# Examples on Direction Cosines and Ratios of Lines Set 1

**Example – 3**

How many lines can we draw that are equally inclined to each of the three coordinate axis?

**Solution:** 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,

\[{l^2} + {m^2} + {n^2} = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \ldots \left( 1 \right)\]

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),

\[\begin{align}&\qquad\;\;\;\left| l \right| = \left| m \right| = \left| n \right| = \frac{1}{{\sqrt 3 }}\\ &\Rightarrow\quad l = \pm \frac{1}{{\sqrt 3 }},\,\,\,m = \pm \frac{1}{{\sqrt 3 }},\,\,\,n = \pm \frac{1}{{\sqrt 3 }} \\ \end{align} \]

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:** Refer to Fig - 4. Note that the *x*, *y* and *z*-components of the segment *AB* are *AD*, *C*B 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

\[\boxed{l = \frac{{{x_2} - {x_1}}}{d},\,\,\,m = \frac{{{y_2} - {y_1}}}{d},\,\,\,n = \frac{{{z_2} - {z_1}}}{d}}\]

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:** 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.

\[\begin{align}& d = \left| {\overrightarrow {AB} \cdot \hat u} \right| \\\\ \,\,\,\, &\;\;= \left| {l\left( {{x_2} - {x_1}} \right) + m\left( {{y_2} - {y_1}} \right) + n\left( {{z_2} - {z_1}} \right)} \right| \\\end{align} \]

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 \({P_1},{P_2},{P_3}.....{P_n}\) to be* n* points in space. The sum of projections of the sequence of segments \({P_1}{P_2},{P_2}{P_3},.....{P_{n - 1}}{P_n}\) 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 \({P_1}{P_2},{P_2}{P_3}.....{P_{n - 1}}{P_n}\) onto *L* is \({Q_1}{Q_2} + {Q_2}{Q_3} + ..... + {Q_{n - 1}}{Q_n} = {Q_1}{Q_n}.\)

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

\[d = \left| {l\left( {{x_2} - {x_1}} \right) + m\left( {{y_2} - {y_1}} \right) + n\left( {{z_2} - {z_1}} \right)} \right|\]

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