I’ve been trying to calculate this myself, but the 3-D sines and cosines have been driving me nuts. So, does anyone know the answer?
Oh, if you want the data,
Voyager 1 is at 120.07 AU, with a heliographic latitude of 34.5, and a heliographic longitude of 184.8.
Voyager 2 is at 98.04 AU, with a heliographic latitude of −29.8, and a heliographic longitude of 228.0.
I get approx 135.581 AU, but I may well have made a mistake, so I’ll show my work.
The approach I took was to position each of them in XYZ cartesian space, where one axis is north-south, another towards-away from the point where the Greenwich meridian meets the equator and the third perpendicular to both of them.
Voyager 1:
North component is equal to 120.07 * Sin(34.5) = 68.0084
Equatorial component is equal to 120.07 * Cos(34.5) = 98.95283
Component towards Greenwich = 98.95283 * Sin(184.8) = -8.28016
Perpendicular component = 98.95283 * Cos(184.8) = -98.6058
Voyager 2:
North component is equal to 98.04 * Sin(-29.8) = -48.7233
Equatorial component is equal to 98.05 * Cos(-29.8) = 85.07573
Component towards Greenwich = 85.07573 * Sin(228) = -63.2236
Perpendicular component = 85.07573 * Cos(228) = -56.9268
Then, you take the difference in each of the XYZ components:
Difference in north: 116.7317 AU
Difference in Greenwich: 54.94343 AU
Difference in perpendicular: 41.679 AU
You can sum up the squares of all three and take the square root, to get the overall distance - that’s a little shortcut through applying the pythag theorem twice.
Overall result = sqrt(13626.3 + 3018.78 + 1737.141) = 135.581
I hope that even if this is wrong, it’s helpful.
Edited to add - I actually made a big mistake when originally working it out that I corrected when typing my numbers into this post - I’d included the sum of the difference in latitudes, instead of north distances.
Just to confirm, I also get 135.581 AU, using a slightly different method.
(I calculated the dot product of the two unit vectors, which will equal the cosine of the angle between them. I then used the Law of Cosines to find the length of the third side of a triangle — which in this case is the very distance we’re after — given the lengths of the other two sides, and the cosine of the angle between those sides.)
Excellent! (High-fives Bytegeist) Math nerds representing! 
Slight error in my notes that does not affect the overall answer - the perpendicular and Greenwich components are mislabeled, because I used the wrong trig functions for each. If the longitude of Voy1 was close to 180 degrees, then it should have a large negative Greenwich component and a small perpendicular component close to zero, not the other way around.
Send more Chuck Berry!: Voyager Golden Record - Wikipedia
Ah, Pythagorus Is there anything you can’t do?
And I dread balancing my checkbook. Way to go guys making some of us feel less than bright! Interesting information.
Thanks so much.
I just want to note that I love belonging to a community where people do this kind of stuff not for money, but just because they want to know. Ever since I posted years ago, this sort of dedication, of love of an answer no matter how inane the question, this stuff is why I come back, and why I tell people I come back. Well done!