Cross product between two lists of vectors

How about this

furious[a_, b_] := Module[{a1, a2, a3, b1, b2, b3, c},
  {a1, a2, a3} = Transpose[a, {2, 3, 4, 1}];
  {b1, b2, b3} = Transpose[b, {2, 3, 4, 1}];
  c = {-a3 b2 + a2 b3, a3 b1 - a1 b3, -a2 b1 + a1 b2};
  Transpose[c, {4, 1, 2, 3}]]

Timing results (from march's answer) for version 10.4.1

list1 = RandomReal[{-1, 1}, {32, 32, 32, 3}];
list2 = RandomReal[{-1, 1}, {32, 32, 32, 3}];

l1 = listCrossMarch1[list1, list2]; // RepeatedTiming // First
l2 = listCrossMarch2[list1, list2]; // RepeatedTiming // First
l3 = listCross[list1, list2]; // RepeatedTiming // First
l4 = furious[list1, list2]; // RepeatedTiming // First

2.67

0.6064

0.0386

0.0015

Interestingly enough, MapThreading Cross works but is much slower:

Using sample lists:

list1 = Array[c, {20, 20, 20, 3}];
list2 = Array[d, {20, 20, 20, 3}];

We can perform this operation in the following two ways, using MapThread:

listCrossMarch1[list1_, list2_] := MapThread[Cross, {list1, list2}, 3]
listCrossMarch2[list1_, list2_] := MapThread[{#1[[2]] #2[[3]] - #1[[3]] #2[[2]], #1[[3]] #2[[1]] - #1[[1]] #2[[3]], #1[[1]] #2[[2]] - #1[[2]] #2[[1]]} &, {list1, list2}, 3]

As we can see below, there is a lot of overhead associated with Cross apparently, since that version is much slower than coding the cross-product explicitly. The MapThread version with the explicit cross-product (rather than Cross) is almost as fast as the OP's version and much cleaner to write down. Simon Woods' answer is the fastest (and also very clean).

l1 = listCrossMarch1[list1, list2]; // AbsoluteTiming // First
l2 = listCrossMarch2[list1, list2]; // AbsoluteTiming // First
l3 = listCross[list1, list2]; // AbsoluteTiming // First
l4 = furious[list1, list2]; // AbsoluteTiming // First
l1 === l2 === l3 === l4
(* 1.436369 *)
(* 0.190366 *)
(* 0.120962 *)
(* 0.084740 *)
(* True *)

---

As suggested by Simon Woods, here are Timings with packed arrays of real numbers. Using

list1 = RandomReal[{-1, 1}, {20, 20, 20, 3}];
list2 = RandomReal[{-1, 1}, {20, 20, 20, 3}];

we do

l1 = listCrossMarch1[list1, list2]; // AbsoluteTiming // First
l2 = listCrossMarch2[list1, list2]; // AbsoluteTiming // First
l3 = listCross[list1, list2]; // AbsoluteTiming // First
l4 = furious[list1, list2]; // AbsoluteTiming // First
(* 0.429275 *)
(* 0.094540 *)
(* 0.008337 *)
(* 0.028592 *)

The OP's messy version is significantly the fastest here! Simon Woods' answer still does a great job, and Cross still has lots of overhead.