1. **Problem statement:** Divide the vector $\vec{v} = 3\vec{i} + 5\vec{j} + \vec{k}$ into three components: one parallel to $\vec{i}$, one parallel to $\vec{i} + 2\vec{j} + 3\vec{k}$, and one perpendicular to both. 2. **Step 1: Define vectors** Let the components be: - $\vec{v}_1 = a\vec{i}$ (parallel to $\vec{i}$) - $\vec{v}_2 = b(\vec{i} + 2\vec{j} + 3\vec{k})$ (parallel to $\vec{i} + 2\vec{j} + 3\vec{k}$) - $\vec{v}_3$ perpendicular to both $\vec{v}_1$ and $\vec{v}_2$ 3. **Step 2: Write the vector sum** $$\vec{v} = \vec{v}_1 + \vec{v}_2 + \vec{v}_3 = a\vec{i} + b(\vec{i} + 2\vec{j} + 3\vec{k}) + \vec{v}_3$$ 4. **Step 3: Use perpendicularity conditions** Since $\vec{v}_3$ is perpendicular to $\vec{v}_1$ and $\vec{v}_2$, it satisfies: $$\vec{v}_3 \cdot \vec{i} = 0$$ $$\vec{v}_3 \cdot (\vec{i} + 2\vec{j} + 3\vec{k}) = 0$$ 5. **Step 4: Express $\vec{v}_3$** Rearranging: $$\vec{v}_3 = \vec{v} - \vec{v}_1 - \vec{v}_2 = (3 - a - b)\vec{i} + (5 - 2b)\vec{j} + (1 - 3b)\vec{k}$$ 6. **Step 5: Apply perpendicularity conditions** - $\vec{v}_3 \cdot \vec{i} = 0 \Rightarrow (3 - a - b) = 0$ so $$a = 3 - b$$ - $\vec{v}_3 \cdot (\vec{i} + 2\vec{j} + 3\vec{k}) = 0$: $$ (3 - a - b) + 2(5 - 2b) + 3(1 - 3b) = 0 $$ Substitute $a = 3 - b$: $$ (3 - (3 - b) - b) + 2(5 - 2b) + 3(1 - 3b) = 0 $$ Simplify: $$ (3 - 3 + b - b) + 10 - 4b + 3 - 9b = 0 $$ $$ 0 + 10 - 4b + 3 - 9b = 0 $$ $$ 13 - 13b = 0 $$ $$ 13b = 13 $$ $$ b = 1 $$ 7. **Step 6: Find $a$** $$ a = 3 - b = 3 - 1 = 2 $$ 8. **Step 7: Find $\vec{v}_3$** $$ \vec{v}_3 = (3 - 2 - 1)\vec{i} + (5 - 2(1))\vec{j} + (1 - 3(1))\vec{k} = 0\vec{i} + 3\vec{j} - 2\vec{k} $$ 9. **Final answer:** $$ \vec{v}_1 = 2\vec{i} $$ $$ \vec{v}_2 = 1(\vec{i} + 2\vec{j} + 3\vec{k}) = \vec{i} + 2\vec{j} + 3\vec{k} $$ $$ \vec{v}_3 = 3\vec{j} - 2\vec{k} $$