Grade by Grade
C. Geometry and Algebra of Vectors · Dot and Cross Products

The Dot Product: Geometric Definition

C2.4About 65 minutes
Learning goals
  • I can define the dot product of two vectors as u · v = |u||v| cos θ.
  • I can calculate a dot product from magnitudes and the angle between the vectors, including exact values.
  • I can use the sign of a dot product to tell whether the angle between two vectors is acute, right or obtuse.
  • I can find the angle between two vectors when I know their dot product and magnitudes.

Words to know

dot product
For vectors u and v with angle θ between them, u · v = |u||v| cos θ. Also called the scalar product.
scalar
A quantity that has size but no direction, such as 20 or −6√3.
angle between two vectors
The angle θ, from 0° to 180°, between the vectors when they are placed tail to tail.
orthogonal
Perpendicular. Two non-zero vectors are orthogonal when the angle between them is 90°.
standard unit vectors
i = [1, 0, 0], j = [0, 1, 0] and k = [0, 0, 1], each of length 1 along an axis.

How much do two vectors agree?

Picture pulling a toboggan across a snowy park with a rope. If the rope is level, all of your pull moves the toboggan forward. If the rope is tilted up, part of your pull lifts the front of the toboggan and only part of it moves the toboggan forward. If you could somehow pull straight up, none of it would move the toboggan forward at all.

We need a way to measure how much one vector acts in the direction of another. That measure is the dot product. It takes two vectors and gives back a single number, a scalar. That is why it is also called the scalar product.

Remember from Vectors in Two-Space and Three-Space: a vector has a magnitude |v| and a direction; you can add vectors, subtract them and multiply them by scalars. Remember from Grade 11 Physics (SPH3U, Work, Energy and Power): work is W = FΔd cos θ. The dot product is the mathematics hiding inside that formula.

The geometric definition

Let u and v be two vectors, and let θ be the angle between them when they are placed tail to tail, with 0° ≤ θ ≤ 180°. The dot product of u and v is

u · v = |u| |v| cos θ

Read u · v as "u dot v". Always write the dot; uv with nothing between them means nothing.

Three things to notice:

  • The answer is a number (a scalar), not a vector. It has no direction.
  • |u| and |v| are never negative, so the sign of u · v comes only from cos θ.
  • |v| cos θ is the length of the "shadow" of v on the line of u (when θ is acute). So the dot product is "the length of one vector times the shadow of the other". You will turn this shadow into the projection in lesson 5.
|v| cos θθuvOu · v = |u| |v| cos θ= |u| × (shadow of v on u)
Put the vectors tail to tail and measure the angle θ between them. The thick bar is the "shadow" of v on the line of u; its length is |v| cos θ. The dot product is |u| times that shadow.
θuv0° ≤ θ < 90°u · v > 0pointing roughly the same wayuvθ = 90°u · v = 0perpendicular (orthogonal)θuv90° < θ ≤ 180°u · v < 0pointing roughly opposite ways
The sign of the dot product tells you about the angle: positive for an acute angle, zero for a right angle, negative for an obtuse angle. The lengths only change the size of the number.
Angle θcos θu · vMeaning
0°1|u||v| (largest possible)same direction
between 0° and 90°positivepositiveacute angle
90°00perpendicular (orthogonal)
between 90° and 180°negativenegativeobtuse angle
180°−1−|u||v| (smallest possible)opposite directions
Because −1 ≤ cos θ ≤ 1, the dot product is always between −|u||v| and |u||v|.
Worked example

Example 1: Using the definition

|u| = 5, |v| = 8 and the angle between u and v is 60°. Find u · v.

  1. Write the definition: u · v = |u||v| cos θ.
  2. Substitute: (5)(8) cos 60°.
  3. cos 60° = 1/2, so u · v = 40 × 1/2 = 20.

u · v = 20. It is positive because 60° is acute.

Worked example

Example 2: An obtuse angle, exactly

|a| = 3, |b| = 4 and the angle between a and b is 150°. Find a · b exactly.

  1. a · b = (3)(4) cos 150°.
  2. 150° is in quadrant II with related acute angle 30°, so cos 150° = −√3/2 (from the CAST rule in Grade 11).
  3. a · b = 12(−√3/2) = −6√3.

a · b = −6√3 ≈ −10.39. The negative sign tells you the angle is obtuse.

Tails together!

The angle in the definition is the angle between the vectors tail to tail. If a diagram shows two vectors joined head to tail (as in a triangle), slide one of them so the tails meet before you read the angle. A common mistake is to use the inside angle of the triangle.

AB moved to B70°110°ABC|AB| = 6|BC| = 5
AB and BC are joined head to tail, so the 70° triangle angle is not the angle between them. Slide AB so its tail is at B (dashed line): the angle between the vectors is 180° − 70° = 110°. (Side lengths are not drawn to scale.)
Worked example

Example 3: Vectors around a triangle

In △ABC, |AB| = 6, |BC| = 5 and the interior angle ∠ABC = 70°. Find AB · BC.

  1. AB ends at B and BC starts at B: they are head to tail.
  2. Move AB so its tail is at B. It now points straight out of the triangle along the extension of AB. The angle between it and BC is 180° − 70° = 110°.
  3. AB · BC = (6)(5) cos 110° ≈ 30(−0.3420) ≈ −10.26.

AB · BC ≈ −10.26. (Using 70° by mistake would give +10.26: the size looks right but the sign is wrong.)

Worked example

Example 4: Working backwards to an angle

u · v = −12, |u| = 4 and |v| = 6. Find the angle between u and v.

  1. Start from u · v = |u||v| cos θ: −12 = (4)(6) cos θ.
  2. cos θ = −12/24 = −1/2.
  3. On 0° ≤ θ ≤ 180° there is only one angle with cosine −1/2: θ = 120°.

θ = 120°. (Since θ is limited to 0° to 180°, cos⁻¹ always gives the one correct angle; there is no second answer to worry about, unlike sin⁻¹ in triangles.)

Dot products of unit vectors

The standard unit vectors i = [1, 0, 0], j = [0, 1, 0] and k = [0, 0, 1] each have length 1 and are perpendicular to each other. So:

  • i · i = (1)(1) cos 0° = 1, and the same for j · j and k · k.
  • i · j = (1)(1) cos 90° = 0, and the same for every pair of different unit vectors.

Also, for any vector, v · v = |v||v| cos 0° = |v|². These facts are the key to the fast Cartesian formula for the dot product in the next lesson.

Just for fun

Why did the two perpendicular vectors never argue?

Their dot product was zero: they had absolutely nothing in common.

Hands-on activity

Pencil vectors and flashlight shadows

You need: two pencils (or a pencil and a ruler), a ruler, a protractor, a sheet of paper, a phone flashlight (or a desk lamp), and a calculator.

  1. Measure each pencil to the nearest millimetre. Call the lengths |u| and |v| (in cm).
  2. Draw a straight line on the paper. Lay the first pencil (u) along it. Put the eraser end of the second pencil (v) at the eraser end of u, so they are tail to tail.
  3. Use the protractor to set v at 30° to u. Hold the flashlight straight above the paper so that v casts a shadow on the line. Mark the ends of the shadow and measure its length.
  4. Compare the shadow with |v| cos 30°. Then compute u · v two ways: |u| × (shadow) and |u||v| cos 30°.
  5. Repeat for 60°, 90° and 120°. For 120°, the shadow falls behind the tail, on the far side from u: count that length as negative.

What to look for: with 18 cm and 15 cm pencils, u · v should be about 234 at 30°, 135 at 60°, 0 at 90° and −135 at 120° (in cm²). Your shadow measurements should land within a few percent. Keep the light high and directly overhead, or the shadow will stretch.

Your turn

1

|u| = 7, |v| = 2 and θ = 0°. Find u · v.

2

|a| = 10, |b| = 3 and θ = 45°. Find a · b exactly.

3

|p| = 6, |q| = 9 and θ = 120°. Find p · q.

4

|u| = 4, |v| = 5 and θ = 90°. Find u · v.

5

u · v = 15, |u| = 5, |v| = 6. Find θ.

6

|a| = 8 and |b| = 2.5. What are the largest and smallest possible values of a · b?

7

Find (i · i) + (j · k).

More practice: warm-up

1

Is the dot product a vector or a scalar?

2

|u| = 3, |v| = 4, θ = 60°. Find u · v.

3

u · v < 0. Is the angle between u and v acute, right or obtuse?

4

|v| = 6. Find v · v.

5

Find j · j and i · k.

More practice: on your own

1

|u| = 12, |v| = 5 and θ = 135°. Find u · v exactly and to two decimal places.

2

|a| = 4, |b| = 7 and a · b = 20. Find θ to the nearest tenth of a degree.

3

△PQR is equilateral with sides 2. Find PQ · PR and PQ · QR.

4

u · v = −18 and |u| = |v| = 6. Describe the vectors.

5

A square ABCD has side 3. Find AB · AC.

Challenge

1

|u| = 3 and |v| = 5. Can u · v = 16? Explain.

2

In a regular hexagon ABCDEF with side 4, find AB · BC and AB · DE.

3

Two unit vectors satisfy a · b = cos 50°. A student says the angle between them must be 50°. Is the student right? Explain.

Check your understanding

Quiz

1.

What kind of quantity is u · v?

2.

|u| = 6, |v| = 5 and θ = 60°. What is u · v?

3.

Two non-zero vectors have a negative dot product. What can you say about the angle between them?

4.

What is v · v equal to?

5.

u · v = −6, |u| = 3 and |v| = 4. What is the angle between u and v?

6.

In equilateral △ABC with side 1, what is AB · BC?

For parents and teachers

Big idea (C2.4): the dot product u · v = |u||v| cos θ measures how much two vectors point the same way. The result is a number, not a vector. Its sign tells you whether the angle is acute (positive), right (zero) or obtuse (negative).

How to help: Ask your teen to explain the "shadow" picture: the dot product is the length of one vector times the shadow of the other. Name two lengths and an angle and ask for the dot product, then ask "Should it be positive or negative?" before they calculate.

Common difficulties:

  • Using the inside angle of a triangle when the vectors are head to tail. The vectors must be placed tail to tail.
  • Writing the answer as a vector, or forgetting to write the dot.
  • Calculator in radian mode (from MHF4U). This unit uses degrees.
  • Thinking the dot product can be bigger than |u||v|. It cannot, because cos θ ≤ 1.