1. **Problem statement:** Find two unit vectors orthogonal to the given vectors $\mathbf{a}$ and $\mathbf{b}$. 2. **Formula and concept:** Two vectors are orthogonal if their dot product is zero: $$\mathbf{u} \cdot \mathbf{a} = 0 \quad \text{and} \quad \mathbf{u} \cdot \mathbf{b} = 0.$$ A vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$ can be found using the cross product: $$\mathbf{u} = \mathbf{a} \times \mathbf{b}.$$ The unit vector is then $$\hat{\mathbf{u}} = \frac{\mathbf{u}}{\|\mathbf{u}\|}.$$ Important rule: The cross product of two vectors is perpendicular to both. 3. **Given vectors:** $$\mathbf{a} = \langle 0, 2, 1 \rangle, \quad \mathbf{b} = \langle 1, 0, -1 \rangle.$$ 4. **Calculate the cross product:** $$\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & 1 \\ 1 & 0 & -1 \end{vmatrix} = \mathbf{i}(2 \cdot (-1) - 1 \cdot 0) - \mathbf{j}(0 \cdot (-1) - 1 \cdot 1) + \mathbf{k}(0 \cdot 0 - 2 \cdot 1)$$ $$= \mathbf{i}(-2) - \mathbf{j}(-1) + \mathbf{k}(-2) = \langle -2, 1, -2 \rangle.$$ 5. **Calculate the magnitude of $\mathbf{u}$:** $$\|\mathbf{u}\| = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$$ 6. **Find the unit vector:** $$\hat{\mathbf{u}} = \frac{1}{3} \langle -2, 1, -2 \rangle = \left\langle -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right\rangle.$$ 7. **The two unit vectors orthogonal to $\mathbf{a}$ and $\mathbf{b}$ are:** $$\hat{\mathbf{u}} = \left\langle -\frac{2}{3}, \frac{1}{3}, -\frac{2}{3} \right\rangle \quad \text{and} \quad -\hat{\mathbf{u}} = \left\langle \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \right\rangle.$$ These are unit vectors because their magnitude is 1 and they are orthogonal to both given vectors by construction.