Multivariable Calculus · Cross product

Cross product

The cross product a × b of two vectors in space is a vector at right angles to both, as long as the area of the parallelogram they span; its three components are the signed areas of that parallelogram’s shadows on the three coordinate planes.

The cross product in this visualization

The cross product of two vectors a and b in space is a vector, a × b. Its length is the area of the parallelogram they span, which for nonzero vectors is ∣a∣ ∣b∣sin⁡θ|a|\,|b|\sin\theta, where θ is the angle between them. When that area is not 0, a × b stands at right angles to both and points along the right thumb when the fingers curl from a toward b. Here you can turn a and b in a room and see the parallelogram’s shadows on the walls, which turn out to be the three components of a × b.

The cross product in coordinates, and its length
a×b=(a2b3−a3b2,  a3b1−a1b3,  a1b2−a2b1),∣a×b∣=∣a∣ ∣b∣sin⁡θa \times b = (a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1), \qquad |a \times b| = |a|\,|b|\sin\theta

Drop the parallelogram straight down onto the floor. Its shadow is the parallelogram spanned by (a1,a2)(a_1, a_2) and (b1,b2)(b_1, b_2), and its area, with a sign, is the 2 × 2 determinant a1b2−a2b1a_1 b_2 - a_2 b_1, which is the third component. The sign is + when the turn from the shadow of a to the shadow of b is counterclockwise seen from above. The two walls work the same way, each seen from the side its axis points to: the shadow on the yz wall gives the first component and the shadow on the zx wall the second. A shadow with a negative area turns clockwise, and the page hatches it.

Why do the shadows make a vector at right angles to the parallelogram? Take a parallelogram with nonzero area. Call its area A and let n be the unit vector at right angles to it, on the side the right-hand rule picks. Seen from a wall with unit normal m, the parallelogram is squashed in one direction only, by the factor ∣n⋅m∣|n\cdot m|. So its shadow on that wall has signed area A (n⋅m)A\,(n\cdot m). Shadow on one wall draws this view with the current numbers. The three walls’ normals are the three axes, so the three shadows together are

The three shadows’ signed areas, as one vector
(A n1,  A n2,  A n3)=A n=a×b(A\,n_1,\; A\,n_2,\; A\,n_3) = A\,n = a \times b

This is a vector along n with length A. The coordinate formula and the geometric description are the same thing.

Swapping the order traces every shadow the other way round, so b × a = −(a × b): the cross product is not commutative. When a and b are parallel, or one of them is zero, the parallelogram collapses to a segment or a point, every shadow has area 0, and a × b = 0. When the area is not 0, Face looks straight along a × b, where the parallelogram shows its true shape, and Edge-on looks along the parallelogram, which becomes a line at right angles to a × b.

The right hand is a convention that matches the room’s right-handed axes, in which the turn from x to y is counterclockwise seen from +z.

Things to notice

  • Swapping the order leaves the parallelogram and its shadows where they are. Only the turns, the hatching and the arrow change.
  • The shadows belong to the walls, not to the camera. Seen from behind a wall, a counterclockwise shadow looks clockwise on the screen, but its sign is read from the side its axis points to.

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