tgt

## 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 $l$$m$$n$ 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 $l$$m$$n$, 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 $1$$2$$3$. 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 $\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}.$