Parallel Lines

Before we proceed towards the concept of parallel lines, we shall first discuss direction ratios and direction cosines of a line.

Direction Cosines and Direction Ratios of a Line Joining two given Points

The direction cosine of a line has the same meaning as the direction cosines of a vector. Any three numbers proportional to direction cosines of a line are called direction ratios. If we have two points say P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), then the direction ratios of the line joining P and Q are

x₂ – x₁ = a(say), y₂ – y₁ = b (say) and z₂ – z₁ = c (say).

Then direction cosines are

l = a/(√a² + b² + c²), m = b/(√a² + b² + c²), n = c/(√a² + b² + c²)

Angle Between Two Lines

Let θ be the angle between two straight lines AB and AC whose direction cosines are given by l₁, m₁, n₁and l₂, m₂, n₂ respectively. Then the angle between the lines AB and AC is given by cos θ = l₁l₂ + m₁m₂ + n₁n₂.

Moreover, if the direction ratios of the two lines are a₁, b₁, c₁and a₂, b₂, c₂, then the angle between the two lines is given by

(a₁a₂+ b₂b₁+ c₁c₂) / √(a₁² + b₁² + c₁²) .√(a₂² + b₂² + c₂²)

The lines can either be parallel, coplanar or perpendicular. Whether they are parallel, coplanar or perpendicular can be decided with the help of their direction cosines.

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Condition of Parallelism

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.

Hence, if we have two lines with dc’s (direction cosines) as l₁, m₁, n₁ and l₂, m₂, n₂ respectively then these two lines will be parallel if  

We try to establish how the above result is obtained:

If the given lines are parallel, then q = 0⁰ i.e. sin θ = 0

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

which is true, only when

l₁m₂ – l₂m₁ = 0, m₁n₂ – m₂n₁ = 0 and n­­₁l₂ – n₂l₁ = 0 ⇒ I₁/l₂ = m₁/m₂ = n₁/n₂

Similarly, a₁/a₂ = b₁/b₂ = c₁/c₂.
 

Condition of Perpendicularity

If the given lines are perpendicular, then θ = 90⁰ i.e. cos θ = 0

⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0 or a₁a₂ + b₁b₂ + c₁c₂ = 0.

• We have, sin²θ = 1 – cos²θ

= (i₁² + m₁² + n₁²)(i₂² + m₂² + n₂²) – (l₁l₂ + m₁m₂ + n₁n₂)²

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

⇒ sin θ = ± √Σ(l₁m₂ – l₂m₁)² .

We now Move on to Some Illustrations Based on the Concepts Discussed Above

Illustration:

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 ⇒ 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

∴ m/n = 1/2 or 2/3

when m/n = 1/2 i.e. n = 2m

∴ l = 5m – 3n = –m or 1/m = –1

Hence, we have l/–1 = m/1 = n/2 = √ (l² + m² + n²) / √ {(–1)² + l² + 2²} = 1/√6

So, direction cosines of one line are –1/√6, 1/√6, 2/√6

Again when m/n = 2/3 or n = 3m/2

∴ l = 5m – 3.3m/2 = m/2 or 1/m = 1/2

Thus, m/n = 2/3 and l/m = 1/2 giving l/1 = m/2 = n/3 = 1/√1² + 2² + 3² = 1/√14

∴ The direction cosines of the other line are 1/√14, 2/√14, 3/√14.

Illustration:

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 (6-2, –7-(-3), –1-1)

(4, – 4, – 2) = (2, – 2, – 1)

Hence, we have a = 2, b = -2, c = -1. So, a² + b² + c² = 9.

Direction cosines are (± 2/3, ± 2/3, ± 1/3).  

Illustration:

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

Solution:

We know that the liens are perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0.

Hence we compute a₁a₂ + b₁b₂ + c₁c₂. We have

a₁ = –1, b₁ = 3, c₁ = 2 and a₂ = 2, b₂ = 2, c₂ = –2

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

This shows that the lines are perpendicular.

Illustration:

(a) Find the equation of the plane passing through the points (2, 1, 0), (5, 0, 1) and (4, 1, 1).

(b) If P is the point (2, 1, 6), then the point Q such that PQ is perpendicular to the plane given above and the mid-point of PQ lies on it. (2003)

Solution:

(a) The equation of the plane passing through (2, 1, 0) is

a(x-2) + b(y-1) + c(z-0) = 0

It also passes through (5, 0, 1) and (4, 1, 1).

Hence, we have 3a – b + c = 0

and 2a + 0b + c = 0

On solving the above equations we get,

a/-1 = b/ -1 = c/2

Hence, the required equation of plane is

-(x-2) – (y-1) + 2(z-0) = 0

- x + 2 – y + 1 + 2z = 0

x + y – 2z = 3.

(b) Let the coordinate of Q be (α, β, γ).

The equation of line PQ is given by

(x-2)/1 = (y-1)/1 = (z-6)/-2

Now, the mid-point of P and Q is ((α+2)/2, (β+1)/2, (γ+6)/2), which lies in a line PQ.

[(α+2)/2 – 2] / 1 = [(β+1)/2 – 1] / 1 = [(γ+6)/2 – 6]/ -2

Simplifying this we get the value as 2.

Hence, α = 6, β = 5 and γ = -2.

Hence, the coordinates of point Q are Q(6, 5, -2).
 

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