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

Properties of the Cross Product

C2.7 C2.8About 65 minutes
Learning goals
  • I can state and verify the anticommutative, distributive and scalar-multiple properties of the cross product.
  • I can show that the cross product of collinear vectors is the zero vector.
  • I can show with an example that the cross product is not associative.
  • I can simplify expressions with cross products and prove that u × v is perpendicular to u and v.

Words to know

anticommutative
Changing the order changes the sign: v × u = −(u × v).
associative
A rule saying grouping does not matter, (ab)c = a(bc). The cross product does not have it.
collinear vectors
Vectors that are parallel; their cross product is 0.
counterexample
One specific example that shows a general statement is false.

The rules change

In lesson 3 the dot product followed almost all of the usual rules of multiplication. The cross product is stranger. Some rules survive, some are reversed, and one fails completely. Knowing exactly which is which lets you simplify expressions like (a + b) × (a − b) without ever writing a component.

PropertyIn symbolsCompare with numbers
Anticommutativeu × v = −(v × u)ab = ba: changed!
Distributiveu × (v + w) = u × v + u × w and (u + v) × w = u × w + v × wsame idea
Scalar multiples(ku) × v = k(u × v) = u × (kv)same idea
A vector with itselfu × u = 0a × a = a²: changed!
Collinear vectorsu × (ku) = 0new
Associative?(u × v) × w ≠ u × (v × w) in generalfails!
Perpendicular to each factoru · (u × v) = 0 and v · (u × v) = 0new
Properties of the cross product of vectors in three-space.
Worked example

Example 1: Anticommutative

Prove that v × u = −(u × v), and check with u = [1, 2, 3] and v = [4, 5, 6].

  1. By the formula, v × u = [v₂u₃ − v₃u₂, v₃u₁ − v₁u₃, v₁u₂ − v₂u₁].
  2. Each component is the corresponding component of u × v = [u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁] with its two terms swapped, so each is the negative. Hence v × u = −(u × v).
  3. Check: u × v = [−3, 6, −3] (lesson 7), and v × u = [(5)(3) − (6)(2), (6)(1) − (4)(3), (4)(2) − (5)(1)] = [3, −6, 3] ✓.

Swapping the order reverses the direction and keeps the length. That matches the right-hand rule: curling from v to u instead of u to v flips your thumb.

Worked example

Example 2: Scalar multiples

Let a = [1, 0, 2] and b = [0, 3, 1]. Compare a × b with (2a) × (3b).

  1. a × b = [(0)(1) − (2)(3), (2)(0) − (1)(1), (1)(3) − (0)(0)] = [−6, −1, 3].
  2. 2a = [2, 0, 4] and 3b = [0, 9, 3]. (2a) × (3b) = [(0)(3) − (4)(9), (4)(0) − (2)(3), (2)(9) − (0)(0)] = [−36, −6, 18].
  3. [−36, −6, 18] = 6[−6, −1, 3].

(2a) × (3b) = 6(a × b). The scalars multiply and come out front, just as the area picture predicts.

abarea |a × b|2a3b2 × 3 = 6 copies(2a) × (3b)= 6(a × b)
Doubling one side and tripling the other makes a parallelogram made of 2 × 3 = 6 copies of the original, so |(2a) × (3b)| = 6|a × b|. The direction does not change, because the new sides lie along the same lines.
Worked example

Example 3: Checking the distributive law

Let u = [1, 0, 2], v = [0, 1, 1] and w = [2, −1, 0]. Verify that u × (v + w) = u × v + u × w.

  1. v + w = [2, 0, 1], and u × (v + w) = [(0)(1) − (2)(0), (2)(2) − (1)(1), (1)(0) − (0)(2)] = [0, 3, 0].
  2. u × v = [(0)(1) − (2)(1), (2)(0) − (1)(1), (1)(1) − (0)(0)] = [−2, −1, 1].
  3. u × w = [(0)(0) − (2)(−1), (2)(2) − (1)(0), (1)(−1) − (0)(2)] = [2, 4, −1].
  4. Sum: [−2 + 2, −1 + 4, 1 − 1] = [0, 3, 0] ✓.

Both sides equal [0, 3, 0].

Worked example

Example 4: Simplifying without components

Simplify (a + b) × (a − b) and (2a + b) × (a + 3b) in terms of a × b.

  1. Expand the first like a binomial, keeping the order of each product: a × a − a × b + b × a − b × b.
  2. a × a = b × b = 0 and b × a = −a × b, so the result is −a × b − a × b = −2(a × b).
  3. Second: 2(a × a) + 6(a × b) + b × a + 3(b × b) = 0 + 6(a × b) − a × b + 0 = 5(a × b).

(a + b) × (a − b) = −2(a × b) and (2a + b) × (a + 3b) = 5(a × b). Geometric meaning: the parallelogram on the diagonals of a parallelogram has twice its area.

Order matters

When you expand a cross product, never change the order inside a product unless you also change the sign. Writing b × a as a × b is the most common error in this lesson.

(i × i) × jxyz0i × i = 00 × j = 0i × (i × j)xyzk−ji × j = ki × k = −j
The cross product is not associative. Grouping the same three vectors differently gives 0 on the left but −j on the right.
Worked example

Example 5: Not associative

Compare (i × i) × j with i × (i × j).

  1. (i × i) × j = 0 × j = 0.
  2. i × (i × j) = i × k = −j (against the cycle i → j → k).

The results are different, so the cross product is not associative. Brackets matter: always say which product to do first.

Worked example

Example 6: The cross product is perpendicular to each factor

Prove that u · (u × v) = 0 for all vectors u and v in three-space. Explain what it shows.

  1. u · (u × v) = u₁(u₂v₃ − u₃v₂) + u₂(u₃v₁ − u₁v₃) + u₃(u₁v₂ − u₂v₁).
  2. Expand: u₁u₂v₃ − u₁u₃v₂ + u₂u₃v₁ − u₁u₂v₃ + u₁u₃v₂ − u₂u₃v₁.
  3. Every term appears once with + and once with −, so the sum is 0. ∎
  4. In the same way, v · (u × v) = 0.

Both dot products are 0, which verifies the key property of the cross product: u × v is orthogonal to both u and v.

Just for fun

Mathematicians call u × v = −(v × u) anticommutative. That makes it the first multiplication you have met that gets upset when you change the order of the guests.

Hands-on activity

Order and grouping with a book

You need: a hardcover book (or a phone in a case) with one clearly marked face, a table, and the pencil-and-clay axes from lesson 7 (or just your right hand).

  1. Set the book flat, cover up, spine toward you. Call the move "rotate 90° clockwise, seen from above" A, and the move "flip it away from you, top edge over the far edge" B.
  2. Do A then B and note the final position of the cover and spine. Return to the start and do B then A. Are the results the same?
  3. Now model the cross product with your right hand and the pencil axes from lesson 7: check that i × j = k but j × i = −k (order matters).
  4. Check that (i × j) × j = k × j = −i, while i × (j × j) = i × 0 = 0 (grouping matters).

What to look for: in step 2 the book ends up in two different positions, just as the cross product gives different answers for different orders. Rotations in three-space and cross products are closely related; in university physics, torque and angular velocity are both described with cross products.

Your turn

1

a × b = [2, −1, 4]. Find b × a.

2

With a × b = [2, −1, 4], find (3a) × b and a × (−2b).

3

Simplify a × (a + b).

4

Simplify (a − b) × (a + b).

5

Compute i × (j × k) and (i × j) × k. Does this prove the cross product is associative?

6

Let u = [1, 2, 0] and v = [3, −1, 4]. Find u × v and verify that v · (u × v) = 0.

More practice: warm-up

1

u × u = ?

2

u × v = [1, 2, 3]. What is v × u?

3

Write (5u) × v in terms of u × v.

4

Is u × 0 a number or a vector, and what is it?

5

Which property of multiplication of numbers fails for the cross product: commutative, distributive or both commutative and associative?

More practice: on your own

1

Verify u × (v + w) = u × v + u × w for u = [1, 1, 1], v = [1, 0, 0] and w = [0, 1, 0].

2

Simplify (3a − b) × (a + 2b).

3

|a × b| = 10. Find |(3a) × (−2b)|.

4

Show that (i × j) × j ≠ i × (j × j).

5

If u × v = u × w and u ≠ 0, must v = w? Explain.

Challenge

1

Prove (ku) × v = k(u × v) using components.

2

Prove that v · (u × v) = 0 using components.

3

Prove: if a + b + c = 0, then a × b = b × c = c × a. What does this say about a triangle?

Check your understanding

Quiz

1.

u × v = [3, −1, 2]. What is v × u?

2.

Which expression equals (2u) × (3v)?

3.

What is u × u for any vector u?

4.

Which statement is false for the cross product?

5.

Simplify (a + b) × (a − b).

6.

For any vectors a and b in three-space, what is a · (a × b)?

For parents and teachers

Big idea (C2.7): the cross product is anticommutative (v × u = −(u × v)), distributive and works with scalar multiples, but it is not associative, and u × u = 0. The cross product of collinear vectors is 0, and u × v is perpendicular to both u and v (so u · (u × v) = 0).

How to help: Ask your teen to expand something like (a + b) × (a + b) out loud. (The answer is 0, which makes sense: a vector crossed with itself.) Listen for whether they keep the order of each product.

Common difficulties:

  • Reordering a product without changing the sign.
  • Writing u × u = |u|², confusing it with the dot product u · u.
  • Assuming brackets do not matter in a triple cross product.
  • "Cancelling" a vector: u × v = u × w does not mean v = w.