Finding the resultant of two forces in 3d

[CivilSigma]

Hi, For the following image, we have to calculate the resultant of those two forces. I have broken them down to their components. Did I do this correctly because when I actually solve for the force I am not getting the correct answer of 940N in the back of my book.

Also, I tried calculating the direction of force P in the x plane , Q in the z plane and when I used the cosine formula I got
cos^2 theta = -0.4 ( a negative number) does that imply that our vector does not exist in the x plane and therefore is 0 for P? and 0 for Q in the z plane?

 

[rude man]

The photograph does not clearly define the vectors, far as I can see.

 

[CivilSigma]

They are not defined by x,y,z co-ordinates but by their coordinate angles?

 

[rude man]

They are not defined by x,y,z co-ordinates but by their coordinate angles?

Yeah, but the angles are not well defined as I see it. It's not clear whether the angles are with the x, y or z axes. Maybe with some effort I could figure it out, but a verbal description would have been a big help.

 

[SteamKing]

Hi, For the following image, we have to calculate the resultant of those two forces. I have broken them down to their components. Did I do this correctly because when I actually solve for the force I am not getting the correct answer of 940N in the back of my book.

Also, I tried calculating the direction of force P in the x plane , Q in the z plane and when I used the cosine formula I got
cos^2 theta = -0.4 ( a negative number) does that imply that our vector does not exist in the x plane and therefore is 0 for P? and 0 for Q in the z plane?

Your calculations show the Qx = Q cos 30. This is incorrect. The projection of Q onto the x-z plane makes an angle of 30 degrees to the x-axis.
The magnitude of the projection of Q onto the x-z plane is not equal to the magnitude of Q.

Your calculation of Qy is correct.

You have a similar situation with the force P. The projection of P onto the x-z plane makes an angle of 25 degrees to the z-axis.
Again, the magnitude of the projection of P onto the x-z plane is not equal to the magnitude of P.

 

[CWatters]

Doesn't look too confusing but I think...

Qx = 450*Cos55*Cos30
Qz = 450*Cos55*Cos60

Haven't checked them all.

Cross posted with the reply by Steamking.