Mathematically known A( 1-2)

Solution: Vector AB = (cosx- 1, t-0) = (cosx- 1, t).

Vector OA=( 1, 0), | vector OA|= 1.

And | vector AB|=√5|OA|=√5.

That is, √ [OSX- 1) 2+t 2] = √ 5.

cos^2x-2cosx+ 1+t^2=5.( 1)

It is also assumed that vector a⊥ vector AB,

∴( 1,2).(cosx- 1，t)=0。

cosx- 1+2t=0。

cosx= 1-2t。 (2)

t^2=4+2cosx-cos^2x.

Substitute formula (2) into (1) and simplify it to 5t 2 = 5, t = 1.

cosx= 1-2t

cosx = 1-2 * 1 =- 1，

Vector ob = (cosx, t) = (- 1. 1). -That's what you want.

If vector A and vector AB are straight lines, then vector a=λ vector AB;

( 1，2)=λ(cosx- 1，t)。

-2λ=- 1.λ=- 1/2，λt=2，t=-4

Vector ab = (cosx- 1, t) = (cosx- 1. -4).

Let f(x)= vector OB. Vector AB

F(x)=(cosx，t)。 (cosx- 1，t)。

=(cosx，-4)。 (cosx- 1。 -4).

=cos^2-cosx+(-4)*(-4).

=cos^2x-cosx+ 16.

=(cosx- 1/2)^2- 1/4+ 16.

=(cos- 1/2)^2+63/4.

When cosx- 1/2 = 0, cosx = 1/2 and x = 60, the function f(x) has a minimum value, and f (x) min = 63/4. -That's what you want. .