1The idea in one paragraph

Multiply two vectors component by component and add: that gives a single number, the scalar product (or dot product) v · w = v₁w₁ + v₂w₂ + v₃w₃. The surprise is that the same number is also |v||w| cos θ, where θ is the angle between the vectors. So the scalar product links the numbers in two vectors to the angle between them, and rearranging gives cos θ = (v · w) ÷ (|v||w|). Two quick tests fall out. If v · w = 0, cos θ = 0 and the vectors are perpendicular; if |v · w| equals |v||w|, cos θ = ±1 and they are parallel. The sign alone tells you whether the angle is acute or obtuse. And because the product obeys the ordinary rules of algebra, you can expand brackets of vectors and prove geometry with them.

2The definition

For v = v₁i + v₂j + v₃k and w = w₁i + w₂j + w₃k,

v · w = v₁w₁ + v₂w₂ + v₃w₃ · in the formula booklet. The answer is a scalar, not a vector.

Worked example 1 (Paper 1). Find (3i + 2j − k) · (i − 4j + 2k).

Pair the i-components, the j-components and the k-components, multiply each pair, and add. A missing component is 0. The result has no i, j or k in it; writing 3i − 8j − 2k as the answer is the most common error on this page, and it scores nothing.

The base vectors show where the definition comes from. i · i = 1 because each has components 1, 0, 0, and i · j = 0 because 1 × 0 + 0 × 1 + 0 × 0 = 0. Expanding (v₁i + v₂j + v₃k) · (w₁i + w₂j + w₃k) term by term, every mixed product such as i · j vanishes, and only v₁w₁ + v₂w₂ + v₃w₃ survives.

3The rules it obeys

The guide lists four properties. Each follows from the definition by writing out components.

The last one is the most useful, and the reason is Pythagoras: v · v = v₁² + v₂² + v₃², which is |v|². Together these let you expand brackets exactly as in ordinary algebra, as long as every product is a dot product.

Worked example 2 (Paper 1). |a| = 3, |b| = 5 and a · b = −4. Find |a − 2b|.

You never needed components. That is the point of the properties: some questions give only lengths and a scalar product.

4The same number from lengths and an angle

The angle between two vectors is measured with both vectors starting at the same point, tail to tail, as in Figure 1. It always lies in the range 0 ≤ θ ≤ π. If the vectors are drawn nose to tail, slide one of them first; the angle at the join is not θ.

v · w = |v||w| cos θ · in the formula booklet, where θ is the angle between v and w.

Why the two forms agree. The guide's TOK question asks why the scalar product is defined as it is, and this is the answer. Draw v and w tail to tail; the third side of the triangle is v − w, as in Figure 2. Find |v − w|² in two ways.

Read backwards, the same two lines prove the cosine rule from the scalar product, which is the guide's enrichment suggestion. Either way, the component definition was chosen because it measures something geometric.

The sign tells you the angle. |v| and |w| are positive, so v · w has the same sign as cos θ. Figure 3 shows the three cases.

• v · w > 0: θ is acute.
• v · w = 0: θ = π/2, a right angle.
• v · w < 0: θ is obtuse.

In Worked example 1, the scalar product −7 says at once that the angle between those vectors is obtuse.

5Finding the angle between two vectors

Rearrange the geometric form:

cos θ = (v₁w₁ + v₂w₂ + v₃w₃) ÷ (|v||w|) · in the formula booklet.

Worked example 3 (Paper 1). Find the exact angle between a = i + j and b = j + k.

Worked example 4 (Paper 2). Find the angle between a = 2i − j + 2k and b = i + 4j − k, in degrees.

Keep the full calculator value of cos θ; rounding it to −0.31 first moves the angle by a tenth of a degree, enough to lose the accuracy mark. And check the answer against the sign: an obtuse answer here agrees with the negative scalar product.

6Angles in a triangle: choose the vectors that start at the vertex

For the angle ABC at the vertex B, both vectors must start at B: use BA⃗ and BC⃗. Using AB⃗ and BC⃗ measures the angle between one side and the extension of the other, which is π − B. Figure 4 shows the difference.

Worked example 5 (Paper 2). A(1, 0, 2), B(3, 1, 0) and C(4, −1, 3). Find angle ABC and the area of triangle ABC.

The area used the SL formula ½ab sin C with the two sides meeting at B. Finding sin B exactly from cos B avoids carrying a rounded angle. (In 3.16 the vector product gives the area more directly.)

7Perpendicular vectors

For non-zero vectors, cos θ = 0 exactly when θ = π/2, so:

v · w = 0 ⇔ v and w are perpendicular (for non-zero v and w).

This is the scalar product's most used fact. It proves a right angle, and it finds an unknown that makes one.

Worked example 6 (Paper 1). Find p such that 2i + pj − k is perpendicular to 3i − 2j + 4k.

The condition "non-zero" matters. 0 · w = 0 for every w, but the zero vector has no direction, so it is not perpendicular to anything in a meaningful sense.

To show a triangle has a right angle at Q, show QP⃗ · QR⃗ = 0; Q4 below does this.

8Parallel vectors

For parallel vectors θ = 0 (same direction) or θ = π (opposite directions), so cos θ = ±1 and

|v · w| = |v||w| for parallel vectors.

Worked example 7 (Paper 1). Show that u = 2i − j + 2k and w = −4i + 2j − 4k are parallel, and say whether they point the same way.

They are parallel and point in opposite directions, because the scalar product is negative. The test from 3.12, w = −2u, reaches the same conclusion faster when the components are simple. Use whichever the question points to; a "hence" after a scalar product calculation wants this one.

9What the number measures, and proofs

The projection. Rearranged, v · w ÷ |v| = |w| cos θ. That is the length of the "shadow" of w on the line of v, shown in Figure 5: how much of w points along v. In physics, the work done by a force F moving an object through a displacement s is F · s, the part of the force along the motion times the distance moved. A force at right angles to the motion does no work, which is the perpendicular test again.

Proofs. The properties turn geometric facts into algebra. The classic example is the angle in a semicircle.

Worked example 8 (Paper 1 or 3). A and B are the ends of a diameter of a circle with centre O, and P is any other point on the circle. Prove that angle APB is a right angle.

PA⃗ and PB⃗ are non-zero (P is neither A nor B) and their scalar product is 0, so they are perpendicular and angle APB = π/2. The proof never used coordinates, so it holds for every circle and every P.