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

Properties of the Dot Product and Orthogonal Vectors

C2.5About 70 minutes
Learning goals
  • I can state and prove the commutative, distributive and scalar-multiple properties of the dot product.
  • I can explain why the dot product has no associative property.
  • I can use u · u = |u|² to find lengths of combinations such as |a + b|.
  • I can test whether vectors are orthogonal and find vectors orthogonal to given vectors.

Words to know

commutative property
Order does not matter: u · v = v · u.
distributive property
u · (v + w) = u · v + u · w.
orthogonal vectors
Vectors at right angles. Non-zero vectors u and v are orthogonal exactly when u · v = 0.
zero vector
The vector 0 with magnitude 0 and no direction; 0 · v = 0 for every v.

Rules you can count on

In algebra you rely on rules like ab = ba and a(b + c) = ab + ac without thinking. Before you can calculate freely with dot products, you need to know which of those rules still hold. Most do, one does not even make sense, and the most useful new fact is a test for right angles: two non-zero vectors are perpendicular exactly when their dot product is 0.

PropertyIn symbols
Commutativeu · v = v · u
Distributiveu · (v + w) = u · v + u · w
Scalar multiples(ku) · v = k(u · v) = u · (kv)
Dot with itselfu · u = |u|²
Zero vector0 · u = 0
Orthogonalityu · v = 0 ⇔ u ⊥ v (u, v non-zero)
Properties of the dot product. They hold in two-space and three-space.

Proving the properties

Each property can be proved from the Cartesian formula. For example, with u = [u₁, u₂, u₃], v = [v₁, v₂, v₃] and w = [w₁, w₂, w₃]:

Commutative: u · v = u₁v₁ + u₂v₂ + u₃v₃ = v₁u₁ + v₂u₂ + v₃u₃ = v · u, because multiplication of numbers is commutative.

Scalar multiples: (ku) · v = (ku₁)v₁ + (ku₂)v₂ + (ku₃)v₃ = k(u₁v₁ + u₂v₂ + u₃v₃) = k(u · v).

You will prove the distributive property in Example 1.

Watch out: no associativity

There is no associative property. The expression (u · v) · w makes no sense: u · v is a number, and you cannot dot a number with a vector. The expression (u · v)w does make sense (a number times a vector), but it is usually different from u(v · w). For example, with u = [1, 0], v = [1, 1] and w = [0, 1]: (u · v)w = 1[0, 1] = [0, 1], but u(v · w) = [1, 0](1) = [1, 0].

Worked example

Example 1: Proving the distributive property

Prove that u · (v + w) = u · v + u · w for vectors in three-space.

  1. v + w = [v₁ + w₁, v₂ + w₂, v₃ + w₃].
  2. u · (v + w) = u₁(v₁ + w₁) + u₂(v₂ + w₂) + u₃(v₃ + w₃).
  3. Expand: u₁v₁ + u₁w₁ + u₂v₂ + u₂w₂ + u₃v₃ + u₃w₃.
  4. Regroup: (u₁v₁ + u₂v₂ + u₃v₃) + (u₁w₁ + u₂w₂ + u₃w₃) = u · v + u · w. ∎

The dot product distributes over addition, just like multiplication of numbers.

Worked example

Example 2: Expanding like a binomial

|a| = 3, |b| = 5 and a · b = 4. Find |a + b| and |2a − b|.

  1. Use |v|² = v · v and expand: |a + b|² = (a + b) · (a + b) = a · a + 2a · b + b · b (because a · b = b · a).
  2. = 9 + 2(4) + 25 = 42, so |a + b| = √42 ≈ 6.48.
  3. |2a − b|² = 4(a · a) − 4(a · b) + b · b = 4(9) − 4(4) + 25 = 45.
  4. |2a − b| = √45 = 3√5.

|a + b| = √42 ≈ 6.48 and |2a − b| = 3√5 ≈ 6.71. You never needed the components!

Orthogonal vectors

If u and v are non-zero, then |u| and |v| are not 0, so u · v = |u||v| cos θ is 0 only when cos θ = 0, that is, when θ = 90°. So:

u · v = 0 if and only if u and v are perpendicular.

Perpendicular vectors are called orthogonal. (By convention, the zero vector is orthogonal to every vector, since 0 · v = 0.) This is the fastest right-angle test in mathematics: no protractor, no slopes, and it works in three-space, where slopes do not exist.

xya = [3, −4][4, 3][−4, −3]Swap the components andchange one sign:[3, −4] · [4, 3] = 12 − 12 = 0[3, −4] · [−4, −3] = −12 + 12 = 0Every multiple t[4, 3] works.
In two-space, [4, 3] and every scalar multiple of it, such as [−4, −3], is orthogonal to a = [3, −4]. They all lie on one line through the origin, at right angles to a.
Worked example

Example 3: Testing for orthogonality

Are a = [2, 1, −3] and b = [4, 1, 3] orthogonal?

  1. a · b = 2(4) + 1(1) + (−3)(3) = 8 + 1 − 9 = 0.

Yes, the dot product is 0, so a ⊥ b.

Worked example

Example 4: Making vectors orthogonal

Find k so that [k, 2, −1] and [3, k, 4] are orthogonal.

  1. Set the dot product to 0: 3k + 2k − 4 = 0.
  2. 5k = 4, so k = 0.8.
  3. Check: [0.8, 2, −1] · [3, 0.8, 4] = 2.4 + 1.6 − 4 = 0. ✓

k = 0.8

Worked example

Example 5: A vector orthogonal to two vectors

Let a = [1, 2, −1] and b = [3, 0, 1]. Find a non-zero vector orthogonal to both. Then show it is also orthogonal to every combination sa + tb.

  1. Let n = [x, y, z]. We need n · a = x + 2y − z = 0 and n · b = 3x + z = 0.
  2. From the second equation, z = −3x. Substitute into the first: x + 2y + 3x = 0, so y = −2x.
  3. Two equations, three unknowns: one variable is free. Choose x = 1: n = [1, −2, −3].
  4. Check: n · a = 1 − 4 + 3 = 0 ✓ and n · b = 3 + 0 − 3 = 0 ✓.
  5. For any scalars s and t, n · (sa + tb) = s(n · a) + t(n · b) = s(0) + t(0) = 0, by the distributive and scalar-multiple properties.

n = [1, −2, −3] (or any non-zero multiple of it). It is orthogonal to a, b, and every vector of the form sa + tb, such as 2a or a − 5b. In lesson 7 the cross product will produce such a vector in one step.

Proving geometry with dot products

The properties let you prove geometric facts without coordinates. Write the sides of a figure as vectors, write what you want to show as a dot product, then expand.

aba + ba − b(a + b) · (a − b)= |a|² − |b|²= 0 when |a| = |b|
In a rhombus the sides a and b have equal length. The diagonals are a + b and a − b, and their dot product is |a|² − |b|² = 0, so the diagonals are perpendicular.
Worked example

Example 6: The diagonals of a rhombus

Prove that the diagonals of a rhombus are perpendicular.

  1. Let two adjacent sides be a and b, starting at the same vertex. In a rhombus, |a| = |b|.
  2. The diagonals are a + b and a − b.
  3. (a + b) · (a − b) = a · a − a · b + b · a − b · b = |a|² − |b|² (the middle terms cancel because the dot product is commutative).
  4. Since |a| = |b|, this equals 0.

The diagonals have a dot product of 0, so they are perpendicular. ∎

Just for fun

A vector asked its orthogonal friend for advice. The friend said: "I'd love to help, but I can't see where you're coming from. We're at completely right angles."

Hands-on activity

Checking square corners with a dot product

You need: a large sheet of paper (or a few sheets taped together), a ruler or tape measure, a pencil, a calculator, and a rectangular object such as a hardcover book or a cereal box.

  1. Mark a point O near the bottom-left of the paper. Draw an x-axis and a y-axis through O using the ruler and the corner of the book to make the right angle.
  2. Put one corner of the book at a point P you choose (not at O), with the book tilted. Mark the corner P and the two neighbouring corners Q and R by pencilling around the book.
  3. Measure the coordinates of P, Q and R in centimetres (x across, y up).
  4. Find PQ and PR in components and compute PQ · PR.
  5. Repeat with a corner of something you suspect is not square (a slice of bread, a leaf, a paper scrap you tore by hand).

What to look for: for the book, the dot product should be close to 0 compared with the size of the products in it. For example, PQ = [14.1, 5.2] and PR = [−5.5, 14.8] give −77.6 + 77.0 = −0.6, which is tiny next to 77 (measuring error). A torn scrap usually gives a dot product far from 0.

Your turn

1

Verify that u · v = v · u for u = [2, −5, 1] and v = [3, 1, 4].

2

|a| = 2, |b| = 4 and a · b = 3. Find (a + b) · (a + b).

3

With the same vectors, find (2a) · (3b).

4

Are [6, −2, 1] and [1, 4, 2] orthogonal?

5

Find k so that [k, −3] and [2, 5] are orthogonal.

6

Give two vectors, not parallel to each other, that are both orthogonal to [1, 1, 1].

7

Explain why (u · v) · w has no meaning.

More practice: warm-up

1

a · b = 7. What is b · a?

2

a · b = 7. What is (−2a) · b?

3

Are [1, 2] and [−2, 1] orthogonal?

4

Are [1, 2, 3] and [3, 2, 1] orthogonal?

5

|v| = 4. What is v · v?

More practice: on your own

1

|a| = 4, |b| = 3 and the angle between them is 60°. Find |a − b|.

2

Expand and simplify (2a + b) · (a − 3b).

3

Find both unit vectors in two-space that are orthogonal to [3, 4].

4

Find a non-zero vector orthogonal to both [1, 0, 2] and [0, 1, −1].

5

Find all k such that [k, k, 2] ⊥ [k, −3, 1].

Challenge

1

Prove the parallelogram law: |a + b|² + |a − b|² = 2|a|² + 2|b|².

2

AB is a diameter of a circle with centre O, and C is any other point on the circle. Use vectors to prove ∠ACB = 90°.

3

If a · b = a · c and a ≠ 0, must b = c? Prove it or give a counterexample.

Check your understanding

Quiz

1.

Which statement is NOT true for all vectors u, v and w?

2.

a · b = 5. What is (2a) · (−3b)?

3.

|a| = 3, |b| = 2 and a · b = 1. What is |a + b|²?

4.

For which k are [k, 4] and [2, −3] orthogonal?

5.

Which vector is orthogonal to [2, −1, 3]?

6.

In the proof that a rhombus has perpendicular diagonals, which fact makes (a + b) · (a − b) equal to 0?

For parents and teachers

Big idea (C2.5): the dot product is commutative, distributes over addition and lets scalars move out front, so (a + b) · (a + b) expands just like (a + b)². There is no associative property. Most importantly, two non-zero vectors are orthogonal (perpendicular) exactly when their dot product is 0.

How to help: Ask your teen to show you the rhombus proof; it is a nice example of proving geometry with algebra. Then name two vectors and ask, "Are these perpendicular?" The answer should come from a quick dot product.

Common difficulties:

  • Writing |a + b| = |a| + |b|. Lengths do not add; you must expand |a + b|² = |a|² + 2a · b + |b|².
  • Forgetting the middle term 2a · b when expanding.
  • "Cancelling" a vector: a · b = a · c does not mean b = c.
  • Thinking there is only one vector orthogonal to a given vector. In three-space there are infinitely many directions.