IB Maths AI HL Vector Equations of Lines Paper 1 & 2 Scalar product ~8 min read

Angle Between Two Lines

The angle between two lines depends only on their direction vectors — the position vectors don’t matter. Use the scalar product: cos θ = (b₁ · b₂) / (|b₁| |b₂|). Take the absolute value on top to get the acute angle directly.

📘 What you need to know

Formula (in booklet): θ = cos⁻¹(b₁ · b₂|b₁| |b₂|)
Use direction vectors only — ignore a₁ and a₂. Two parallel lines have angle 0 even if they’re far apart.
Two pairs of angles are formed at any intersection: θ and 180° − θ (one acute, one obtuse). They add to 180°.
Sign of the scalar product:
· b₁ · b₂ > 0 ⇒ θ is acute
· b₁ · b₂ < 0 ⇒ θ is obtuse
· b₁ · b₂ = 0 ⇒ lines perpendicular
To always get the acute angle, put |…| around the scalar product: cos θ = |b₁ · b₂| / (|b₁| |b₂|).
Scalar product: b₁ · b₂ = l₁l₂ + m₁m₂ + n₁n₂.
Magnitude: |b| = √(l² + m² + n²).
Radians or degrees: check what the question asks for and set your GDC accordingly.

The formula and why |·| gives the acute angle

When two lines cross, they make four angles in two pairs: two acute angles (size θ) and two obtuse angles (size 180° − θ). The scalar product formula gives whichever angle corresponds to the chosen direction vectors. Flip one direction vector (multiply by −1) and the angle flips between θ and 180° − θ. Putting absolute value on the dot product chops off any negative sign — you always land on the acute angle.

The acute angle θ and obtuse angle 180° − θ both lie between the two lines. The sign of b₁ · b₂ tells you which one cos⁻¹ returns; use absolute value to lock onto the acute one.

Angle between two lines cos θ = b₁ · b₂|b₁| |b₂| or, for the acute angle, cos θ = |b₁ · b₂||b₁| |b₂| obtuse angle = 180° − θ (degrees) or π − θ (radians)

Special cases — parallel and perpendicular

Two quick checks from the direction vectors alone:

Parallel: b₁ is a scalar multiple of b₂ (i.e. b₁ = k b₂). Then θ = 0 (same direction) or θ = 180° (opposite). Perpendicular: b₁ · b₂ = 0. Then θ = 90°.

🧭 Recipe — angle between two lines

Pick out the direction vectors b₁ and b₂ from each line equation.
Compute the scalar product: b₁ · b₂ = l₁l₂ + m₁m₂ + n₁n₂.
Compute the magnitudes: |b₁| = √(l₁² + m₁² + n₁²), same for |b₂|.
For acute angle: take |dot product| on top, then θ = cos⁻¹(|b₁·b₂|/(|b₁||b₂|)).
For obtuse angle: compute the acute first, then subtract from 180° (or π).
Set GDC mode — radians or degrees to match the question.

Worked examples

WE 1

Acute angle in radians (from the PDF)

Find the acute angle, in radians, between l₁: r = (203) + λ(1−4−3) and l₂: r = (1−43) + μ(−325).

scalar product b₁·b₂ = (1)(−3) + (−4)(2) + (−3)(5) = −3 − 8 − 15 = −26 magnitudes |b₁| = √(1 + 16 + 9) = √26 |b₂| = √(9 + 4 + 25) = √38 acute angle: use |dot product| cos θ = 26 / (√26 · √38) θ = cos⁻¹(26 / √988) θ ≈ 0.597 radians (3 sf) sign of dot product was negative → obtuse first; absolute value flips to acute.

WE 2

Angle in degrees, acute case

Find the angle, in degrees, between lines with direction vectors b₁ = (212) and b₂ = (122).

dot product b₁·b₂ = 2(1) + 1(2) + 2(2) = 8 magnitudes |b₁| = √(4 + 1 + 4) = 3 |b₂| = √(1 + 4 + 4) = 3 angle cos θ = 8 / (3 · 3) = 8/9 θ = cos⁻¹(8/9) ≈ 27.3° dot product positive → already acute, no |…| needed.

WE 3

Perpendicular lines

Show that lines with directions b₁ = (23−1) and b₂ = (1−1−1) are perpendicular.

compute b₁·b₂ = 2(1) + 3(−1) + (−1)(−1) = 2 − 3 + 1 = 0 dot product = 0 ⇒ θ = 90° lines are perpendicular ✓ no need to compute magnitudes — zero dot product is enough.

WE 4

Obtuse angle requested

Lines have direction vectors b₁ = (110) and b₂ = (−101). Find the obtuse angle between the two lines in degrees.

dot product b₁·b₂ = 1(−1) + 1(0) + 0(1) = −1 magnitudes |b₁| = √2, |b₂| = √2 acute angle first (use |−1|) cos θ = 1 / (√2 · √2) = 1/2 θ = 60° obtuse = 180° − 60° obtuse angle = 120°

WE 5

Find the unknown that makes lines perpendicular

Find the value of k such that the lines with direction vectors b₁ = (2k−1) and b₂ = (32k) are perpendicular.

perpendicular ⇒ b₁·b₂ = 0 2(3) + k(2) + (−1)(k) = 0 6 + 2k − k = 0 6 + k = 0 k = −6 just set the dot product to zero — no magnitudes needed.

WE 6

Identify parallel vs intersecting lines

Two lines have direction vectors b₁ = (2−46) and b₂ = (−12−3). State the relationship between the directions.

check if scalar multiple b₁ = (2, −4, 6) = −2 · (−1, 2, −3) = −2 · b₂ b₁ is a scalar multiple of b₂ ⇒ parallel directions are parallel (angle = 0° or 180°) parallel directions don’t mean parallel lines — the lines could be identical. Need to also check a point doesn’t lie on both for “skew parallel” to make sense (actually parallel + share a point → same line).

💡 Top tips

Only direction vectors matter — ignore the a₁, a₂ position vectors entirely.
Take |b₁·b₂| at the top to lock onto the acute angle. Save 180° − θ for when the obtuse is specifically asked for.
For perpendicular, just check b₁ · b₂ = 0 — no need for magnitudes or arccos.
For parallel, check whether one direction vector is a scalar multiple of the other.
GDC angle mode — set radians vs degrees to match the question before pressing cos⁻¹.

⚠ Common mistakes

Using the position vectors a₁, a₂ in the formula — only direction vectors go in.
Forgetting absolute value when the acute angle is requested and the dot product is negative — gives the obtuse angle by mistake.
Wrong angle unit — answering in radians when degrees was asked (or vice versa).
Missing the magnitudes — without dividing by |b₁||b₂|, cos θ won’t be in [−1, 1] and cos⁻¹ errors out.
Assuming parallel directions ⇒ parallel lines — parallel directions can mean the same line. Check whether they share a point.

Next up — Shortest Distance Between a Point and a Line. Drop a perpendicular from the point to the line; that perpendicular’s length is the shortest distance. Find it by parameterising the foot of the perpendicular F and setting FP · b = 0 to solve for λ.

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