1. **Problem Statement:** We are given vectors and asked to find their norms (lengths), check if they are unit vectors, and normalize a vector. 2. **Formula for norm:** The norm (or length) of a vector $\mathbf{u} = (u_1, u_2, \ldots, u_n)$ is given by: $$||\mathbf{u}|| = \sqrt{u_1^2 + u_2^2 + \cdots + u_n^2}$$ 3. **Part (a):** Given $\mathbf{u} = (1, -2, -4, 5, 3)$, calculate $||\mathbf{u}||^2$ by squaring each component and summing: $$||\mathbf{u}||^2 = 1^2 + (-2)^2 + (-4)^2 + 5^2 + 3^2 = 1 + 4 + 16 + 25 + 9 = 55$$ Then take the square root to find the norm: $$||\mathbf{u}|| = \sqrt{55}$$ 4. **Part (b):** Given $\mathbf{v} = (1, -3, 4, 2)$ and $\mathbf{w} = \left(\frac{1}{2}, -\frac{1}{6}, \frac{5}{6}, \frac{1}{6}\right)$, calculate their norms: $$||\mathbf{v}|| = \sqrt{1^2 + (-3)^2 + 4^2 + 2^2} = \sqrt{1 + 9 + 16 + 4} = \sqrt{30}$$ $$||\mathbf{w}|| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{6}\right)^2 + \left(\frac{5}{6}\right)^2 + \left(\frac{1}{6}\right)^2} = \sqrt{\frac{1}{4} + \frac{1}{36} + \frac{25}{36} + \frac{1}{36}} = \sqrt{\frac{9}{36} + \frac{1}{36} + \frac{25}{36} + \frac{1}{36}} = \sqrt{\frac{36}{36}} = 1$$ Since $||\mathbf{w}|| = 1$, $\mathbf{w}$ is a unit vector. 5. **Normalize $\mathbf{v}$:** To normalize $\mathbf{v}$, divide each component by $||\mathbf{v}||$: $$\hat{\mathbf{v}} = \frac{\mathbf{v}}{||\mathbf{v}||} = \left(\frac{1}{\sqrt{30}}, \frac{-3}{\sqrt{30}}, \frac{4}{\sqrt{30}}, \frac{2}{\sqrt{30}}\right)$$ This is the unique unit vector in the same direction as $\mathbf{v}$. 6. **Summary:** - $||\mathbf{u}|| = \sqrt{55}$ - $||\mathbf{v}|| = \sqrt{30}$ - $||\mathbf{w}|| = 1$ (unit vector) - Normalized $\mathbf{v}$ is $\hat{\mathbf{v}} = \left(\frac{1}{\sqrt{30}}, \frac{-3}{\sqrt{30}}, \frac{4}{\sqrt{30}}, \frac{2}{\sqrt{30}}\right)$ This completes the solution.