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Third Force for a Vertical Resultant in 3D

ID: h9rlraa4cf

Problem: \(F_1 = 2\,\text{kN}\) has \(\theta = 120^\circ\) in the \(xy\) plane and \(\phi = 60^\circ\) from \(+z\). \(F_2 = 3\,\text{kN}\) has \(\theta_x = 120^\circ\), \(\theta_y = 60^\circ\), \(\theta_z = 45^\circ\). A third force \(\mathbf{F}_3\) is added so that the resultant of all three becomes \(\{1.2\,\mathbf{k}\}\,\text{kN}\). Find \(\mathbf{F}_3\), its magnitude and its direction angles.

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7 Answer Field(s)
3/3 Attempts
#1
F_{3x}
\text{kN}
#2
F_{3y}
\text{kN}
#3
F_{3z}
\text{kN}
#4
F_3
\text{kN}
#5
\theta_x
^{\circ}
#6
\theta_y
^{\circ}
#7
\theta_z
^{\circ}
Scientific notation supported (e.g. 2e4, 2e-4)

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