∵ plane ABC⊥ plane B 1BCC 1

∴ AD⊥ plane B 1BCC 1

∴ ∠AB 1D is the angle formed by AB 1 and plane B 1BCC 1

Let b1d ∩ BC1= e.

Let B 1B=a, then BC= (root number 2) a. ..

∫b 1B/BD = b 1c 1/b 1B，∠C 1B 1B=∠DBB 1

Triangle C 1B 1B∽ triangle DBB 1, so ∠ bb1d = ∠ BC1b1.

∫∠bb 1D+∠db 1c 1 = 90

∴ BC1b1+∠ db1c1= 90, so ∠ b 1ec 1 = 90.

Let the angle formed by AB 1 and C 1B be α, which is given by the three cosine theorem.

cosα= cos∠b 1ec 1×∠ab 1D = 0

∴ the angle between ab1and C 1B is 90.