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. θacute = 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 1Find 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 2Acute 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 3Find 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 4Show 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 5Angle 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 6Find 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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