IB Maths AI HL Vector Properties Paper 1 & 2 cos θ = v · w / (|v||w|) ~8 min read

Angle Between Two Vectors & Perpendicular Vectors

Rearranging the geometric formula for the scalar product gives a direct way to compute the angle between any two vectors: cos θ = (v · w) / (|v||w|). Combine that with the magnitude formula and you have a one-step angle-finder. A special case follows immediately: v · w = 0 means the vectors are perpendicular (and the converse holds too) — one line of arithmetic decides perpendicularity, with no calculator needed.

📘 What you need to know

Angle formula: cos θ = (v · w) / (|v||w|) — given in the formula booklet (geometry & trigonometry section).
Procedure: compute the scalar product, compute both magnitudes, divide, then take arccos.
Sign of v · w reveals the angle type: positive → acute (θ < 90°), zero → right (θ = 90°), negative → obtuse (θ > 90°).
Perpendicular test: non-zero vectors v and w are perpendicular if and only if v · w = 0. Compute the dot product and check the value.
Finding an unknown component for perpendicularity: set v · w = 0, expand, and solve a linear equation in the unknown.
Angle at a vertex: for the angle at A in triangle ABC, compute the displacement vectors AB and AC first, then apply the angle formula to those.

Computing the angle

The recipe is short: dot, two magnitudes, divide, arccos. The dot product carries the geometric information (positive ↔ acute, negative ↔ obtuse), and dividing by the magnitudes normalises the result into a cosine. One small care point: many questions give vectors in column form, others in i–j–k form, and a few define points from which you must first compute the displacement vectors before applying the formula. Identify the form early and translate as needed.

The angle θ between v and w is computed from the scalar product divided by the product of the magnitudes; taking arccos returns the angle. The special case v · w = 0 fixes θ = 90° directly — no calculator step needed.

Angle between two vectors cos θ = v · w|v| |w| · perpendicular: v · w = 0 scalar product over product of magnitudes; arccos gives θ

Perpendicularity and unknowns

The perpendicular condition v · w = 0 is the simplest test in the entire vector chapter: compute one dot product, check for zero. The flip side — “find the value of the unknown that makes these perpendicular” — turns into solving a small linear equation. Expand the dot product symbolically (with the unknown in place), set the whole expression equal to zero, and solve. The unknown may appear in either or both vectors; the principle is the same.

Sanity check: before reaching for arccos, look at the sign of v · w. A positive dot product means the angle must be acute (under 90°); a negative one means obtuse (over 90°); zero means exactly 90°. If your final θ doesn’t match the sign, retrace the arithmetic.

🧭 Recipe — angle & perpendicularity

Compute v · w using the component formula (multiply matching components, then sum).
Compute |v| and |w| separately, using the magnitude formula.
Divide: cos θ = (v · w) / (|v||w|).
Apply arccos on the calculator (mode set to degrees) to obtain θ.
For perpendicularity: set the dot product equal to zero and solve for any unknown component instead.

Worked examples

WE 1

Angle between two 3D vectors

Find the angle between v = (3, 0, 4)^T and w = (1, 2, 2)^T, giving your answer to 3 s.f.

scalar product v · w = 3·1 + 0·2 + 4·2 = 11 magnitudes |v| = √(9+0+16) = √25 = 5 |w| = √(1+4+4) = √9 = 3 divide and take arccos cos θ = 11/(5·3) = 11/15 θ = arccos(11/15) ≈ 42.833° θ ≈ 42.8° (3 s.f.) positive v · w ⇒ acute angle, matches answer.

WE 2

2D angle

Find the angle between p = (3, 4)^T and q = (−1, 2)^T, giving your answer to 3 s.f.

two-component dot product p · q = 3·(−1) + 4·2 = −3 + 8 = 5 magnitudes |p| = √(9+16) = √25 = 5 |q| = √(1+4) = √5 divide and take arccos cos θ = 5/(5·√5) = 1/√5 ≈ 0.4472 θ = arccos(1/√5) ≈ 63.435° θ ≈ 63.4° (3 s.f.) positive dot product ⇒ acute angle; same formula, just two components instead of three.

WE 3

Test for perpendicularity

Determine whether the vectors a = (3, 2, −1)^T and b = (1, −2, −1)^T are perpendicular.

compute a · b a · b = 3·1 + 2·(−2) + (−1)·(−1) = 3 − 4 + 1 = 0 a · b = 0 ⇒ perpendicular yes — a and b are perpendicular no need for magnitudes or arccos; the zero dot product settles it directly.

WE 4

Find unknown for perpendicular condition

Find the value of k such that the vectors u = (4, k, −2)^T and v = (3, 5, k)^T are perpendicular.

set u · v = 0 u · v = 4·3 + k·5 + (−2)·k = 12 + 5k − 2k = 12 + 3k solve 12 + 3k = 0 3k = −12 k = −4 verify: u = (4, −4, −2), v = (3, 5, −4); u · v = 12 − 20 + 8 = 0 ✓.

WE 5

Obtuse angle — negative dot product

Find the angle between a = (2, 1, −2)^T and b = (−1, 2, 2)^T, giving your answer to 3 s.f.

scalar product (negative ⇒ obtuse) a · b = 2·(−1) + 1·2 + (−2)·2 = −2 + 2 − 4 = −4 magnitudes |a| = √(4+1+4) = √9 = 3 |b| = √(1+4+4) = √9 = 3 divide and take arccos cos θ = −4/(3·3) = −4/9 θ = arccos(−4/9) ≈ 116.388° θ ≈ 116° (3 s.f.) negative cosine gives an angle between 90° and 180° — matches the obtuse expectation.

WE 6

Angle at a vertex of a triangle

Three points have coordinates A(1, 3, −1), B(3, 4, 1), and C(3, 5, 0). Find the angle BAC (the angle at vertex A), giving your answer to 3 s.f.

find displacement vectors from A AB = b − a = (2, 1, 2)^T AC = c − a = (2, 2, 1)^T scalar product AB · AC = 4 + 2 + 2 = 8 magnitudes |AB| = √(4+1+4) = 3 |AC| = √(4+4+1) = 3 divide and take arccos cos(BAC) = 8/(3·3) = 8/9 BAC = arccos(8/9) ≈ 27.266° BAC ≈ 27.3° (3 s.f.) always use displacement vectors *from* the vertex (here A) — not the position vectors of B and C.

💡 Top tips

For “are they perpendicular?”, compute only the dot product and check for zero — no need for magnitudes or arccos.
Use the sign of v · w as a check: an “obtuse” answer (over 90°) requires a negative dot product; reconcile this before reading off your calculator.
Calculator mode: ensure degrees vs radians matches what the question asks for. The default in IB exam answers is degrees, to 3 s.f.
Triangle problems: the angle at vertex A uses AB and AC — both start at A. Never use position vectors directly.
For unknown components, set v · w = 0 immediately when you read “perpendicular” — that’s the only equation you’ll need.

⚠ Common mistakes

Forgetting the denominator: writing cos θ = v · w without dividing by |v||w|.
Using position vectors for an angle at a vertex: the angle BAC needs AB and AC, not OB and OC.
Missing the negative sign in cos θ for obtuse angles, then reporting an acute angle by accident.
Using sin instead of cos — the dot product formula uses cos θ, not sin θ. (Sin appears in the vector product, coming next sub-topic.)
Treating “zero dot product” as a coincidence — it’s a definite proof of perpendicularity (for non-zero vectors), worth a full mark.

Next up — The Vector Product (also called the cross product). It produces a vector perpendicular to both inputs, with magnitude |v||w| sin θ. Where the scalar product detects perpendicularity by giving zero, the vector product gives zero for parallel vectors instead — a complementary diagnostic that unlocks area and volume calculations later in the chapter.

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