Saturday, 9 August 2014

CHAPTER 3 - Worked Out Examples

 Example: 3
 How many lines can we draw that are equally inclined to each of the three coordinate axis?
 Solution: 3
Intuitively, we can expect the answer to be $8$, one for each of the $8$ octants. Lets try to derive this answer rigorously.
Assume the direction cosines of the lines to be $l$$m$$n$. Thus,
 ${l^2} + {m^2} + {n^2} = 1$ $\ldots(1)$
But since the lines are equally inclined to the three axes, we have $\left| l \right| = \left| m \right| = \left| n \right|.$ This gives using $(1)$
 $\left| l \right| = \left| m \right| = \left| n \right| = \dfrac{1}{{\sqrt 3 }}$ $\Rightarrow \,\,\,\,\, l = \pm \dfrac{1}{{\sqrt 3 }},\,\,\,m = \pm \dfrac{1}{{\sqrt 3 }},\,\,\,n = \pm \dfrac{1}{{\sqrt 3 }}$
It is obvious that $8$ combinations of $l$$m$$n$ are possible. Hence, $8$ lines can be drawn which are equally inclined to the axes.
 Example: 4
 Find the direction cosines of the line segment joining $A({x_1},\,{y_1},\,{z_1})$ and $B({x_2},\,{y_2},\,{z_2}).$
 Solution: 4
Refer to Fig – $4$. Note that the $x$$y$ and $z$ -components of the segment $AB$are $AD$$CB$ and $DC$ respectively. If the direction cosines of $AB$ are $l$$m$$n$and the length of $AB$ is $d$, we have
 $ld = {x_2} - {x_1},\,\,\,md = {y_2} - {y_1},\,\,nd = {z_2} - {z_1}$
Thus, the direction cosines of $AB$ are given by
 ${l = \dfrac{{{x_2} - {x_1}}}{d},\,\,\,m = \dfrac{{{y_2} - {y_1}}}{d},\,\,\,n = \dfrac{{{z_2} - {z_1}}}{d}}$
This result is quite important and will be used frequently in subsequent discussions.
 Example: 5
 Find the projection of the line segment joining the points $A({x_1},{y_1},{z_1})$ and $B({x_2},{y_2},{z_2})$ onto a line with direction cosines $l$, $m$, $n$.
 Solution: 5
Let us first consider a vector approach to this problem. The vector $\overrightarrow {AB}$ can be written as
 $\overrightarrow {AB} = \left( {{x_2} - {x_1}} \right)\hat i + \left( {{y_2} - {y_1}} \right)\hat j + \left( {{z_2} - {z_1}} \right)\hat k$
A unit vector $\hat u$ along the line with direction cosines $l$$m$$n$ will be
 $\hat u = l\hat i + m\hat j + n\hat k$
Therefore, the projected length of $\overrightarrow {AB}$ upon this line will be.
 $d = \left| {\overrightarrow {AB} \cdot \hat u} \right|$ $= \left| {l\left( {{x_2} - {x_1}} \right) + m\left( {{y_2} - {y_1}} \right) + n\left( {{z_2} - {z_1}} \right)} \right|$
This assertion can also be proved without resorting to the use of vectors. For this, we first understand the projection of a sequence of line segments on a given line.
Assume ${P_1},{P_2},{P_3}\ldots {P_n}$ to be $n$ points in space. The sum of projections of the sequence of segments ${P_1}{P_2},{P_2}{P_3},\ldots {P_{n - 1}}{P_n}$ onto a fixed line $L$ will be the same as the projection of ${P_1}{P_n}$ onto $L$. This should be obvious from the following diagram:
The projection of the segment ${P_1}{P_n}$ onto $L$ is ${Q_1}{Q_n}.$ The sum of projections of segments ${P_1}{P_2},{P_2}{P_3}\ldots {P_{n - 1}}{P_n}$ onto $L$ is ${Q_1}{Q_2} + {Q_2}{Q_3} + \ldots + {Q_{n - 1}}{Q_n} = {Q_1}{Q_n}.$
We use this fact in our original problem as follows:
The projection $d$ of $AB$ onto any line $L$ (with direction cosines say $l$$m$,$n$ ) will be sum of projections of $AC$$CD$$DB$ onto $L$. Since $AC$$CD$, and $DB$ are $l\left( {{x_2} - {x_1}} \right),\,m\left( {{y_2} - {y_1}} \right)$ and $n\left( {{z_2} - {z_1}} \right)$ respectively, we get the total projection of $AB$ onto $L$ as
 $d = \left| {l\left( {{x_2} - {x_1}} \right) + m\left( {{y_2} - {y_1}} \right) + n\left( {{z_2} - {z_1}} \right)} \right|$