The direction of line AB is defined by the position vector r_AB, which can be found from vector addition,

     r_A + r_AB = r_B
     r_AB = r_B - r_A

           = (39 - 175)i + (70 - 0)j + (29 - 0)k

           = -136i + 70j + 29k

The unit vector in the direction of r_AB can be determined by dividing r_AB by its magnitude r_AB,

     r_AB = (-1,36² + 70² + 29²)^0.5 = 155.7 m

     u_AB = r_AB/r_AB

            = -0.8736i + 0.4496j + 0.1863k

Add the two forces applied to the trolley to find the total force,

     F_T = F₁ + F₂

          = -1,863i - 437.7j - 83.2k N

The scalar component of the total force parallel to the direction u_AB can be found by using the dot product,

    F_T|| = F_T • u_AB

       = -1,863 (-0.8736) - 437.7 (0.4496) - 83.2 (0.1863)

       = 1,415 N

This is just the magnitude of the total force acting in the direction of the cable AB. This can also be represented as a vector by multiplying this scalar component by the unit vector u_AB, giving,

     F_T|| = F_T|| u_AB

         = -1,236i + 636.3j + 263.7k N

The magnitude or F_T|| has not changed but is now represented as a vector.