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Saturday, 9 August 2014

CHAPTER 10 - Worked Out Examples

     Example: 18   

Find the distance of the point A(1, - 2,3) from the plane x - y + z = 5measured parallel to the line \dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{{ - 6}}.
Solution: 18

The direction cosines of the line parallel to whom we wish to measure the distance, can be evaluated to be
\dfrac{2}{7},\dfrac{3}{7}, - \dfrac{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 + \dfrac{{2r}}{7},\,\, - 2 + \dfrac{{3r}}{7},\,\,3 - \dfrac{{6r}}{7}} \right)
If this point lies on the given plane, we have
\left( {1 + \dfrac{{2r}}{7}} \right) - \left( { - 2 + \dfrac{{3r}}{7}} \right) + \left( {3 - \dfrac{{6r}}{7}} \right) = 5
 \Rightarrow  \,\,\,\, r = 1
Thus, the required distance is 1 unit.
     Example: 19     

Find a set of direction ratios of the line
{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}
Solution: 19

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
{\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
Thus, the line of intersection will be parallel to {\vec n_1} \times {\vec n_2}, i.e. to
\left| {\begin{array}{*{20}{c}}  {\hat i}\,\,\,\,{\hat j}\,\,\,\,{\hat k}\\  {{a_1}}\,\,\,\,{{b_1}}\,\,\,\,{{c_1}}\\  {{a_2}}\,\,\,\,{{b_2}}\,\,\,\,{{c_2}}  \end{array}} \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 \dfrac{{x - 1}}{2} = \dfrac{{y - 3}}{4} = \dfrac{{z - 5}}{5}
Solution: 20

In terms of a parameter \lambda , the equation of the plane that we require can be written as
\left( {2x + y - z - 3} \right) + \lambda \left( {5x - 3y + 4z + 9} \right) = 0
 \Rightarrow  \,\,\,\, \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(1)
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
2\left( {2 + 5\lambda } \right) + 4\left( {1 - 3\lambda } \right) + 5\left( {4\lambda  - 1} \right) = 0
 \Rightarrow  \,\,\,\, 3 + 18\lambda  = 0
 \Rightarrow  \,\,\,\, \lambda  =  - \dfrac{1}{6}
Using this value of \lambda  in (1)  we get the required equation of the plane as 7x + 9y - 10z = 27.
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