Reducing allocations when subtracting vectors and multiplying by scalars

General Usage Performance
,
2

That is a benchmarking artifact, you are benchmarking the creation of the arrays. If you interpolate them, it has only one allocation:

julia> @btime foo($([1.0, 1.0, 1.0]), $([2.0, 2.0, 2.0]))
  32.036 ns (1 allocation: 112 bytes)
3-element Vector{Float64}:
 -3.0
 -3.0
 -3.0

If you use static arrays, you can go to zero allocations:

julia> using StaticArrays

julia> function foo(x::AbstractVector{Float64}, y::AbstractVector{Float64})
           (x .- y) .* 3.0
       end
foo (generic function with 2 methods)

julia> @btime foo($(SVector{3,Float64}(1.0, 1.0, 1.0)), $(SVector{3,Float64}(2.0, 2.0, 2.0)))
  0.015 ns (0 allocations: 0 bytes)
3-element SVector{3, Float64} with indices SOneTo(3):
 -3.0
 -3.0
 -3.0

(which can be very useful if your problem deals with small arrays, as in the example).

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