1. **Problem statement:** Show that for any vector $\vec{v}$, the dot product $\vec{v} \cdot \vec{v} = |\vec{v}|^2$. 2. **Step 1:** Recall the definition of the dot product for a vector $\vec{v} = (v_1, v_2, v_3)$: $$\vec{v} \cdot \vec{v} = v_1^2 + v_2^2 + v_3^2$$ 3. **Step 2:** Recall the magnitude (length) of $\vec{v}$ is: $$|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$$ 4. **Step 3:** Squaring the magnitude gives: $$|\vec{v}|^2 = v_1^2 + v_2^2 + v_3^2$$ 5. **Step 4:** Comparing the expressions, we see: $$\vec{v} \cdot \vec{v} = |\vec{v}|^2$$ --- 6. **Problem statement:** Given $\vec{v} = \mathbf{i} + \mathbf{j} - \mathbf{k}$ and $\vec{w} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$, find: 7. **4.2.1 Find the unit vector in the direction of $\vec{v}$.** 8. **Step 1:** Calculate the magnitude of $\vec{v}$: $$|\vec{v}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$$ 9. **Step 2:** The unit vector $\hat{v}$ is: $$\hat{v} = \frac{1}{|\vec{v}|} \vec{v} = \frac{1}{\sqrt{3}} (\mathbf{i} + \mathbf{j} - \mathbf{k}) = \frac{1}{\sqrt{3}}\mathbf{i} + \frac{1}{\sqrt{3}}\mathbf{j} - \frac{1}{\sqrt{3}}\mathbf{k}$$ --- 10. **4.2.2 Find the component of $\vec{w}$ parallel to $\vec{v}$.** 11. **Step 1:** The component of $\vec{w}$ parallel to $\vec{v}$ is the projection of $\vec{w}$ onto $\vec{v}$: $$\text{proj}_{\vec{v}} \vec{w} = \left( \frac{\vec{w} \cdot \vec{v}}{|\vec{v}|^2} \right) \vec{v}$$ 12. **Step 2:** Calculate the dot product $\vec{w} \cdot \vec{v}$: $$\vec{w} \cdot \vec{v} = (2)(1) + (1)(1) + (2)(-1) = 2 + 1 - 2 = 1$$ 13. **Step 3:** Recall $|\vec{v}|^2 = 3$ from earlier. 14. **Step 4:** Calculate the scalar multiplier: $$\frac{\vec{w} \cdot \vec{v}}{|\vec{v}|^2} = \frac{1}{3}$$ 15. **Step 5:** Multiply by $\vec{v}$: $$\text{proj}_{\vec{v}} \vec{w} = \frac{1}{3} (\mathbf{i} + \mathbf{j} - \mathbf{k}) = \frac{1}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{1}{3} \mathbf{k}$$ **Final answers:** - $\vec{v} \cdot \vec{v} = |\vec{v}|^2$ - Unit vector in direction of $\vec{v}$ is $\frac{1}{\sqrt{3}}\mathbf{i} + \frac{1}{\sqrt{3}}\mathbf{j} - \frac{1}{\sqrt{3}}\mathbf{k}$ - Component of $\vec{w}$ parallel to $\vec{v}$ is $\frac{1}{3} \mathbf{i} + \frac{1}{3} \mathbf{j} - \frac{1}{3} \mathbf{k}$