Let f 1 = (cos θ, sin θ) and F2 = (a, b).

Then F3 = (-a-cos θ, -b-sin θ).

|F2|=(√6+√2)/2→a^2+b^2=2+√3.

∫Acosθ+BSINθ= f 1 product of quantity with F2 = | f1| F2 | cos45 = (1+√ 3)/2.

∴|f3|^2=(-a-cosθ)^2+(-b-sinθ)^2

=(a^2+b^2)+2(acosθ+bsinθ)+ 1

=(2+√3)+( 1+√3)+ 1

=4+2√3,

∴|F3|= 1+√3.

note:

The quantum product of F3 and F 1 = cos θ (-a-cos θ)+sin θ (-b-sin θ) =-(acos θ+bsin θ)-1= (-1-√ 3)/2,

∴cos<； F3，f 1 & gt； = (product of F3 and F 1)/(| f/kloc-0 /|| F3 |) = [(-1-√ 3)/2]/(1+√ 3) =-1/.

∴<； F3，f 1 & gt； = 120 .

To sum up, the size of F3 is (1+√ 3) n, and the included angle between F3 and F 1 is 120.

Note: the square root of the condition "| F2 | = 6+ radical number 2/2" should be "| F2 | = (√ 6+√ 2)/2", that is, the molecular knockout "()". This made me calculate twice more.