Angle between two lines
Two roads meet at a crossing. The angle you turn depends only on the directions of the roads, not on where they begin. So we compare the direction ratios.
The angle between two lines is the angle between their direction vectors. If the ratios are (a₁, b₁, c₁) and (a₂, b₂, c₂), the acute angle θ satisfies cos θ as below. In the vector form the same number is |b₁ · b₂|/(|b₁| |b₂|).
Formula: cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / (√(a₁² + b₁² + c₁²) · √(a₂² + b₂² + c₂²))
Rule 1: Dot product over product of lengths — Take the modulus for the acute angle.
- Easy: Ratios (1, 1, 0) and (0, 1, 1) give cos θ = 1/2, so θ = 60°.
- Medium: Ratios (2, 3, 6) and (1, 2, 2) give cos θ = 20/21.
- Exam-type: The x-axis and the line with ratios (1, 1, 1) make cos θ = 1/√3.
Rule 2: Special angles — Right angle if the dot product is zero.
- Easy: Ratios (1, 2, 2) and (2, 1, −2) give 2 + 2 − 4 = 0, so the angle is 90°.
- Medium: Ratios (3, 4, 0) and (4, 3, 0) give cos θ = 24/25.
- Exam-type: Ratios (1, −1, 0) and (0, 1, −1) give cos θ = 1/2, so θ = 60°.
Use the direction ratios only, not the points. The modulus gives the acute angle.
Forgetting the square roots in the denominator.
Dot over lengths, then cos inverse.
Practice questions
- 1Find the angle between lines with direction ratios (1, 1, 0) and (0, 1, 1).
- 2Find cos θ for direction ratios (2, 3, 6) and (1, 2, 2).
- 3Find the angle between the x-axis and a line with ratios (1, 1, 1).
- 4Find the angle between r = λ(i + j) and r = μ(j + k).
- 5Find the angle between direction ratios (1, 2, 2) and (2, 1, −2).
- 6Find the angle between direction ratios (1, −1, 0) and (0, 1, −1).
- 7Find cos θ for direction ratios (3, 4, 0) and (4, 3, 0).
- 8Find the angle between the lines x/1 = y/0 = z/1 and x/1 = y/1 = z/0.
Show answers (try them yourself first)
- 1cos θ = 1/(√2 · √2) = 1/2, so 60°
- 2|2 + 6 + 12|/(7 × 3) = 20/21
- 3cos θ = 1/√3, so θ = cos⁻¹(1/√3), about 54.7°
- 460°
- 5cos θ = 0, so θ = 90°
- 660°
- 724/25
- 8cos θ = 1/(√2 · √2) = 1/2, so 60°
Parallel and perpendicular lines
Railway tracks stay at the same distance because their directions match. A road crossing at a right angle has a direction with zero dot product.
Two lines are parallel if their direction ratios are proportional, and perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0. Use these two conditions to find unknown constants.
Formula: Parallel: a₁/a₂ = b₁/b₂ = c₁/c₂. Perpendicular: a₁a₂ + b₁b₂ + c₁c₂ = 0
Rule 1: Check the condition — Compare ratios or compute the dot product.
- Easy: (2, 3, 6) and (3, −2, 0) are perpendicular, since 6 − 6 + 0 = 0.
- Medium: (1, 2, 3) and (2, 4, 6) are parallel.
- Exam-type: (1, 2, 3) and (2, 3, 4) are neither, since the dot product is 20 and the ratios differ.
Rule 2: Find the unknown — Put the condition in as an equation.
- Easy: (1, k, 2) is perpendicular to (3, 1, −1) when 3 + k − 2 = 0, so k = −1.
- Medium: (2, k, 4) is parallel to (1, 3, 2) when k/3 = 2, so k = 6.
- Exam-type: The cosines (1/3, 2/3, 2/3) and (2/3, 1/3, −2/3) give 2/9 + 2/9 − 4/9 = 0, so these lines are perpendicular.
Parallel means proportional, perpendicular means the dot product is zero.
Using the perpendicular condition for parallel lines or the reverse.
Parallel is a multiple, perpendicular is zero.
Practice questions
- 1Are the lines with ratios (2, 3, 6) and (3, −2, 0) perpendicular?
- 2Are the lines with ratios (1, 2, 3) and (2, 4, 6) parallel?
- 3Are the lines with ratios (1, 2, 3) and (2, 3, 4) parallel or perpendicular?
- 4Find k so that (1, k, 2) is perpendicular to (3, 1, −1).
- 5Find k so that (2, k, 4) is parallel to (1, 3, 2).
- 6Are the lines with cosines (1/3, 2/3, 2/3) and (2/3, 1/3, −2/3) perpendicular?
- 7Is the line through the origin and (4, 6, 12) parallel to (x − 1)/2 = (y − 1)/3 = (z − 1)/6?
- 8Are the x-axis and a line with ratios (0, 3, 4) perpendicular?
Show answers (try them yourself first)
- 1Yes, 6 − 6 + 0 = 0
- 2Yes, the second is twice the first
- 3Neither: the dot product is 20 and the ratios are not proportional
- 43 + k − 2 = 0, so k = −1
- 52/1 = k/3 = 4/2 gives k = 6
- 6Yes, 2/9 + 2/9 − 4/9 = 0
- 7Yes, since (4, 6, 12) = 2 (2, 3, 6)
- 8Yes, since 1 × 0 + 0 × 3 + 0 × 4 = 0