In the triangular pyramid p-abc, △ PAC and △ PBC are equilateral triangles with side length √ 2, ab = 2, O and D are the midpoint of AB and Pb respectively 1) Verification: plane PAB ⊥ plane ABC (2) Find the volume of p-abc of triangular pyramid (3); OD is parallel to PAC

In the triangular pyramid p-abc, △ PAC and △ PBC are equilateral triangles with side length √ 2, ab = 2, O and D are the midpoint of AB and Pb respectively 1) Verification: plane PAB ⊥ plane ABC (2) Find the volume of p-abc of triangular pyramid (3); OD is parallel to PAC

(1) Proof: link Po, Co
Because in the triangle PAB, PA = Pb = √ 2, O is the midpoint of ab
So Po ⊥ ab
And ab = 2, so PA & # 178; + Pb & # 178; = AB & # 178;
Then in the right triangle PAB, Po = 1 / 2 * AB = 1
Similarly, from AC = BC = √ 2, we know Pa & # 178; + Pb & # 178; = AB & # 178; that is, the triangle ABC is a right triangle
Then: CO = 1 / 2 * AB = 1
Because PC = √ 2, Po & # 178; + CO & # 178; = PC & # 178;
Then: Po ⊥ OC
From the above, Po ⊥ ab
So Po ⊥ plane ABC
Because Po is in plane PAB
So plane PAB ⊥ plane ABC
(2) From (1) we know Po ⊥ AB, Co ⊥ ab
Then ab ⊥ plane POC
So v-pyramid p-abc = 1 / 3 * AB * s triangle POC
=1/3*2*1/2*1*1
=1/3
(3) Because o and D are the midpoint of AB and Pb respectively
So od / / PA
PA is in the plane PAC, OD is not in the plane PAC
Therefore, we can know from the judgment theorem of parallel lines and planes
Od parallel to PAC

In the triangular pyramid p-abc, if ∠ ABC = 90 ° Pb, plane ABC, ab = BC = 22, Pb = 2, then the distance from point B to plane PAC is___ ．

In △ PAC, PC = PA = 23, AC = 4, s △ PAC = 12 × 4 × 22 = 42, if the distance from point B to plane PAC is h, then 13 × 12 × 22 × 2 = 13 × 42h,  H = 2