Prove (a×∇)×r = −2a

Here a is constant, r = xi + yj + zk, and ∇ acts only on r. Write the result as an expansion of the vector triple product.

Method 1: triple product expansion

For vectors, (P × Q) × R = Q(P · R) − P(Q · R). Take P = a, Q = ∇, R = r. Because ∇ is differentiating r (a is constant):

(a × ∇) × r = ∇(a · r) − a(∇ · r)

• ∇(a · r) with a constant: a · r = a₁x + a₂y + a₃z, so ∇(a · r) = a₁i + a₂j + a₃k = a.
• ∇ · r = ∂x/∂x + ∂y/∂y + ∂z/∂z = 3.

Therefore (a × ∇) × r = a − 3a = −2a.

Method 2: components (confirmation)

Let a = (a₁, a₂, a₃). Then

a × ∇ = (a₂∂z − a₃∂y) i + (a₃∂x − a₁∂z) j + (a₁∂y − a₂∂x) k = (D₁, D₂, D₃).

The x-component of (a × ∇) × r is D₂ z − D₃ y:

D₂ z = (a₃∂x − a₁∂z) z = −a₁
D₃ y = (a₁∂y − a₂∂x) y = a₁
x-component = −a₁ − a₁ = −2a₁

The y and z components follow cyclically: −2a₂ and −2a₃.

Hence (a × ∇) × r = −2a.