Let's solve this step by step. ### **Step 1: Find the position vectors of points A and B.** From the figure: - \( A(0, 0, 0.5) \) - \( B(0.6, 0.2, 0.2) \) ### **Step 2: Find the vector from A to B (\(\vec{AB}\)).** \[ \vec{AB} = (0.6 - 0, 0.2 - 0, 0.2 - 0.5) = (0.6, 0.2, -0.3) \] ### **Step 3: Find the unit vector along \(\vec{AB}\).** First calculate the magnitude of \(\vec{AB}\): \[ |\vec{AB}| = \sqrt{(0.6)^2 + (0.2)^2 + (-0.3)^2} = \sqrt{0.36 + 0.04 + 0.09} = \sqrt{0.49} = 0.7 \] Unit vector: \[ \hat{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \left(\frac{0.6}{0.7}, \frac{0.2}{0.7}, \frac{-0.3}{0.7}\right) = (0.857, 0.286, -0.429) \] ### **Step 4: Find the y-component of the force** The y-component of the force is the y-component of the unit vector times the magnitude of the force: \[ F_y = F \cdot \text{(y-component of unit vector)} \] Given \( F = 100\,N \): \[ F_y = 100 \times 0.286 = 28.6\,N \] --- ## **Final Answer** \[ \boxed{28.6\,N} \] This is the magnitude of the force in the y-direction (to three significant figures).

The y-component of the force is found by multiplying the total force by the sine of the angle it makes with the xz-plane, determined from the unit vector along its direction.

The y-component of the force is obtained by resolving the force vector along the y-axis using its direction cosines, which are derived from the unit vector along the line of action.