\, 2n =  - l\,\,\,\,{\rm{or}}\,\,\,\,n =  - l
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 {\cos ^{ - 1}}\left( {\dfrac{1}{6}} \right).
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