### 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: I know that a unit vector is a vector with length 1, and that two vectors are orthogonal if their dot product is zero.
To find a vector that is orthogonal to both of the given vectors, I can use the cross product, which returns a vector that is perpendicular to the plane spanned by the two vectors.
The cross product of $\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and $\begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}$ is $\begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix}$, so I can apply this formula to the given vectors.
The cross product of $\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix}$ is $\begin{pmatrix} 1 \cdot 2 - 0 \cdot 0 \\ 0 \cdot 1 - 1 \cdot 2 \\ 1 \cdot 0 - 1 \cdot 1 \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ -1 \end{pmatrix}$.
This vector is orthogonal to both of the given vectors, but it is not a unit vector, because its length is $\sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3$.
To make it a unit vector, I need to divide it by its length, which will scale it down to have length 1.