ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 9 - Introduction to Straight Lines

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 $\dfrac{1}{{\sqrt {14} }},\dfrac{2}{{\sqrt {14} }},\dfrac{3}{{\sqrt {14} }}.$

Saturday, 9 August 2014

CHAPTER 9 - Introduction to Straight Lines

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