I)

According to the problem, when the quadrilateral OABP is a parallelogram, there are

OB=OA+OP……*

Again? OP=mOA+OB, bringing in * type.

∴? OB=(m+ 1)OA+OB

∴? m=- 1

2)

When m=2, OP = 20a+OB.

It is easy to know that lOPl is the largest when OA and OB are in the same direction and the smallest when they are in the opposite direction.

∴? lOPl & ltmax & gt=3，lOPl & ltmin & gt= 1

∴? lOPl *

(3)

According to OP=mOA+OB

∴? BP=OP-OB=mOA

Namely. BP∑OA

And by OA ob =- 1/3.

∴? cos∠AOB=- 1/3

(1) When ∠ BPO = 90, P is the intersection of BP and Y axis, and at this time it is OD⊥OA.

∴? OA OD=0

② When ∠ POB = 90, ∠ AOD = ∠ AOB-90.

∴? cos∠AOD=sin∠AOB=2√2/3

And ∠ b = (180-∠ AOB)/2 = 90-1/2 ∠ AOB.

∴? sinB = cos 1/2∠AOB =√[( 1+cos∠AOB)/2]=√3/3，cosB=√6/3

∴? tanB=sinB/cosB=√2/2

∴? lODl=lOBltanB=√2/2

∴? OA OD = lOAl lODl cos∠AOD = 1×√2/2×2√2/3 = 2/3

Note: Because the unit circle is too small, I used a circle with a radius of 6 instead.