1. **Problem Statement:** We are given points in 3D space: $A=(0,150,10)$, $B=(180,150,10)$, $C=(180,450,10)$, $D=(280,500,10)$, $E=(280,500,62)$. Find vectors $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CD}$, and $\overrightarrow{DE}$. Then sum these vectors, and compare the sum to vector $\overrightarrow{EA}$. 2. **Calculate each vector as difference of coordinates:** - $\overrightarrow{AB} = B - A = (180-0, 150-150, 10-10) = (180, 0, 0)$ - $\overrightarrow{BC} = C - B = (180-180, 450-150, 10-10) = (0, 300, 0)$ - $\overrightarrow{CD} = D - C = (280-180, 500-450, 10-10) = (100, 50, 0)$ - $\overrightarrow{DE} = E - D = (280-280, 500-500, 62-10) = (0, 0, 52)$ 3. **Sum the vectors:** $$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} = (180,0,0) + (0,300,0) + (100,50,0) + (0,0,52) = (180+0+100+0, 0+300+50+0, 0+0+0+52) = (280, 350, 52)$$ 4. **Compute $\overrightarrow{EA}$ directly:** $$\overrightarrow{EA} = A - E = (0-280, 150-500, 10-62) = (-280, -350, -52)$$ 5. **Comparison:** The vector sum from A to E following the path is $ (280, 350, 52) $, which is the exact negative of $\overrightarrow{EA} = (-280, -350, -52)$. This shows that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} + \overrightarrow{DE} = -\overrightarrow{EA}$, confirming the path sums correctly and $\overrightarrow{EA}$ points from E back to A along the direct path. **Final answers:** - $\overrightarrow{AB} = (180, 0, 0)$ - $\overrightarrow{BC} = (0, 300, 0)$ - $\overrightarrow{CD} = (100, 50, 0)$ - $\overrightarrow{DE} = (0, 0, 52)$ - Sum: $(280, 350, 52)$ - $\overrightarrow{EA} = (-280, -350, -52)$ They are opposites, confirming vector addition and path properties.