We will denote by #vecF_1# the vector F_1.
We are given that, magnitude of #vecF_1=||vecF_1||=500 N#, &, its direction is due East, i.e., the #+ve# direction of the X-axis. The unit vector along this direction is #hati=(1,0)#.
Recall that a #vecx!=vec0# can completely be described as #vecx=||vecx||hatx#, where, #hatx# is a unit vector in the direction of #vecx#.
Hence, #vecF_1=||vecF_1||hati=500(1,0)=(500,0)#.
Similarly, #vecF_2=(0,250)#
Therefore, #vecF_2-vecF_1=vecF#, say #=(0,250)-(500,0)=(-500,250)=250(-2,1)#
#||vecF||=250sqrt{(-2)^2+1^2}=250sqrt5 N#
To find the direction of #vecF#, suppose that #vecF# is making an angle #theta# due East, i.e., with #hati#.
Then, #vecF.hati=||vecF||*||hati||*costheta#
#:. (-500,250).(1,0)=250sqrt5*1*costheta#
#:.-500=250sqrt5*costheta#
#:. costheta=-500/(250sqrt5)=-2/sqrt5#
#:. theta=cos^-1(-2/sqrt5)=pi-cos^-1(2/sqrt5)#
Thus, #vecF_2-vecF_1=(-500,250)# has magnitude #250sqrt5N# and it acts at an angle #theta=pi-cos^-1(2/sqrt5)# due East.
Hope, this will be of Help! Enjoy maths.!
