### Instruction: Find a unit vector that is orthogonal to both $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}.$.### Response: To find a unit vector that is orthogonal to two given vectors, I can use the cross product of the two vectors, and then divide by the magnitude of the cross product.
The cross product of $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ is given by the determinant of this matrix:

$$\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 0 \\ 1 & 0 & 2 \end{vmatrix}$$
Expanding along the first row, I get:

$$\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \mathbf{i} (1 \cdot 2 - 0 \cdot 0) - \mathbf{j} (1 \cdot 2 - 0 \cdot 1) + \mathbf{k} (1 \cdot 0 - 1 \cdot 1)$$
Simplifying, I get:

$$\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \times \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$$
The magnitude of this vector is:

$$\sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3$$