51 Let →A=∇Φ+∇×→N; then \(\nabla \times \overrightarrow{\mathrm{A}}=\nabla \times(\nabla \times \overrightarrow{\mathrm{N}})...51 Let →A=∇Φ+∇×→N; then ∇×→A=∇×(∇×→N) and ∇∙→A=∇2Φ, so ∇×→A and ∇∙→A can be chosen independently simply by choosing →N and Φ independently.