1. **State the problem:** We need to find the components of the force exerted by cable BG on the frame at point B, given the tension in cable BG is 555 N. 2. **Identify coordinates:** Point G is at $(-0.56, y_G, 1.48)$ meters, where $y_G$ is unknown but irrelevant since the frame lies in the x-z plane and cable BG is attached at B. Point B lies on the frame; since AB and BC are 0.8 m each and the frame lies in the x-z plane, we can find B's coordinates. 3. **Find coordinates of B:** Since A lies on the z-axis and D on the x-axis, and AB = BC = 0.8 m, point B is at $(0,0,0.8)$ meters (assuming A at origin for simplicity). 4. **Vector from B to G:** $$\vec{BG} = (x_G - x_B, y_G - y_B, z_G - z_B) = (-0.56 - 0, y_G - 0, 1.48 - 0.8) = (-0.56, y_G, 0.68)$$ Since the problem states G is 0.56 m from y-z plane in negative x direction and 1.48 m above x-z plane, and B lies in x-z plane at $y=0$, $y_G$ is 0. So, $$\vec{BG} = (-0.56, 0, 0.68)$$ 5. **Calculate magnitude of vector BG:** $$|\vec{BG}| = \sqrt{(-0.56)^2 + 0^2 + 0.68^2} = \sqrt{0.3136 + 0 + 0.4624} = \sqrt{0.776} \approx 0.88$$ 6. **Find unit vector along BG:** $$\hat{u}_{BG} = \frac{1}{0.88}(-0.56, 0, 0.68) = (-0.636, 0, 0.773)$$ 7. **Calculate force components:** Force vector $$\vec{F}_{BG} = Tension \times \hat{u}_{BG} = 555 \times (-0.636, 0, 0.773) = (-353, 0, 429)$$ N 8. **Final answer:** - Component in x-direction: $-353$ N - Component in y-direction: $0$ N - Component in z-direction: $429$ N These are the components of the force exerted by cable BG on the frame at B.