1. **State the problem:** We are given a vector $\mathbf{v} = \begin{pmatrix}4 \\ -2 \\ 1\end{pmatrix}$ and need to find a unit vector $\mathbf{u}$ in the same direction as $\mathbf{v}$. 2. **Find the magnitude of $\mathbf{v}$:** The magnitude (or length) is given by $$ \|\mathbf{v}\| = \sqrt{4^2 + (-2)^2 + 1^2} = \sqrt{16 + 4 + 1} = \sqrt{21}. $$ 3. **Find the unit vector $\mathbf{u}$:** A unit vector in the same direction is $$ \mathbf{u} = \frac{1}{\|\mathbf{v}\|} \mathbf{v} = \frac{1}{\sqrt{21}} \begin{pmatrix}4 \\ -2 \\ 1\end{pmatrix} = \begin{pmatrix}\frac{4}{\sqrt{21}} \\ \frac{-2}{\sqrt{21}} \\ \frac{1}{\sqrt{21}}\end{pmatrix}. $$ 4. **Interpretation:** This vector $\mathbf{u}$ has length 1 and points in the same direction as $\mathbf{v}$. **Final answer:** $$ \mathbf{u} = \begin{pmatrix}\frac{4}{\sqrt{21}} \\ \frac{-2}{\sqrt{21}} \\ \frac{1}{\sqrt{21}}\end{pmatrix}. $$