A force of 8 N makes an angle of π/4 radian with the y-axis, pointing to the right. The force acts against the movement of an object along the straight line connecting (1, 3) to (4, 5).a. Find a formula for the force vector Fb. Find the angle θ between the displacement direction D = (4 − 1)i + (5 − 3)j and the force direction F. (Round your answer to one decimal place.)c. The work done is F · D or, equivalently, ||F||||D||cos(θ). Compute the work from both formulas and compare

Answer:a) F= 4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j b) Θ=11.3c) The work done is 20[tex]\sqrt{2}[/tex]Step-by-step explanation:a) ||F||=8, α=π/4Fx=||F||·sin(π/4)=8·[tex]\frac{\sqrt{2}}{2}[/tex]Fy=||F||·cos(π/4)=8·[tex]\frac{\sqrt{2}}{2}[/tex]F=Fx i + Fy j = 4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j b) We can find the value of Θ using the equation:cos(Θ)=[tex]\frac{F.D}{||F||||D||}[/tex]where:D= 3 i + 2 j F=4[tex]\sqrt{2}[/tex] i + 4[tex]\sqrt{2}[/tex] j The dot product is defined as the sum of the products of the components of each vector as:F · D= [tex](4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}[/tex]||F||= 8||D||= [tex]\sqrt{3^2+2^2} =\sqrt{13}[/tex]Hence:Θ=arccos([tex]\frac{20\sqrt{2} }{8\sqrt{13} }[/tex])Θ=arccos(0.981)Θ= 11.3°c) Work is equal to: F · D= [tex](4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}[/tex]=28.3Other way of obtainig the work is: ||F||||D||cos(Θ)where:||F||= 8||D||= [tex]\sqrt{3^2+2^2} =\sqrt{13}[/tex]Θ=11.3°So,  ||F||||D||cos(Θ)=8×[tex]\sqrt{13}[/tex]×cos(11.3°)=28.3