Lines in Space

DIRECTION COSINES OF A LINE

If $\alpha$, $\beta$, $\gamma$ be the angles which a given directed line makes with the positive directions of the co-ordinate axes, then cos$\alpha$, cos$\beta$, cos$\gamma$ are called the direction cosines of the given line and are generally denoted by l, m, n respectively.

Thus, l = cos$\alpha$, m = cos$\beta$ and n = cos$\gamma$

By the definition it follows that the direction cosine of the axis of x are respectively cos0°, cos90°, cos 90° i.e. (1, 0, 0).

Similarly direction cosines of the axes of y and z are respectively (0, 1, 0) and (0, 0, 1).

Relation between the Direction Cosines:

Let OP be any line through the origin O which has direction cosines l, m, n.

Let P $\equiv$ (x, y, z) and OP = r

Then OP² = x² + y² + z² = r² …. (1)

From P draw PA, PB, PC perpendicular on the coordinate axes, so that

OA = x, OB = y, OC = z.

Also, $\angle$POA = $\alpha$, $\angle$POB = $\beta$ and $\angle$POC = $\gamma$.

From triangle AOP, l = cos$\alpha$ = $\frac{x}{r}$ $\Rightarrow$ x = lr

Similarly y = mr and z = nr

Hence from (1)

r²(l² + m² + n²) = x² + y² + z² = r² $\Rightarrow$ l² + m² + n² = 1

Note:

$\bullet$ If the coordinates of any point P be (x, y, z) and l, m, n be the direction cosines of the line OP, O being the origin, then (lr, mr, nr) will give us the co-ordinates of a point on the line OP which is at a distance r from (0, 0, 0).

Direction Ratios:

If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios. They are also called direction numbers or direction components.

Hence by definition, we have

$\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k$ (say)

$\Rightarrow$ l = ak, m = bk, n = ck $\Rightarrow$ k²(a² + b² + c²) = l² + m² + n² = 1

$\Rightarrow$ k = $\pm$ $\frac{1}{\sqrt{a^2+b^2+c^2}}$ = $\pm$ $\frac{1}{\sqrt{\sum a^2}}$

$\therefore$ l = $\pm$ $\frac{a}{\sqrt{\sum a^2}}$. Similarly m = $\pm$ $\frac{b}{\sqrt{\sum a^2}}$ and n = $\pm$ $\frac{c}{\sqrt{\sum a^2}}$

where the same sign either positive or negative is to be chosen throughout.

Example: If 2, – 3, 6 be the direction ratios, then the actual direction cosines are $\frac{2}{7},\frac{-3}{7},\frac{6}{7}$.

Note:

$\bullet$ Direction cosines of a line are unique but direction ratios of a line in no way unique but can be infinite.

Parallel Lines:

Since parallel lines have the same direction, it follows that the direction cosines of two or more parallel straight lines are the same. So in case of lines, which do not pass through the origin, we can draw a parallel line passing through the origin and direction cosines of that line can be found.

Illustration 1: Find the direction cosines of two lines which are connected by the relations l–5m + 3n = 0 and 7l² + 5m² – 3n² = 0.

Solution: The given relations are

l–5m + 3n = 0 $\Rightarrow$ l = 5m – 3n ……(1)

and 7l² + 5m² – 3n² = 0 ……(2)

Putting the value of l from (1) in (2), we get

7(5m – 3n)² + 5m² – 3n² = 0

or, 180m² – 210mn + 60n² = 0 or, (2m – n)(3m – 2n) = 0

$\therefore$ $\frac{m}{n}=\frac{1}{2}$ or $\frac{2}{3}$

when $\frac{m}{n}=\frac{1}{2}$ i.e. n = 2m

$\therefore$ l = 5m – 3n = –m or $\frac{l}{m}$ = –1

thus $\frac{m}{n}=\frac{1}{2}$ and $\frac{l}{m}$ = –1 giving $\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}$

or, $\frac{l}{-1}=\frac{m}{1}=\frac{n}{2}=\frac{\sqrt{(l^2+m^2+n^2)}}{\sqrt{\{{(-1)}^2+1^2+2^2\}}}=\frac{1}{\sqrt{6}}$

So, direction cosines of one line are –$\frac{1}{\sqrt{6}}$, $\frac{1}{\sqrt{6}}$, $\frac{2}{\sqrt{6}}$

Again when $\frac{m}{n}=\frac{2}{3}$ or n = $\frac{3m}{2}$

$\therefore$ l = 5m – 3.$\frac{3m}{2}=\frac{m}{2}$ or $\frac{l}{m}=\frac{1}{2}$

Thus, $\frac{m}{n}=\frac{2}{3}$ and $\frac{l}{m}=\frac{1}{2}$ giving $\frac{l}{1}=\frac{m}{2}=\frac{n}{3}=\frac{1}{\sqrt{1^2+2^2+3^2}}=\frac{1}{\sqrt{14}}$

$\therefore$ The direction cosines of the other line are $\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},\frac{3}{\sqrt{14}}$.

Direction Cosine of a Line joining two given Points:

The direction ratios of line PQ joining P (x₁, y₁, z₁) and Q(x₂, y₂, z₂) are x₂ – x₁ = a(say), y₂ – y₁ = b (say) and z₂ – z₁ = c (say).

Then direction cosines are

l = $\frac{(x_2-x_1)}{\sqrt{\Sigma {\left(x_2-x_1\right)}^2}}$, m = $\frac{(y_2-y_1)}{\sqrt{\Sigma {\left(x_2-x_1\right)}^2}}$, n = $\frac{(z_2-z_1)}{\sqrt{\Sigma {\left(x_2-x_1\right)}^2}}$

Illustration 2: Find the direction ratios and direction cosines of the line joining the points A(6, –7, –1) and B(2, –3, 1).

Solution: Direction ratios of AB are (4, – 4, – 2) = (2, – 2, – 1)

a² + b² + c² = 9

Direction cosines are $\left(\pm \frac{2}{3},\mp \frac{2}{3},\mp \frac{1}{3}\right)$.

Angle between two Lines:

Let $\theta$ be the angle between two straight lines AB and AC whose direction cosines are given whose direction cosines are l₁, m₁, n₁ and l₂, m₂, n₂ respectively, is given by cos$\theta$ = l₁l₂ + m₁m₂ + n₁n₂

If direction ratios of two lines are a₁, b₁, c₁ and a₂, b₂, c₂ are given, then angle between two lines is given by cos $\theta$ = $\frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{a_1^2+b_1^2+c_1^2}\cdot \sqrt{a_2^2+b_2^2+c_2^2}}$

Particular Results:

$\bullet$ We have, sin²$\theta$ = 1 – cos²$\theta$

= $(l_1^2+m_1^2+n_1^2)(l_2^2+m_2^2+n_2^2)$– (l₁l₂ + m₁m₂ + n₁n₂)²

= (l₁m₂ – l₂m₁)² + (m₁n₂ – m₂n₁)² + (n₁l₂ – n₂l₁)²

$\Rightarrow$ sin$\theta$ = $\pm$ $\sqrt{\sum {\left(l_1{m}_2-l_2{m}_1\right)}^2}$.

$\bullet$Condition of perpendicularity:

If the given lines are perpendicular, then $\theta$ = 90⁰ i.e. cos $\theta$ = 0

$\Rightarrow$ l₁l₂ + m₁m₂ + n₁n₂ = 0 or a₁a₂ + b₁b₂ + c₁c₂ = 0 .

$\bullet$ Condition of parallelism:

If the given lines are parallel, then $\theta$ = 0⁰ i.e. sin $\theta$ = 0

$\Rightarrow$ (l₁m₂ – l₂m₁)² + (m₁n₂ – m₂n₁)² + (n­­₁l₂ – n₂l₁)² = 0

which is true, only when

l₁m₂ – l₂m₁ = 0, m₁n₂ – m₂n₁ = 0 and n­­₁l₂ – n₂l₁ = 0 $\Rightarrow$ $\frac{l_1}{l_2}=\frac{m_1}{m_2}=\frac{n_1}{n_2}$

Similarly, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.

Illustration 3: Show that two lines having direction ratios –1, 3, 2 and 2, 2, –2 are perpendicular.

Solution: a₁a₂ + b₁b₂ + c₁c₂ = (–1)(2) + (3)(2) + (2)(–2) = –2 + 6 – 4 = 0

$\therefore$ lines are perpendicular.

Projection of a Line:

Projection of the line joining two point P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) on another line whose direction cosines are l, m, n is

AB = l(x₂ – x₁) + m(y₂ – y₁) + n(z₂ – z₁)

Perpendicular Distance of a Point from a Line:

Let AB is straight line passing through point A (a, b, c) and having direction cosines l, m, n.

AN = projection of line AP on straight line AB

= l(x – a) + m(y – b) + n(z – c)

and AP = $\sqrt{{(x-a)}^2+{(y-b)}^2+{(z-c)}^2}$

$\therefore$ perpendicular distance of point P

PN = $\sqrt{A{P}^2-A{N}^2}$

Illustration 4: Find out perpendicular distance of point P (0, –1, 3) from straight line passing through A (1, –3, 2) and having direction ratios 1, 2, 2.

Solution: Direction cosines of the line is $\frac{1}{\sqrt{1^2+2^2+2^2}},\frac{2}{\sqrt{1^2+2^2+2^2}},\frac{2}{\sqrt{1^2+2^2+2^2}}$ i.e. $\frac{1}{3},\frac{2}{3},\frac{2}{3}$.

$\therefore$ PN = l(x – a) + m(y – b) + n(z – c) = $\frac{1}{3}$(0 – 1) + $\frac{2}{3}$(–1 + 3) + $\frac{2}{3}$(3 – 2) = $\frac{5}{3}$

AP = $\sqrt{{(0-1)}^2+{(-1+3)}^2+{(3-2)}^2}=\sqrt{6}$

$\therefore$ Perpendicular distance PN = $\sqrt{A{P}^2-P{N}^2}=\sqrt{6-\frac{25}{9}}=\frac{\sqrt{29}}{3}$.

Area of a Triangle

_{$\Delta$ x} $\frac{1}{2}\left|\begin{array}{c}{y}_1{z}_{1}1\\ y_2{z}_{2}1\\ y_3{z}_{3}1\end{array}\right|$

_$\Delta$y $\frac{1}{2}\left|\begin{array}{c}{x}_1{z}_{1}1\\ x_2{z}_{2}1\\ x_3{z}_{3}1\end{array}\right|$

_$\Delta$z = $\frac{1}{2}\left|\begin{array}{c}{x}_1{y}_{1}1\\ x_2{y}_{2}1\\ x_3{y}_{3}1\end{array}\right|$

So, area of DABC is given by the relation $\Delta$² = ${\Delta }_x^2+{\Delta }_y^2+{\Delta }_z^2$

EQUATION OF STRAIGHT LINE IN DIFFERENT FORMS

Symmetrical Form:

$\bullet$ Equation of straight line passing through point P (x₁, y₁, z₁) and whose direction cosines are l, m, n is $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$

$\bullet$ Equation of straight line passing through two points P (x₁, y₁, z₁) and Q (x₂, y₂, z₂) is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$.

Note:

$\bullet$ The general coordinates of a point on a line is given by (x₁ + lr, y₁ + mr, z₁ + nr) where r is distance between point (x₁, y₁, z₁) and point whose coordinates is to be written.

Illustration 5. Find the equations of the straight lines through the point (a, b, c) which are

(a) parallel to z-axis

(b) perpendicular to z-axis

Solution (i) Equation of straight lines parallel to z-axis have $\alpha$ = 90⁰, $\beta$ = 90⁰, $\gamma$ = 0⁰ $\Rightarrow$ l = 0, m = 0, n = 1

Therefore equation of straight line is parallel to z-axis and passing through (a, b, c) is $\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

(ii) equation of straight lines perpendicular to z-axis

let they make $\alpha$, $\beta$ angle with x and y axes respectively.

Then equation of straight lines perpendicular to z axis and passing through (a, b, c) is $\frac{x-a}{\cos \alpha }=\frac{y-b}{\sin \alpha }=\frac{z-c}{0}$ $\Rightarrow$ $\frac{x-a}{l}=\frac{y-b}{m}=\frac{z-c}{0}$.

Shortest Distance between two non Intersecting Line:

Two lines are called non intersecting lines if they do not lie in the same plane. The straight line which is perpendicular to each of non-intersecting lines is called the line of shortest distance. And length of shortest distance line intercepted between two lines is called length of shortest distance.

Method: Let the equation of two non-intersecting lines be

$\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ = r₁ (say) ……(1)

And $\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2}=\frac{z-z_2}{n_2}$ = r₂ (say) ……(2)

Any point on line (1) is P (x₁ + l₁r₁, y₁ + m₁r₁, z₁ + n₁r₁) and on line (2) is Q (x₂ + l₂r₂, y₂ + m₂r₂, z₂ + n₂r₂).

Let PQ be the line of shortest distance. Its direction ratios will be

[(l₁r₁ + x₁– x₂– l₂r₂), (m₁r₁ + y₁– y₂– m₂r₂), (n₁r₁ + z₁– z₂– n₂r₂)]

This line is perpendicular to both given line. By using condition of perpendicularity we obtain 2 equations in r₁ and r₂.

So by solving these, values of r₁ and r₂ can be found. And subsequently point P and Q can be found. The distance PQ is shortest distance.

The shortest distance can be found by PQ = $\left|\begin{array}{c}{x}_1-x_2{y}_1-y_2{z}_1-z_2\\ l_1{m}_1{n}_1\\ l_2{m}_2{n}_2\end{array}\right|$

.

Note: $\bullet$ If any straight line is given in general form then it can be transformed into symmetrical form and we can further proceed.

Illustration 6: Find the shortest distance between the lines $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$, $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$. Also find the equation of line of shortest distance.

Solution: Given lines are $\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}$ = r₁ (say) ……(1)

$\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-6}{4}$ = r₂ (say) ……(2)

Any point on line (1) is P (3r₁ + 3, 8 – r₁, r₁ + 3) and on line (2) is

Q (–3 – 3r₂, 2r₂ – 7, 4r₂ + 6).

If PQ is line of shortest distance, then direction ratios of PQ

= (3r₁ + 3) – (–3 – 3r₂), (8 – r₁) – (2r₂ – 7), (r₁+ 3) – (4r₂ + 6)

i.e. 3r₁ + 3r₂ + 6, –r₁ – 3r₂ + 15, r₁ – 4r₂ – 3

As PQ is perpendicular to liens (1) and (2)

$\therefore$ 3(3r₁ + 3r₂ + 6) – 1(–r₁ – 2r₂ + 15) + 1(r₁ – 4r₂ + 3) = 0

$\Rightarrow$ 11r₁ + 7r₂ = 0 ……(3)

and –3(3r₁ + 3r₂ + 6) + 2(–r₁ – 2r₂ + 15) + 4(r₁ – 4r₂ + 3) = 0

i.e. 7r₁ + 11r₂ = 0 ……(4)

On solving equations (3) and (4), we get r₁ = r₂= 0.

So, point P (3, 8,3) and Q (–3, –7, 6)

$\therefore$ Length of shortest distance PQ = $\sqrt{\{{(-3-3)}^2+{(-7-8)}^2+{(6-3)}^2\}}=3\sqrt{30}$

Direction ratios of shortest distance line is 2, 5, –1

$\therefore$ Equation of shortest distance line $\frac{x-3}{2}=\frac{y-8}{5}=\frac{z-3}{-1}$.