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 and , it is .
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: . For the second, start at place 3: . For the third, start at place 1: . 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 . The second is . The third is . So a × b = (0, 0, 5): no part along the floor at all, and 5 straight up.
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 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. . Multiplied out, that is , 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: , 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 , 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 , 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.The moment is r × F, with r = 301 first and F = 02010 second. The order matters: F × r is the opposite vector.
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.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.
The components of r × F: 0 × 10 − 1 × 20 = −20, 1 × 0 − 3 × 10 = −30 and 3 × 20 − 0 × 0 = 60. 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.
(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.The size of the turning effect is the magnitude of the moment: √(−20)2 + (−30)2 + 602 = √400 + 900 + 3600 = √4900 = 70 N m.
The turning effect is its magnitude: √400 + 900 + 3600 = √4900 = 70 N m. 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.
(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 →