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

The Cross Product

C2.6About 70 minutes
Learning goals
  • I can compute the cross product of two vectors in three-space.
  • I can verify that u × v is perpendicular to both u and v.
  • I can use the right-hand rule to predict the direction of a cross product.
  • I can use the cross product to find vectors and unit vectors orthogonal to two given vectors.

Words to know

cross product
For vectors in three-space, u × v = [u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁], a vector perpendicular to both u and v. Also called the vector product.
right-hand rule
Point the fingers of your right hand along u and curl them toward v; your thumb points along u × v.
right-handed coordinate system
Axes arranged so that i × j = k: fingers along x curling to y give a thumb along z.
normal vector
A vector perpendicular to a surface or to every vector in a plane.

A product that makes a new direction

The dot product takes two vectors and gives a number. The cross product takes two vectors in three-space and gives a vector, and that new vector is perpendicular to both of the vectors you started with.

Why would you want that? In lesson 3 you found a vector orthogonal to two given vectors by solving two equations in three unknowns. That is slow. The cross product does it in one step. Vectors perpendicular to two given directions show up everywhere: the axis a wrench turns a bolt around, the "normal" that describes how a plane is tilted (you will use these in Lines and Planes), and the way 3-D graphics software decides which side of a surface faces the light.

Definition

For u = [u₁, u₂, u₃] and v = [v₁, v₂, v₃], the cross product is

u × v = [u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁]

Read u × v as "u cross v". Notes:

  • It is defined only in three-space. (A 2-D vector [a, b] can be treated as [a, b, 0].)
  • The result is a vector, also called the vector product.
  • Each component skips one index: the x-component uses only the 2nd and 3rd components, the y-component uses the 3rd and 1st, and the z-component uses the 1st and 2nd. The pattern 2-3, 3-1, 1-2 cycles.
u:v:−143−221−14u₂, v₂u₃, v₃u₁, v₁u₂, v₂x: (−1)(−2) − (3)(4) = −10y: (3)(1) − (2)(−2) = 7z: (2)(4) − (−1)(1) = 9u × v = [−10, 7, 9]solid line: multiply, then subtractthe dashed-line product
A memory aid for u = [2, −1, 3] and v = [1, 4, −2]: list the components in the order 2nd, 3rd, 1st, 2nd. Each neighbouring 2 × 2 block gives one component: down-right product minus up-right product.
Worked example

Example 1: Computing a cross product

Find u × v for u = [2, −1, 3] and v = [1, 4, −2], and check that the answer is perpendicular to both.

  1. x-component: u₂v₃ − u₃v₂ = (−1)(−2) − (3)(4) = 2 − 12 = −10.
  2. y-component: u₃v₁ − u₁v₃ = (3)(1) − (2)(−2) = 3 + 4 = 7.
  3. z-component: u₁v₂ − u₂v₁ = (2)(4) − (−1)(1) = 8 + 1 = 9.
  4. Check with dot products: [−10, 7, 9] · [2, −1, 3] = −20 − 7 + 27 = 0 ✓ and [−10, 7, 9] · [1, 4, −2] = −10 + 28 − 18 = 0 ✓.

u × v = [−10, 7, 9], which is perpendicular to both u and v.

Always check

After every cross product, dot your answer with both original vectors. If either dot product is not 0, there is an arithmetic slip, usually a sign in the y-component. It takes ten seconds and catches nearly every mistake.

Which perpendicular direction?

In three-space there are two directions perpendicular to the plane of u and v: "up" and "down". The formula always picks the one given by the right-hand rule. Our axes are set up to be right-handed: if your right-hand fingers point along the positive x-axis and curl toward the positive y-axis, your thumb points along the positive z-axis. (You used right-hand rules for magnetic fields in Grade 11 Physics; this is the same hand, used for a different purpose.)

uvu × vv × uRight-hand rule1. Point the fingers of yourright hand along u.2. Curl them toward v(through the smaller angle).3. Your thumb points alongu × v.Swap the order: the thumbflips, so v × u = −(u × v).
u × v is perpendicular to the plane that contains u and v (the shaded parallelogram). The right-hand rule picks which of the two perpendicular directions it points in. Reversing the order reverses the direction.
Worked example

Example 2: The standard unit vectors

Find i × j and j × i.

  1. i × j = [1, 0, 0] × [0, 1, 0] = [(0)(0) − (0)(1), (0)(0) − (1)(0), (1)(1) − (0)(0)] = [0, 0, 1] = k.
  2. j × i = [0, 1, 0] × [1, 0, 0] = [(1)(0) − (0)(0), (0)(1) − (0)(0), (0)(0) − (1)(1)] = [0, 0, −1] = −k.
  3. Right-hand rule check: fingers along x, curl toward y, thumb up the z-axis ✓.

i × j = k and j × i = −k. Order matters for the cross product.

ijkclockwiseWith the arrows (positive):i × j = k, j × k = i, k × i = jAgainst the arrows (negative):j × i = −k, k × j = −i, i × k = −jAny vector with itself: i × i = 0
Going around the circle in the arrow direction (i to j to k), each cross product gives the next unit vector. Going against the arrows gives the negative.
Worked example

Example 3: Two vectors in the xy-plane

Find [1, 2, 0] × [3, 1, 0]. Describe the result.

  1. x: (2)(0) − (0)(1) = 0. y: (0)(3) − (1)(0) = 0. z: (1)(1) − (2)(3) = 1 − 6 = −5.
  2. The answer is [0, 0, −5].

[0, 0, −5]: straight down the z-axis. Both vectors lie flat in the xy-plane, so anything perpendicular to both must be vertical. It points down because turning from [1, 2, 0] to [3, 1, 0] is clockwise when seen from above.

Worked example

Example 4: A unit vector perpendicular to two vectors

Find a unit vector perpendicular to both a = [1, 0, 1] and b = [0, 2, 1].

  1. a × b = [(0)(1) − (1)(2), (1)(0) − (1)(1), (1)(2) − (0)(0)] = [−2, −1, 2].
  2. Check: [−2, −1, 2] · [1, 0, 1] = 0 ✓; [−2, −1, 2] · [0, 2, 1] = 0 ✓.
  3. |[−2, −1, 2]| = √(4 + 1 + 4) = 3.
  4. Divide by the length: [−2/3, −1/3, 2/3].

[−2/3, −1/3, 2/3], or its opposite [2/3, 1/3, −2/3]: both are unit vectors perpendicular to a and b.

Worked example

Example 5: Faster than lesson 3

In lesson 3 you solved equations to find a vector orthogonal to a = [1, 2, −1] and b = [3, 0, 1]. Find one with the cross product.

  1. x: (2)(1) − (−1)(0) = 2. y: (−1)(3) − (1)(1) = −4. z: (1)(0) − (2)(3) = −6.
  2. a × b = [2, −4, −6] = 2[1, −2, −3].

[2, −4, −6], a multiple of the answer [1, −2, −3] found in lesson 3, in a fraction of the time.

Just for fun

Why is the cross product so good at parties?

It always finds a way to stand out from both of the people it is talking to: perpendicular to each of them.

Hands-on activity

Build a right-handed axis set

You need: three pencils or pens (ideally three different colours), a ball of play dough, modelling clay or a large eraser, a marker, small paper labels, tape, and two straws or extra pencils.

  1. Push the three pencils into the clay so they meet at one point and are perpendicular to each other, like the corner of a room.
  2. Label one pencil x and another y. Use your right hand: fingers along x, curl toward y. Label the pencil your thumb points along as z.
  3. Put the model on the table and use it to check i × j = k, j × k = i and k × i = j with your right hand. Then check j × i = −k (your thumb points the opposite way).
  4. Take Example 3's vectors [1, 2, 0] and [3, 1, 0]. Hold two straws (or more pencils) along them on the table, apply the right-hand rule, and confirm the cross product points down.

What to look for: if you place the model so that x points toward you and y points to your right, z must point up. If your z points down, you built a left-handed set: swap the x and y labels and try again.

Your turn

1

Find [1, 2, 3] × [4, 5, 6].

2

Find [3, 0, −1] × [2, 1, 4].

3

Check your answer to the previous question with dot products.

4

Find j × i and k × j.

5

Find [2, −3, 1] × [−4, 6, −2]. What do you notice about the two vectors?

6

Find (2i + j) × (i − k).

7

Is u × v a scalar or a vector? Is u · v?

More practice: warm-up

1

k × i = ?

2

i × i = ?

3

i × k = ?

4

Find [0, 0, 2] × [0, 3, 0].

5

In which space is the cross product defined?

More practice: on your own

1

Find a vector perpendicular to both [2, 1, −1] and [1, 3, 2].

2

Find both unit vectors perpendicular to [1, 1, 0] and [0, 1, 1].

3

A(1, 0, 0), B(0, 2, 0) and C(0, 0, 3) are points. Find AB × AC.

4

Show that [a, b, 0] × [c, d, 0] = [0, 0, ad − bc].

5

Find k so that [1, k, 2] × [2, 4, 4] = 0.

Challenge

1

Find all vectors of length 6 that are perpendicular to both [1, 2, 2] and [2, 1, −2].

2

Explain why there is no cross product for two vectors in two-space that gives a vector in the same plane.

3

Find all vectors w with i × w = k.

Check your understanding

Quiz

1.

What is [1, 0, 0] × [0, 1, 0]?

2.

What is [1, 2, 3] × [4, 5, 6]?

3.

Which statement about u × v is always true?

4.

What is j × i?

5.

What is [2, 4, 6] × [1, 2, 3]?

6.

A classmate claims [1, 3, 5] × [2, 0, 1] = [3, −9, −6]. Which quick check shows the answer is wrong?

For parents and teachers

Big idea (C2.6): the cross product u × v = [u₂v₃ − u₃v₂, u₃v₁ − u₁v₃, u₁v₂ − u₂v₁] of two vectors in three-space is a vector perpendicular to both of them, pointing in the direction given by the right-hand rule. Reversing the order reverses the direction.

How to help: The formula has many signs, so practice matters more than memorizing. Ask your teen to show you the "always check" habit: dot the answer with both original vectors and make sure both give 0. The pencil-and-clay axes activity helps with the right-hand rule.

Common difficulties:

  • Sign errors in the y-component (it is u₃v₁ − u₁v₃, not u₁v₃ − u₃v₁).
  • Using the left hand, or curling the fingers from v to u.
  • Thinking u × v = v × u. The order matters.
  • Trying to take the cross product of two-space vectors without adding a 0 component.