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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 lmn. 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 lmn  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 xy and z -components of the segment ABare ADCB and DC respectively. If the direction cosines of AB are lmnand 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 lmn.
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 lmn 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 lm,n ) will be sum of projections of ACCDDB onto L. Since ACCD, 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|
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