Winter vacation homework Problem of Mathematics in Senior One (Key Question 2)

Proof: (1) Because PA⊥ plane ABC, BC belongs to plane ABC.

So PA⊥BC

And BC⊥AB, AB crosses pa = a.

So BC⊥ plane PAB, and AE belongs to plane PAB:

So BC⊥AE

AE⊥PB lead cross BC = B

So AE⊥ plane PBC, and AE belongs to plane AEF.

So airplane AEF⊥ airplane PBC

(2) From (1), AE is the height of the triangular pyramid P-AEF based on PEF.

From the data in the question, we can get PB=2 root number 2, PC=2 root number 3, AE= root number 2, AF = (square root of 2 6)/3,

PE=PB/2= radical number 2, PF= radical number (PA 2-AF 2) = (2 radical number 3)/3.

The cosine theorem in triangle PBC can derive the square root of cosine angle BPC= 6/3, and then sine angle BPC = 3/3.

So the area of triangle PEF =(PE*PFsin angle BPC)/2= root number 2/3.

So the volume of the triangular pyramid P-AEF = (root number 2/3)*AE/3=2/9.