The Vector Product

A third vector, perpendicular to both.

A product that is a vector

The dot product of two vectors is a number. The vector product of two vectors in three dimensions is a new vector, written a × b and read “a cross b”, which is why it is also called the cross product.

For a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃), it is a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁).

Each component leaves out its own place and uses the other two, taken round the cycle 1, 2, 3, 1. For the first component, start at place 2: a₂b₃ − a₃b₂. For the second, start at place 3: a₃b₁ − a₁b₃. For the third, start at place 1: a₁b₂ − a₂b₁. Each time, a comes from the starting place and b from the next place round, and the product taken away swaps those two places.

Two flat arrows

Take a = (2, 1, 0) and b = (1, 3, 0). Both have a third component of 0, so both lie flat in the floor, the plane z = 0.

The first component is a₂b₃ − a₃b₂ = 1 × 0 − 0 × 3 = 0. The second is a₃b₁ − a₁b₃ = 0 × 1 − 2 × 0 = 0. The third is a₁b₂ − a₂b₁ = 2 × 3 − 1 × 1 = 6 − 1 = 5. So a × b = (0, 0, 5): no part along the floor at all, and 5 straight up.

ab45°

a = (2, 1, 0) and b = (1, 3, 0), seen from above. They meet at 45°, and a × b = (0, 0, 5) points straight up out of the page. Its length 5 is |a| |b| sin 45° = √5 × √10 × 1/√2 = 5.

aba × bθ = 60°|a × b| = |a||b| sin θ = 5.2

a × b stands perpendicular to both, as long as the parallelogram's area |a||b| sin θ

Swing b until it is parallel to a

a and b lie in the shaded plane, with lengths 3 and 2 and an angle of 60° between them. a × b stands at right angles to the plane, with length 3 × 2 × sin 60° = 5.20 to 2 decimal places. Swing b round: past 180° the product points down instead, and when b lies along the line of a, at 180° or 360°, the product is zero.

At right angles to both

The point of the vector product is its direction: a × b is perpendicular to a and to b. The dot product checks it. For the flat pair, a · (a × b) = 2 × 0 + 1 × 0 + 0 × 5 = 0 and b · (a × b) = 1 × 0 + 3 × 0 + 0 × 5 = 0. Both are zero, so (0, 0, 5) is perpendicular to each.

This holds for every a and b. a · (a × b) = a₁(a₂b₃ − a₃b₂) + a₂(a₃b₁ − a₁b₃) + a₃(a₁b₂ − a₂b₁). Multiplied out, that is a₁a₂b₃ − a₁a₃b₂ + a₂a₃b₁ − a₁a₂b₃ + a₁a₃b₂ − a₂a₃b₁, and the six terms cancel in pairs, leaving 0. The same happens with b.

A full example

Take a = (2, −1, 3) and b = (1, 4, −2). The first component is (−1) × (−2) − 3 × 4 = 2 − 12 = −10. The second is 3 × 1 − 2 × (−2) = 3 + 4 = 7. The third is 2 × 4 − (−1) × 1 = 8 + 1 = 9. So a × b = (−10, 7, 9).

Check with dot products: a · (a × b) = −20 − 7 + 27 = 0 and b · (a × b) = −10 + 28 − 18 = 0. A check that does not come out to 0 means a slip, most often a sign in the middle component.

The order matters

Swap the two vectors and each component subtracts the same two products the other way round, so every sign changes: b × a = −(a × b). For the flat pair, (1, 3, 0) × (2, 1, 0) = (0, 0, 1 × 1 − 3 × 2) = (0, 0, −5), the same length pointing straight down.

The right-hand rule gives the direction. Point the fingers of your right hand along a and curl them toward b: your thumb points along a × b. With the unit vectors, i × j = k, j × k = i and k × i = j, while j × i = −k.

Parallel vectors give zero

A vector crossed with itself is the zero vector. In a × a, each component subtracts a product from itself: a₂a₃ − a₃a₂ = 0, and the same in the other two places.

The same is true for any two parallel vectors. (2, 1, 0) × (4, 2, 0) = (0, 0, 2 × 2 − 1 × 4) = (0, 0, 0). The length of a × b is |a| |b| sin θ, and sin 0° = sin 180° = 0, so two vectors along one line have no vector product.

Sums and scalars

The vector product distributes over addition: a × (b + c) = a × b + a × c. Check it with a = (2, 1, 0), b = (1, 3, 0) and c = (3, 1, 0). Then b + c = (4, 4, 0) and a × (b + c) = (0, 0, 2 × 4 − 1 × 4) = (0, 0, 4). Separately, a × b = (0, 0, 5) and a × c = (0, 0, 2 × 1 − 1 × 3) = (0, 0, −1), which add to (0, 0, 4).

A scalar can be taken outside: (2a) × b = 2(a × b). The only rule that does not carry over from ordinary multiplication is the order.

The usual mistakes

Giving a number. The vector product has three components; a single number is the dot product.

Getting the middle component the wrong way round. It is a₃b₁ − a₁b₃, starting from place 3; the dot-product check catches the slip.

Adding the two products. For the flat pair, 2 × 3 + 1 × 1 = 7 is not the third component; it is 2 × 3 − 1 × 1 = 5.

Ignoring the order. b × a is the opposite vector to a × b, not the same one.

The turning effect of a force

In mechanics, a force F acting at a point with position vector r from a pivot has a moment, its turning effect about the pivot, equal to r × F, in that order. The moment points along the axis the force turns about, and its length is the size of the turning effect. In the application below, a bolt holds a sign against the wind.

Worked example: The Turning Effect of the Wind on the Bolt That Holds a Sign

Question A large sign hangs from a bracket fixed to a wall by a single bolt at O. The wind pushes on the sign at the point A, where OA = r = 301 m, with a force F = 02010 N. The moment of the force about O is r × F. (a) Find the moment as a column vector. (b) The bolt can resist a moment of magnitude up to 80 N m. Is the bolt strong enough, and with how much to spare?

  1. 1.The moment is r × F, with r = 301 first and F = 02010 second. The order matters: F × r is the opposite vector.

    wallr, the bracketFOAr =301, F =02010the moment is r × F, in that order
    wallr, the bracketFOAr =301, F =02010the moment is r × F, in that order
    The bracket runs from the bolt O to A: r = 301 m. The wind pushes at A with F = 02010 N, drawn at 1 m for every 10 N.
  2. 2.Work out the components. The first is 0 × 10 − 1 × 20 = −20, the second is 1 × 0 − 3 × 10 = −30, and the third is 3 × 20 − 0 × 0 = 60.

    wallrFOAfirst: 0 × 10 − 1 × 20 = −20second: 1 × 0 − 3 × 10 = −30third: 3 × 20 − 0 × 0 = 60
    wallrFOAfirst: 0 × 10 − 1 × 20 = −20second: 1 × 0 − 3 × 10 = −30third: 3 × 20 − 0 × 0 = 60
    The components of r × F: 0 × 10 − 1 × 20 = −20, 1 × 0 − 3 × 10 = −30 and 3 × 20 − 0 × 0 = 60.
  3. 3.(a) The moment is −20−3060 N m. Check: r · (r × F) = −60 + 0 + 60 = 0 and F · (r × F) = 0 − 600 + 600 = 0, so the moment is perpendicular to both the bracket and the force.

    wallrFr × FOAr × F =−20−3060N mcheck: r · M = −60 + 0 + 60 = 0
    wallrFr × FOAr × F =−20−3060N mcheck: r · M = −60 + 0 + 60 = 0
    (a) The moment is −20−3060 N m, drawn at 1 m for every 20 N m. It is perpendicular to both the bracket and the force: the axis the wind tries to turn the sign about.
  4. 4.The size of the turning effect is the magnitude of the moment: √(−20)2 + (−30)2 + 602 = √400 + 900 + 3600 = √4900 = 70 N m.

    wallrF70 N mOA400 + 900 + 3600 = 4900√4900= 70 N m
    wallrF70 N mOA400 + 900 + 3600 = 4900√4900= 70 N m
    The turning effect is its magnitude: √400 + 900 + 3600 = √4900 = 70 N m.
  5. 5.(b) 70 N m is less than 80 N m, so the bolt is strong enough, with 80 − 70 = 10 N m to spare.

    wallrF70 N mOA70 N m is less than 80 N mthe bolt holds, with 10 N m to spare
    wallrF70 N mOA70 N m is less than 80 N mthe bolt holds, with 10 N m to spare
    (b) 70 N m is less than the 80 N m the bolt can resist, so it holds with 10 N m to spare.

Answer: (a) −20−3060 N m; (b) yes: the moment has magnitude 70 N m, which leaves 10 N m to spare

Common mistakes

  • Multiplying the length of the bracket by the size of the force, √10 × √500 ≈ 70.7 N m. That is the moment only when the force is at right angles to the bracket; here it is not quite, and the vector product takes the angle into account.
  • Working out F × r instead of r × F. The magnitude is the same, but every component changes sign, so the moment points along the axis the wrong way round.

More planes and the vector product problems, worked step by step →

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