A force$=(2\mathrm{.}10$N$)+(2\mathrm{.}90$N$)+(8\mathrm{.}90$N$)$F$=(2\mathrm{.}10$N$)$i$+(2\mathrm{.}90$N$)$j$+(8\mathrm{.}90$N$)$k$$acts on a$7\mathrm{.}80$kg$$mobile object that moves from an initial position of$=(6\mathrm{.}70$m$)+(4\mathrm{.}10$m$)+(2\mathrm{.}70$m$)$d$$i$=(6\mathrm{.}70$m$)$i$+(4\mathrm{.}10$m$)$j$+(2\mathrm{.}70$m$)$k$$to a final position of$=(3\mathrm{.}50$m$)+(1\mathrm{.}80$m$)+(7\mathrm{.}40$m$)$d$$f$=(3\mathrm{.}50$m$)$i$+(1\mathrm{.}80$m$)$j$+(7\mathrm{.}40$m$)$k$$in$9\mathrm{.}10$s$.$ Find$($a$)$the work done on the object by the force in the$9\mathrm{.}10$s$$interval$,($b$)$the average power due to the force during that interval, and$($c$)$the angle between vectors$$d$$i$$and$$d$$f$.$