Prove ∇×(a×r) = 2a

Let a = (a₁, a₂, a₃) be constant and r = xi + yj + zk.

Compute a × r

a × r = | i   j   k  |
        | a₁  a₂  a₃ |
        | x   y   z  |
      = (a₂z − a₃y) i + (a₃x − a₁z) j + (a₁y − a₂x) k

Take the curl

∇ × (a × r) = | i      j      k    |
              | ∂/∂x   ∂/∂y   ∂/∂z |
              | a₂z−a₃y  a₃x−a₁z  a₁y−a₂x |

• i component: ∂(a₁y − a₂x)/∂y − ∂(a₃x − a₁z)/∂z = a₁ + a₁ = 2a₁
• j component: ∂(a₂z − a₃y)/∂z − ∂(a₁y − a₂x)/∂x = a₂ + a₂ = 2a₂
• k component: ∂(a₃x − a₁z)/∂x − ∂(a₂z − a₃y)/∂y = a₃ + a₃ = 2a₃

So ∇ × (a × r) = 2a₁i + 2a₂j + 2a₃k.

Hence ∇ × (a × r) = 2a.

(Check with the identity ∇×(a×r) = a(∇·r) − (a·∇)r = 3a − a = 2a.) This also shows that the velocity v = ω × r of a rigid rotation has curl v = 2ω.