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Saturday, 9 August 2014

CHAPTER 9 - Introduction to Straight Lines

In this section, we’ll discuss how to write the equation for a straight line in coordinate form. There are essentially two different ways of doing so:
UNSYMMETRICAL FORM OF THE EQUATION OF A LINE
A line can be defined as the intersection of two planes. Thus, the equations of two planes considered together represents a straight line. For example, the set of equations
{a_1}x + {b_1}y + {c_1}z + {d_1} = 0
{a_2}x + {b_2}y + {c_2}z + {d_2} = 0
represents the straight line formed by the intersection of these two planes.
Recall that the planes will intersect only if they are non-parallel, i.e., only if
{a_1}:{b_1}:{c_1} \ne {a_2}:{b_2}:{c_2}
SYMMETRICAL FORM OF THE EQUATION OF A LINE:
Consider a line with direction cosines lmn and passing through the point A({x_1},{y_1},{z_1}). For any point P\left( {x,y,z} \right) on this line, the set of numbers \left\{ {\left( {x - {x_1}} \right),\left( {y - {y_1}} \right),\left( {z - {z_1}} \right)} \right\} must be proportional to the direction cosines, as has already been discussed. Thus, the equation of this line can be written as
\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}
Extending this, we can write the equation of the line passing through A\left( {{x_1},{y_1},{z_1}} \right) and B\left( {{x_2},{y_2},{z_2}} \right) as
\dfrac{{x - {x_1}}}{{{x_2} - {x_1}}} = \dfrac{{y - {y_1}}}{{{y_2} - {y_1}}} = \dfrac{{z - {z_1}}}{{{z_2} - {z_1}}}
Note that for any point P\left( {x,y,z} \right) at a distance r from A\left( {{x_1},{y_1},{z_1}} \right) along the line with direction cosines lmn, we have
\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n} = r
Thus, the coordinates of P can be written as
{x = {x_1} + lr,\,\,\,\,\,y = {y_1} + mr,\,\,\,\,\,z = {z_1} + nr}
This is a useful fact and we’ll be using it frequently.
     Example: 17      

Find the direction cosines of the line 6x - 2 = 3y + 1 = 2z - 2.
Solution: 17

We have ,
6\left( {x - \dfrac{1}{3}} \right) = 3\left( {y + \dfrac{1}{3}} \right) = 2\left( {z - 1} \right)
 \Rightarrow  \,\,\,\, \dfrac{{x - \dfrac{1}{3}}}{1} = \dfrac{{y + \dfrac{1}{3}}}{2} = \dfrac{{z - 1}}{3}
Comparing this with the symmetrical form of the equation of a line, we can say that the direction ratios of this line are proportional to 123. Thus, the direction cosines are
l = \dfrac{1}{{\sqrt {{1^2} + {2^2} + {3^2}} }} = \dfrac{1}{{\sqrt {14} }},\,\,\,\,\,m = \dfrac{2}{{\sqrt {14} }},\,\,\,\,\,n = \dfrac{3}{{\sqrt {14} }}
\Rightarrow \,\,\,\, The direction cosines are \dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}.
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