1. Chapter at a Glance
An aircraft flies along a straight path in the sky; a drone must keep a safe distance from it. How do we describe that path, find the angle between two flight paths, or the shortest distance between them? With vectors and coordinates in space. This chapter describes lines in three dimensions by their direction cosines and ratios, writes their equations in vector and Cartesian form, and finds the angle between two lines and the shortest distance between skew lines.
- Direction cosines and direction ratios of a line.
- Direction ratios of the line through two points.
- Equation of a line: through a point with given direction; through two points (vector and Cartesian forms).
- Angle between two lines; perpendicular and parallel lines.
- Skew lines and the shortest distance between two lines (skew and parallel).
2. Direction Cosines and Ratios
If a line makes angles α, β, γ with the positive x, y, z axes, its direction cosines are l = cos α, m = cos β, n = cos γ, with l² + m² + n² = 1. Any numbers a, b, c proportional to l, m, n are direction ratios:
Line through P(x₁, y₁, z₁) and Q(x₂, y₂, z₂): direction ratios x₂ − x₁, y₂ − y₁, z₂ − z₁.
Find the direction cosines of the line through (−2, 4, −5) and (1, 2, 3).
Solution: DRs 3, −2, 8; √(9 + 4 + 64) = √77. DCs 3/√77, −2/√77, 8/√77.
Show that (2, 3, 4), (−1, −2, 1), (5, 8, 7) are collinear.
Solution: DRs of AB: −3, −5, −3; of BC: 6, 10, 6 = −2 × (−3, −5, −3). Proportional and sharing B → collinear.
3. Equation of a Line
| Given | Vector form | Cartesian form |
|---|---|---|
| Point a⃗ (x₁, y₁, z₁), direction b⃗ (a, b, c) | r⃗ = a⃗ + λb⃗ | (x − x₁)/a = (y − y₁)/b = (z − z₁)/c |
| Two points a⃗, b⃗ | r⃗ = a⃗ + λ(b⃗ − a⃗) | (x − x₁)/(x₂ − x₁) = (y − y₁)/(y₂ − y₁) = (z − z₁)/(z₂ − z₁) |
Find the line through (5, 2, −4) parallel to 3î + 2ĵ − 8k̂, in both forms.
Solution: r⃗ = 5î + 2ĵ − 4k̂ + λ(3î + 2ĵ − 8k̂); (x − 5)/3 = (y − 2)/2 = (z + 4)/(−8).
4. Angle Between Two Lines
For lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂):
- Perpendicular ⇔ a₁a₂ + b₁b₂ + c₁c₂ = 0.
- Parallel ⇔ a₁/a₂ = b₁/b₂ = c₁/c₂.
Find the angle between (x + 3)/3 = (y − 1)/5 = (z + 3)/4 and (x + 1)/1 = (y − 4)/1 = (z − 5)/2.
Solution: cos θ = |3 + 5 + 8|/(√50 √6) = 16/√300 = 16/(10√3) = 8√3/15 → θ = cos⁻¹(8√3/15).
5. Shortest Distance Between Two Lines
Lines in space can intersect, be parallel, or be skew (neither parallel nor intersecting — like a flyover road and the road below it). The shortest distance is along the common perpendicular.
Skew lines r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂:
Parallel lines r⃗ = a⃗₁ + λb⃗ and r⃗ = a⃗₂ + μb⃗:
If d = 0 for non-parallel lines, they intersect.
Find the shortest distance between r⃗ = î + ĵ + λ(2î − ĵ + k̂) and r⃗ = 2î + ĵ − k̂ + μ(3î − 5ĵ + 2k̂).
Solution: b⃗₁ × b⃗₂ = î(−2 + 5) − ĵ(4 − 3) + k̂(−10 + 3) = 3î − ĵ − 7k̂, magnitude √59. a⃗₂ − a⃗₁ = î − k̂. Dot = 3 + 7 = 10. d = 10/√59.
Find the distance between r⃗ = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂) and r⃗ = 3î + 3ĵ − 5k̂ + μ(2î + 3ĵ + 6k̂).
Solution: a⃗₂ − a⃗₁ = 2î + ĵ − k̂. b⃗ × (a⃗₂ − a⃗₁) = î(−3 − 6) − ĵ(−2 − 12) + k̂(2 − 6) = −9î + 14ĵ − 4k̂, magnitude √293. |b⃗| = 7. d = √293/7.
6. Solved Problems (Board Pattern)
Find k so that (x − 1)/(−3) = (y − 2)/(2k) = (z − 3)/2 and (x − 1)/(3k) = (y − 1)/1 = (z − 6)/(−5) are perpendicular.
Solution: −9k + 2k − 10 = 0 → k = −10/7.
Find the vector and Cartesian equations of the line through (3, −2, −5) and (3, −2, 6).
Solution: Direction 0î + 0ĵ + 11k̂. r⃗ = 3î − 2ĵ − 5k̂ + λ(11k̂); Cartesian: x = 3, y = −2 (the line is parallel to the z-axis).
Find the shortest distance between (x + 1)/7 = (y + 1)/(−6) = (z + 1)/1 and (x − 3)/1 = (y − 5)/(−2) = (z − 7)/1.
Solution: b⃗₁ × b⃗₂ = î(−6 + 2) − ĵ(7 − 1) + k̂(−14 + 6) = −4î − 6ĵ − 8k̂, magnitude √116 = 2√29. a⃗₂ − a⃗₁ = 4î + 6ĵ + 8k̂. Dot = −16 − 36 − 64 = −116. d = 116/(2√29) = 2√29.
7. Master Formula Table
| Result | Formula |
|---|---|
| Direction cosines | l² + m² + n² = 1 |
| DCs from DRs | a/√(a² + b² + c²), … |
| Line (point, direction) | r⃗ = a⃗ + λb⃗; (x − x₁)/a = (y − y₁)/b = (z − z₁)/c |
| Line (two points) | r⃗ = a⃗ + λ(b⃗ − a⃗) |
| Angle | cos θ = |b⃗₁·b⃗₂|/(|b⃗₁||b⃗₂|) |
| Skew lines | |(b⃗₁ × b⃗₂)·(a⃗₂ − a⃗₁)|/|b⃗₁ × b⃗₂| |
| Parallel lines | |b⃗ × (a⃗₂ − a⃗₁)|/|b⃗| |
8. Master Summary Table
| Pair of lines | Test | Shortest distance |
|---|---|---|
| Parallel | Proportional DRs | |b⃗ × (a⃗₂ − a⃗₁)|/|b⃗| |
| Intersecting | Not parallel, d = 0 | 0 |
| Skew | Not parallel, d ≠ 0 | Triple-product formula |
| Perpendicular | Σa₁a₂ = 0 | — |
9. Mind Map
- DCs, DRs
- l² + m² + n² = 1
- a⃗ + λb⃗
- Symmetric form
- cos θ formula
- ⊥ and ∥ tests
- Skew lines
- Parallel lines
10. Tips and Tricks
- Before using the line formulas, rewrite each equation so the coefficients of x, y, z are 1: (2x − 1)/4 → (x − ½)/2.
- Read the point from the numerators (with signs flipped) and the direction from the denominators.
- If the shortest distance comes out 0, the lines intersect.
- Taking (x + 3)/2 as passing through (3, …) — it passes through (−3, …).
- Using the skew-lines formula for parallel lines (b⃗₁ × b⃗₂ = 0⃗).
- Forgetting the absolute value in cos θ (we want the acute angle).
11. Important One-liners
- Direction cosines of the x-axis are 1, 0, 0.
- Skew lines are neither parallel nor intersecting.
- Two lines are perpendicular iff a₁a₂ + b₁b₂ + c₁c₂ = 0.
- The shortest distance between skew lines lies along their common perpendicular.
- A line is fixed by a point and a direction.
12. Exam-pattern Questions with Detailed Answers
The direction cosines of the y-axis are:
- 1, 0, 0
- 0, 1, 0
- 0, 0, 1
- 1, 1, 1
Show answer
It makes 90°, 0°, 90° with the axes.
Answer: (B)
If a line makes angles 90°, 135°, 45° with the axes, find its direction cosines.
Show answer
0, −1/√2, 1/√2.
Answer: 0, −1/√2, 1/√2
Write the Cartesian equation of the line r⃗ = (2î − ĵ + 4k̂) + λ(î + ĵ − 2k̂).
Show answer
(x − 2)/1 = (y + 1)/1 = (z − 4)/(−2).
Answer: (x − 2)/1 = (y + 1)/1 = (z − 4)/(−2)
Show that the lines (x − 5)/7 = (y + 2)/(−5) = z/1 and x/1 = y/2 = z/3 are perpendicular.
Show answer
7 − 10 + 3 = 0 ✓.
Answer: Proved
Find the angle between the lines r⃗ = 3î + 2ĵ − 4k̂ + λ(î + 2ĵ + 2k̂) and r⃗ = 5î − 2ĵ + μ(3î + 2ĵ + 6k̂).
Show answer
cos θ = (3 + 4 + 12)/(3 × 7) = 19/21 → θ = cos⁻¹(19/21).
Answer: cos⁻¹(19/21)
Find the equation of the line through (1, 2, 3) parallel to the line (x − 4)/2 = (y + 1)/(−3) = (z − 7)/6.
Show answer
(x − 1)/2 = (y − 2)/(−3) = (z − 3)/6.
Answer: (x − 1)/2 = (y − 2)/(−3) = (z − 3)/6
Find the direction cosines of the line (x − 2)/2 = (2y − 5)/(−3) = (z + 1)/(−1).
Show answer
(2y − 5)/(−3) = (y − 5/2)/(−3/2). DRs 2, −3/2, −1 → 4, −3, −2; √29. DCs 4/√29, −3/√29, −2/√29.
Answer: 4/√29, −3/√29, −2/√29
Show that the points (1, 2, 3), (3, 4, 7) and (−3, −2, −5) are collinear.
Show answer
DRs AB: 2, 2, 4; AC: −4, −4, −8 = −2 × AB. Collinear.
Answer: Proved
Find the shortest distance between r⃗ = (î + 2ĵ + k̂) + λ(î − ĵ + k̂) and r⃗ = 2î − ĵ − k̂ + μ(2î + ĵ + 2k̂).
Show answer
b⃗₁ × b⃗₂ = î(−2 − 1) − ĵ(2 − 2) + k̂(1 + 2) = −3î + 3k̂, magnitude 3√2. a⃗₂ − a⃗₁ = î − 3ĵ − 2k̂. Dot = −3 − 6 = −9. d = 9/(3√2) = 3/√2.
Answer: 3/√2
Show that the lines (x − 1)/2 = (y − 2)/3 = (z − 3)/4 and (x − 4)/5 = (y − 1)/2 = z/1 intersect.
Show answer
b⃗₁ × b⃗₂ = î(3 − 8) − ĵ(2 − 20) + k̂(4 − 15) = −5î + 18ĵ − 11k̂. a⃗₂ − a⃗₁ = 3î − ĵ − 3k̂. Dot = −15 − 18 + 33 = 0 → shortest distance 0, and they are not parallel → they intersect.
Answer: Proved (they meet at (−1, −1, −1))
Find the shortest distance between the parallel lines r⃗ = î + ĵ + λ(2î − ĵ + k̂) and r⃗ = 2î + ĵ − k̂ + μ(4î − 2ĵ + 2k̂).
Show answer
Take b⃗ = 2î − ĵ + k̂ (|b⃗| = √6). a⃗₂ − a⃗₁ = î − k̂. b⃗ × (î − k̂) = î(1 − 0) − ĵ(−2 − 1) + k̂(0 + 1) = î + 3ĵ + k̂, magnitude √11. d = √11/√6.
Answer: √(11/6)
Find the vector equation of the line through (1, 2, −4) perpendicular to both (x − 8)/3 = (y + 19)/(−16) = (z − 10)/7 and (x − 15)/3 = (y − 29)/8 = (z − 5)/(−5).
Show answer
Direction = (3, −16, 7) × (3, 8, −5) = î(80 − 56) − ĵ(−15 − 21) + k̂(24 + 48) = 24î + 36ĵ + 72k̂ ∥ 2î + 3ĵ + 6k̂. r⃗ = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂).
Answer: r⃗ = î + 2ĵ − 4k̂ + λ(2î + 3ĵ + 6k̂)
13. Quick Revision Summary
- DCs: l² + m² + n² = 1; DRs proportional; through two points: differences.
- Line: r⃗ = a⃗ + λb⃗ ⇔ (x − x₁)/a = (y − y₁)/b = (z − z₁)/c.
- Angle: cos θ = |b⃗₁·b⃗₂|/(|b⃗₁||b⃗₂|); ⊥ if dot = 0; ∥ if DRs proportional.
- Skew lines: d = |(b⃗₁ × b⃗₂)·(a⃗₂ − a⃗₁)|/|b⃗₁ × b⃗₂|.
- Parallel lines: d = |b⃗ × (a⃗₂ − a⃗₁)|/|b⃗|.