Practice Maths

The Dot Product

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Key Terms

Dot product (geometric)
a · b = |a| |b| cos θ where θ is the angle between the vectors.
Dot product (component)
a · b = a1b1 + a2b2 (for 2D vectors).
Perpendicular
a ⊥ b if and only if a · b = 0.
Parallel
a ∥ b if and only if |a · b| = |a| |b| (i.e. cos θ = ±1).
Scalar projection
The projection of a onto b = (a · b) / |b|.
Vector projection
projba = [(a · b) / |b|²] b.

The Dot Product — Definition and Properties

Let a = a1i + a2j and b = b1i + b2j.

Algebraic form:   a · b = a1b1 + a2b2

Geometric form:   a · b = |a||b| cos θ   (where θ is the angle between a and b, 0 ≤ θ ≤ π)

Angle formula:   cos θ = (a · b) / (|a||b|)

Perpendicularity test: a ⊥ b if and only if a · b = 0

Self-dot product: a · a = |a|2

Sign of the dot product: positive ⇒ acute angle; zero ⇒ right angle; negative ⇒ obtuse angle.

Projections

Scalar projection of a onto b (the signed length of the shadow of a in the direction of b):

compb(a) = (a · b) / |b|

Vector projection of a onto b (the component of a in the direction of b, as a vector):

projb(a) = (a · b) / |b|2 × b

Geometrically: drop a perpendicular from the tip of a onto the line of b. The foot of that perpendicular is the tip of the vector projection.

Worked Example 1 — Computing the Dot Product and Angle

Find a · b and the angle between a = 3i + 4j and b = i + 2j.

Step 1: a · b = 3 × 1 + 4 × 2 = 3 + 8 = 11

Step 2: |a| = √(9 + 16) = 5.   |b| = √(1 + 4) = √5.

Step 3: cos θ = 11 / (5√5) = 11√5 / 25.

Step 4: θ = arccos(11√5/25) ≈ 10.3°

Worked Example 2 — Testing Perpendicularity

Are a = 3i − 2j and b = 4i + 6j perpendicular?

a · b = 3(4) + (−2)(6) = 12 − 12 = 0

Since a · b = 0, yes — the vectors are perpendicular. ✓

Worked Example 3 — Vector Projection

Find the vector projection of a = 5i + 2j onto b = 3i + 4j.

Step 1: a · b = 5(3) + 2(4) = 15 + 8 = 23.

Step 2: |b|2 = 32 + 42 = 9 + 16 = 25.

Step 3: projb(a) = (23/25)(3i + 4j) = (69/25)i + (92/25)j

Hot Tip — What the Dot Product Measures: The dot product measures how much two vectors “point in the same direction.” If θ = 0° (same direction), a · b = |a||b| (maximum). If θ = 90°, a · b = 0. If θ = 180° (opposite directions), a · b = −|a||b| (minimum). The sign tells you whether the angle is acute, right, or obtuse.

The Geometric Meaning of the Dot Product

When two vectors a and b are placed tail-to-tail, they form an angle θ between them. The dot product a · b = |a||b|cosθ captures how much the vectors “cooperate” in the same direction.

Think of it this way: |b|cosθ is the length of the shadow that b casts onto the line of a. Multiplying this shadow length by |a| gives the dot product. It is the product of one vector’s magnitude and the other’s component along it.

Computing the Dot Product Algebraically

In component form, a · b = a1b1 + a2b2. This is simply: multiply the i-components together, multiply the j-components together, and add. The dot product produces a scalar (a plain number), not a vector.

Key properties: a · b = b · a (commutative); a · (b + c) = a · b + a · c (distributive).

Finding the Angle Between Vectors

Rearranging the geometric form: cos θ = (a · b) / (|a||b|). Since 0 ≤ θ ≤ 180°, the angle is uniquely determined by its cosine. If the cosine is positive, the angle is acute; if zero, the angle is 90°; if negative, the angle is obtuse.

The Perpendicularity Test

Vectors a and b are perpendicular (orthogonal) if and only if a · b = 0. This is one of the most powerful and frequently used tools in geometry proofs. Given a vector a = a1i + a2j, any vector perpendicular to it has the form a2i − a1j (swap components and negate one).

Scalar and Vector Projections

The scalar projection compb(a) = (a · b)/|b| is the signed length of the projection. It is positive if a has a component in the same direction as b, and negative if in the opposite direction.

The vector projection projb(a) = ((a · b)/|b|2)b is that same projection as an actual vector pointing in the direction of b.

Applications: resolving forces into components, finding the closest point on a line to a given point, and calculating work done (work = force · displacement = |F||d|cosθ).

Classifying Triangles Using the Dot Product

Given a triangle with vertices P, Q, R, you can classify each angle by computing the dot product of the two sides meeting at that vertex. If positive, the angle is acute; if zero, it is a right angle; if negative, it is obtuse. This is far more efficient than computing the actual angles.

Mastery Practice

  1. Fluency

    Q1 — Calculate Dot Products

    Given a = 2i + 3j and b = 4i − j, find:   (a) a · b    (b) a · a    (c) b · b

  2. Fluency

    Q2 — Test Perpendicularity

    Determine whether each pair of vectors is perpendicular:   (a) a = 3i − 2j and b = 2i + 3j    (b) a = 4i + j and b = 2i − 8j    (c) a = 5i and b = 3j

  3. Fluency

    Q3 — Find the Angle Between Vectors

    Find the exact angle between a = i and b = i + j.

  4. Fluency

    Q4 — Find k for Perpendicularity

    Find the value of k so that a = ki + 3j and b = 2i − 4j are perpendicular.

  5. Understanding

    Q5 — Scalar Projection

    Find the scalar projection of a = 7i + 2j onto b = 3i + 4j.

  6. Understanding

    Q6 — Vector Projection

    Find the vector projection of a = 7i + 2j onto b = 3i + 4j.

  7. Understanding

    Q7 — Angle in a Triangle

    Points A = (1, 0), B = (4, 0), C = (1, 3) form a triangle. Find the angle at vertex A using the dot product.

  8. Understanding

    Q8 — Dot Product Equation

    Given |a| = 5, |b| = 4, and a · b = 10, find the angle between a and b.

  9. Problem Solving

    Q9 — Classifying a Triangle

    Triangle PQR has vertices P = (0, 0), Q = (4, 0), R = (1, 3). Use dot products to classify each angle as acute, right, or obtuse, and hence classify the triangle.

  10. Problem Solving

    Q10 — Rhombus Diagonals are Perpendicular

    ABCD is a rhombus with A = (0, 0), B = (3, 1), C = (4, 4), D = (1, 3). (a) Verify it is a rhombus by checking all four sides are equal. (b) Prove the diagonals AC and BD are perpendicular using the dot product.

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