## Step 1: **Find the Unit Vector Along Cable \( AC \)** ### **Position Vectors** - \( \vec{A} \) (origin): \( (, , 6) \) - \( \vec{C} \): \( (3, , ) \) ### **Cable \( AC \) Vector** \[ \vec{AC} = \vec{C} - \vec{A} = (3 - )\hat{i} + ( - )\hat{j} + ( - 6)\hat{k} = 3\hat{i} - 6\hat{k} \] ### **Magnitude of \( \vec{AC} \)** \[ |\vec{AC}| = \sqrt{3^2 + ^2 + (-6)^2} = \sqrt{9 + 36} = \sqrt{45} = 6.708 \] ### **Unit Vector Along \( AC \)** \[ \vec{u}_{AC} = \frac{\vec{AC}}{|\vec{AC}|} = \frac{3}{6.708}\hat{i} + \hat{j} + \frac{-6}{6.708}\hat{k} = .447\hat{i} + \hat{j} - .894\hat{k} \] #### **Explanation** - We calculated the direction vector for cable \( AC \), its length, and normalized it to get the unit vector in the direction from \( A \) to \( C \). --- ## Step 2: **Project Force \( \vec{F} \) onto Cable \( AC \)** ### **Force Vector \( \vec{F} \)** Given: \( \vec{F} = \{200, 500, 300\} \) N ### **Projected Component Formula** The magnitude of the projected component of \( \vec{F} \) along \( AC \) is: \[ F_{AC} = \vec{F} \cdot \vec{u}_{AC} \] ### **Dot Product Calculation** \[ F_{AC} = (200)(.447) + (500)() + (300)(-.894) \] \[ F_{AC} = 89.4 + - 268.2 = -178.8 \text{ N} \] #### **Explanation** - We took the dot product of the force vector with the unit vector along the cable to find the component of force along \( AC \). - The negative sign indicates the component is opposite to the direction from \( A \) to \( C \). --- ## **Final Answer** ### **The magnitude of the projected component of the force \( \vec{F} \) along cable \( AC \) is:** \[ \boxed{178.8 \text{ N}} \] *(The direction is from \( C \) to \( A \), since the result is negative for \( A \) to \( C \)).*