The plate is suspended using the three cables which exert the forces shown, express each force as a cartesian vector

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