solution:
write the coordinates of the points A and B
A= (0,0,14)ft
B= (5,-6,0)ft
write the position vector of a with recpect to b
R_{BA}=(0-5)i+(0-(-6))j+(14-0)k
=-5i+6k+14k
calculation of a magnitude of the vector R_{BA}
\left | R_{BA} \right |=\sqrt{(-5)^2+6^2+14^2}
=\sqrt{25+36+196}
=16.031ft
calculation of the unit vector
\frac{-5i+6k+14k}{16.031}
calculation of the force
F_{BA}=R_{BA}U_{BA}
here, F_{BA} is the magnitude of the force.
substitute 350lb for F_{BA} and (\frac{-5i+6k+14k}{16.031}) for U_{BA}.
F_{BA} = 350 x (\frac{-5i+6k+14k}{16.031})
=-109.2i+131j+305.7k lb
