Let's solve **part (a)** and **(b)** step by step as requested. --- ## **Step 1: Find the unit vector along the AB line** ### **Coordinates:** - \( A (1, 2, ) \) - \( B (4, 6, 2) \) ### **Vector AB:** \[ \vec{AB} = (4-1)\mathbf{i} + (6-2)\mathbf{j} + (2-)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \] ### **Magnitude of AB:** \[ |\vec{AB}| = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \] ### **Unit Vector along AB:** \[ \mathbf{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{3}{\sqrt{29}}\mathbf{i} + \frac{4}{\sqrt{29}}\mathbf{j} + \frac{2}{\sqrt{29}}\mathbf{k} \] **Explanation:** We obtained the direction vector from A to B, found its magnitude, and divided each component by the magnitude to get the unit vector. --- ## **Step 2: Calculate the moment of the force \(\vec{F}\) about point B** ### **Given Force:** \[ \vec{F} = 222\mathbf{i} - 222\mathbf{j} + 220\mathbf{k} \] ### **Position vector from B to the point of force application (Assume force is applied at A):** \[ \vec{r}_{BA} = \vec{r}_A - \vec{r}_B = (1-4)\mathbf{i} + (2-6)\mathbf{j} + (-2)\mathbf{k} = -3\mathbf{i} - 4\mathbf{j} - 2\mathbf{k} \] ### **Moment about B:** \[ \vec{M}_B = \vec{r}_{BA} \times \vec{F} \] #### **Cross Product Calculation:** \[ \vec{M}_B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -4 & -2 \\ 222 & -222 & 220 \end{vmatrix} \] \[ = \mathbf{i} \left((-4)(220) - (-2)(-222)\right) - \mathbf{j} \left((-3)(220) - (-2)(222)\right) + \mathbf{k} \left((-3)(-222) - (-4)(222)\right) \] \[ = \mathbf{i} \left(-880 - 444\right) - \mathbf{j} \left(-660 + 444\right) + \mathbf{k} \left(666 + 888\right) \] \[ = \mathbf{i} (-1324) - \mathbf{j} (-216) + \mathbf{k} (1554) \] \[ = -1324\mathbf{i} + 216\mathbf{j} + 1554\mathbf{k} \] **Explanation:** We found the position vector from B to A, then used the cross product with the given force to compute the moment about point B. --- ## **Final Answers** ### **(a) Unit vector along AB:** \[ \boxed{ \mathbf{u}_{AB} = \frac{3}{\sqrt{29}}\mathbf{i} + \frac{4}{\sqrt{29}}\mathbf{j} + \frac{2}{\sqrt{29}}\mathbf{k} } \] ### **(b) Moment of force about point B:** \[ \boxed{ \vec{M}_B = -1324\mathbf{i} + 216\mathbf{j} + 1554\mathbf{k}\ \text{(N·m)} } \] --- If you want to proceed to part **(c)**, let me know!

Let's solve **all parts** step by step as requested. --- ## **Part (a): Determine the unit vector along the AB line** ### **Coordinates:** - \( A (1, 2, 0) \) - \( B (4, 6, 2) \) ### **Vector AB:** \[ \vec{AB} = (4 - 1)\mathbf{i} + (6 - 2)\mathbf{j} + (2 - 0)\mathbf{k} = 3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k} \] ### **Magnitude of AB:** \[ |\vec{AB}| = \sqrt{3^2 + 4^2 + 2^2} = \sqrt{9 + 16 + 4} = \sqrt{29} \] ### **Unit Vector along AB:** \[ \mathbf{u}_{AB} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{3}{\sqrt{29}}\mathbf{i} + \frac{4}{\sqrt{29}}\mathbf{j} + \frac{2}{\sqrt{29}}\mathbf{k} \] **Explanation:** We computed the direction vector from A to B, determined its magnitude, and normalized it to find the unit vector. --- ## **Part (b): Calculate the moment of the force \(\vec{F}\) about point B** ### **Given Force:** \[ \vec{F} = 222\mathbf{i} - 222\mathbf{j} + 220\mathbf{k} \] ### **Position vector from B to the point of force application (assume force is applied at A):** \[ \vec{r}_{BA} = \vec{r}_A - \vec{r}_B = (1 - 4)\mathbf{i} + (2 - 6)\mathbf{j} + (0 - 2)\mathbf{k} = -3\mathbf{i} - 4\mathbf{j} - 2\mathbf{k} \] ### **Moment about B:** \[ \vec{M}_B = \vec{r}_{BA} \times \vec{F} \] #### **Cross Product Calculation:** \[ \vec{M}_B = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & -4 & -2 \\ 222 & -222 & 220 \end{vmatrix} \] Calculating the determinant gives: \[ \vec{M}_B = \mathbf{i}((-4)(220) - (-2)(-222)) - \mathbf{j}((-3)(220) - (-2)(222)) + \mathbf{k}((-3)(-222) - (-4)(222)) \] Calculating each component: - \( \mathbf{i} \) component: \[ -880 - 444 = -1324 \] - \( \mathbf{j} \) component: \[ -660 + 444 = -216 \implies \text{(change sign)} = 216 \] - \( \mathbf{k} \) component: \[ 666 + 888 = 1554 \] ### **Final Moment:** \[ \vec{M}_B = -1324\mathbf{i} + 216\mathbf{j} + 1554\mathbf{k} \] **Explanation:** We computed the position vector from B to A, then calculated the moment using the cross product of this vector and the force vector. --- ## **Part (c): Calculate the moment of the force \(\vec{F}\) about the AB line** ### **Projection of Force onto AB:** To find the moment about the line, we need to project the force onto the unit vector along AB. \[ \text{Projection of } \vec{F} \text{ onto } \mathbf{u}_{AB} = \left(\vec{F} \cdot \mathbf{u}_{AB}\right) \mathbf{u}_{AB} \] Calculating the dot product: \[ \vec{F} \cdot \mathbf{u}_{AB} = (222)(\frac{3}{\sqrt{29}}) + (-222)(\frac{4}{\sqrt{29}}) + (220)(\frac{2}{\sqrt{29}}) \] \[ = \frac{666 - 888 + 440}{\sqrt{29}} = \frac{218}{\sqrt{29}} \] Now multiply by the unit vector: \[ \text{Projection} = \frac{218}{\sqrt{29}} \left(\frac{3}{\sqrt{29}}\mathbf{i} + \frac{4}{\sqrt{29}}\mathbf{j} + \frac{2}{\sqrt{29}}\mathbf{k}\right) = \frac{218}{29}(3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) \] ### **Moment about the AB line:** The moment about the line is given by: \[ \vec{M}_{AB} = \vec{M}_B - \text{Projection} \] Calculating it: \[ \vec{M}_{AB} = (-1324\mathbf{i} + 216\mathbf{j} + 1554\mathbf{k}) - \frac{218}{29}(3\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}) \] Calculating each component individually: - \( \mathbf{i} \): \[ -1324 - \frac{654}{29} \approx -1324 - 22.55 \approx -1346.55 \] - \( \mathbf{j} \): \[ 216 - \frac{872}{29} \approx 216 - 30.1 \approx 185.9 \] - \( \mathbf{k} \): \[ 1554 - \frac{436}{29} \approx 1554 - 15.06 \approx 1538.94 \] ### **Final Moment about AB line in Cartesian Vector Form:** \[ \vec{M}_{AB} \approx -1346.55\mathbf{i} + 185.9\mathbf{j} + 1538.94\mathbf{k} \] --- ## **Final Answers** ### **(a) Unit vector along AB:** \[ \boxed{ \mathbf{u}_{AB} = \frac{3}{\sqrt{29}}\mathbf{i} + \frac{4}{\sqrt{29}}\mathbf{j} + \frac{2}{\sqrt{29}}\mathbf{k} } \] ### **(b) Moment of force about point B:** \[ \boxed{ \vec{M}_B = -1324\mathbf{i} + 216\mathbf{j} + 1554\mathbf{k}\ \text{(N·m)} } \] ### **(c) Moment of the force about the AB line:** \[ \boxed{ \vec{M}_{AB} \approx -1346.55\mathbf{i} + 185.9\mathbf{j} + 1538.94\mathbf{k}\ \text{(N·m)} } \]