IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read HL only

Angle Between Two Lines

The angle between two lines is the angle between their direction vectors — found from the scalar product. Two lines crossing produce two angles (one acute, one obtuse), and the sign of b₁·b₂ tells you which one your formula gave.

📘 What you need to know

Formula: θ = cos⁻¹(b₁ · b₂|b₁| |b₂|) — uses the direction vectors only.
Two angles at the intersection: an acute θ and an obtuse 180° − θ (or π − θ).
Sign of dot product: positive → formula returns acute; negative → formula returns obtuse.
For the acute angle, take the absolute value: θ = cos⁻¹(|b₁ · b₂||b₁| |b₂|).
Perpendicular lines: b₁ · b₂ = 0.
Lines don’t have to intersect for the angle to make sense — use the directions even if skew.
Anchor points don’t matter — only directions affect the angle.

The angle formula

Angle between two lines cos θ = b₁ · b₂|b₁| |b₂|

It’s the same formula as the angle between two vectors — applied to the directions b₁ and b₂ of the lines. Anchors play no role.

Two angles, one formula

Positive dot product

b₁ · b₂ > 0

formula gives the acute angle directly

Negative dot product

b₁ · b₂ < 0

formula gives the obtuse angle — subtract from 180° (or π) for the acute

For the acute angle every time: drop the sign with absolute value. θ_acute = cos⁻¹(|b₁·b₂| / (|b₁||b₂|)) — works no matter which way the directions point.

🧭 Recipe — angle between two lines

Read off the direction vectors b₁ and b₂ (ignore anchor points).
Compute the scalar product b₁·b₂.
Compute magnitudes |b₁| and |b₂|.
Decide: for the acute angle, use |b₁·b₂| in the numerator.
Apply: θ = cos⁻¹(numerator ÷ (|b₁||b₂|)). Convert units (degrees/radians) as the question asks.

Worked examples

WE 1

Find the acute angle in degrees

Find the acute angle, in degrees, between the lines r₁ = (1, 2, 3) + λ(2, 1, −1) and r₂ = (4, 0, −2) + μ(1, 3, 2).

Step 1: Scalar product of direction vectors b₁·b₂ = (2)(1) + (1)(3) + (−1)(2) = 2 + 3 − 2 = 3 Step 2: Magnitudes |b₁| = √(4+1+1) = √6 |b₂| = √(1+9+4) = √14 Step 3: Apply the formula cos θ = 3 / (√6 · √14) = 3/√84 ≈ 0.3273 θ ≈ 70.9° positive dot product → the formula already gave the acute angle

WE 2

Acute angle in radians (negative dot product)

Find the acute angle, in radians, between the lines with direction vectors b₁ = 3i − 2j + k and b₂ = i + 4j − 3k.

Step 1: Scalar product b₁·b₂ = (3)(1) + (−2)(4) + (1)(−3) = 3 − 8 − 3 = −8 Step 2: Magnitudes |b₁| = √(9+4+1) = √14 |b₂| = √(1+16+9) = √26 Step 3: Use absolute value for the acute angle cos θ = |−8| / (√14 · √26) = 8/√364 ≈ 0.4193 θ ≈ 1.14 radians absolute value flips the sign — saves an extra “180° − …” step

WE 3

Find both angles between two lines

Two lines have direction vectors b₁ = (2, 0, −1) and b₂ = (1, 2, 3). Find both angles formed where the lines meet, in degrees.

Step 1: Scalar product and magnitudes b₁·b₂ = 2 + 0 − 3 = −1 |b₁| = √5; |b₂| = √14 Step 2: Direct formula → obtuse (negative dot product) cos θ = −1/√70 ≈ −0.1195 θ ≈ 96.9° (obtuse) Step 3: Acute = 180° − obtuse 180° − 96.9° = 83.1° Acute ≈ 83.1°, Obtuse ≈ 96.9° they sum to 180° — that’s how supplementary angles at an intersection work

WE 4

Show two lines are perpendicular

Show that the lines with direction vectors b₁ = 2i + j − 2k and b₂ = i + 4j + 3k are perpendicular.

Compute the scalar product b₁·b₂ = (2)(1) + (1)(4) + (−2)(3) = 2 + 4 − 6 = 0 b₁·b₂ = 0 → lines are perpendicular no need to compute the angle — zero dot product is the test

WE 5

Angle between a line through two points and a given direction

Line l₁ passes through A(1, 0, −2) and B(3, 4, 1). Line l₂ has direction vector d = −i + 2j − k. Find the acute angle between l₁ and l₂, in degrees.

Step 1: Direction of l₁ is AB AB = B − A = (2, 4, 3) Step 2: Scalar product with d AB·d = (2)(−1) + (4)(2) + (3)(−1) = −2 + 8 − 3 = 3 Step 3: Magnitudes |AB| = √(4+16+9) = √29; |d| = √(1+4+1) = √6 Step 4: Apply formula (positive dot product → acute) cos θ = 3 / (√29 · √6) = 3/√174 ≈ 0.2274 θ ≈ 76.9° when only points are given, compute AB first to get the line’s direction

WE 6

Find an unknown so two lines are perpendicular

The lines with direction vectors b₁ = 3i + 2j + kk and b₂ = 4i − j + 2k are perpendicular. Find the value of k.

Step 1: Set b₁·b₂ = 0 (3)(4) + (2)(−1) + (k)(2) = 0 Step 2: Simplify and solve 12 − 2 + 2k = 0 10 + 2k = 0 2k = −10 k = −5 verify: (3)(4) + (2)(−1) + (−5)(2) = 12 − 2 − 10 = 0 ✓

💡 Top tips

Anchors are irrelevant for the angle — use directions only.
Use absolute value to get the acute angle every time without case-splitting.
Read the units carefully — “degrees” or “radians” — and set your calculator accordingly.
Lines don’t need to intersect: skew lines still have a well-defined angle from their directions.
Round only at the end; carry √ values through to avoid rounding error.

⚠ Common mistakes

Forgetting the absolute value when the question asks for the acute angle — yields the obtuse one when dot product is negative.
Using anchors (a₁, a₂) in the formula instead of directions.
Calculator in wrong mode — degrees vs radians mismatch.
Concluding “perpendicular” based on a small dot product instead of zero.
Sign errors in the scalar product with negative components — write each multiplication out explicitly.

Next: Shortest Distance Between a Point and a Line. The shortest distance is always the perpendicular distance — drop a perpendicular from the point to the line and find its length. Two main methods: parameterise and minimise via dot product, or use the vector product as a one-shot formula.

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