Decrypting a vigenere cipher with a reused one-time pad I understand that when a plaintext is encrypted with a key of the same length, it is referred to as a one-time pad. Now when this key is reused to encrypt another plaintext, there is potential for decryption of the original ciphertexts. 

Following on from the question asked here, Taking advantage of one-time pad key reuse?, the ciphertexts there were retrieved via a Vernam cipher:-

$ciphertext = plaintext \oplus key$ 

which gave rise to the following relationship between the two:-

$ciphertext_1 \oplus ciphertext_2 = plaintext_1 \oplus plaintext_2$

If instead of a Vernam cipher the encryption is carried out via a Vigenere cipher, where I have two plaintexts and a key, all three of which are the same length - is it possible to draw any relationships like the one above, from the resulting ciphertexts, when I do not know what the plaintexts and key were?
The crib-drag attack applies with a different operator.

p1 = AtextencryptedByFirstPassword
p  = HelloHelloHelloHelloHelloHell
c1 = hxpihlrncmwxpopfjtcgatldgdsco

p2 = ASecondEncryptedTextBySecondPassword
p  = HelloHelloHelloHelloHelloHelloHelloH
c2 = hwpncuhpyqycaeskxpihicdpqvroaozwhzfk


When you $c_1 - c_2 \bmod 26$ you will have the same problem as many-time-pad. This subtraction will remove the key. 

 c1[i] - c2[i] = p1[i] - p2[i]


But this time instead of $\oplus$ we will have plus modulo 26. 

 c1 - c2 = azafvjqcwecflqdfowgbijsmkszmm