I just need an explanation as to how to do this problem. The solution is already

Question

I just need an explanation as to how to do this problem.  The solution is already given.  Mainly the circled items.  

The magnitudes of the two force vectors are |FA | = 140 lb and |FB| = 100 lb. Determine the magnitude of the sum of the forces FA + FB. Solution: We have the vectors FA = 140 lb[cos40degree sin50degree]i + [sin40degree]j +[cos40degree cos50degree]k) FA = (82.2i + 90.0j + 68.9k) lb FB = 100 lb([- cos 60degree sin 30degree]i+ [sin 60degree]j + [cos 60degree cos 30degree]k) FB = (-25.0i + 86.6j + 43.3k) lb Adding and taking the magnitude we have FA + FB = (57.2i + 176.6j + 112.2k) lb |FA| + |FB| = (57.2 lb)2 +(176.6 lb)2 +(112.2 lb)2 = 217 lb |FA +FB| = 217 lb

Explanation / Answer

The length of projection of Fa on x-z plane is Fa*Cos40 (because it makes 40 deg with x-z plane.). The length of projection of Fa on y axis is therefore, Fa*Sin40.

Further resolving this Fa*Cos40 into x and z directions. z-direction becomes (Fa*Cos40)*cos50 (because angle with z is 50 deg.) and x-direction component becomes (Fa*Cos40)*sin50.

Thus, Fa = FaCos40*sin50 i + Fa*sin40 j + Fa*cos40*cos50 k

Similarly for Fb. Projection of Fb on x-z plane is Fb*cos60. Resolving it in -x and z directions respectively gives Fb*cos60*sin30 and Fb*cos60*cos30.

Component in +x direction would therefore be -Fb*cos60*sin30.

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