### Instruction: Let $\mathbf{a} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}.$  Find the vector $\mathbf{v}$ that satisfies $\mathbf{v} \times \mathbf{a} = \mathbf{b} \times \mathbf{a}$ and $\mathbf{v} \times \mathbf{b} = \mathbf{a} \times \mathbf{b}.$.### Response: This problem involves finding a vector that has the same cross product with two given vectors as those vectors have with each other.
One way to approach this is to use the fact that the cross product is distributive over addition, meaning that $(\mathbf{u} + \mathbf{v}) \times \mathbf{w} = \mathbf{u} \times \mathbf{w} + \mathbf{v} \times \mathbf{w}.$
If I let $\mathbf{v} = \mathbf{a} + \mathbf{b},$ then I can check if it satisfies the given conditions.
First, I compute $\mathbf{v} \times \mathbf{a} = (\mathbf{a} + \mathbf{b}) \times \mathbf{a} = \mathbf{a} \times \mathbf{a} + \mathbf{b} \times \mathbf{a}.$
Since the cross product of any vector with itself is zero, $\mathbf{a} \times \mathbf{a} = \mathbf{0},$ and I get $\mathbf{v} \times \mathbf{a} = \mathbf{b} \times \mathbf{a},$ as desired.
Next, I compute $\mathbf{v} \times \mathbf{b} = (\mathbf{a} + \mathbf{b}) \times \mathbf{b} = \mathbf{a} \times \mathbf{b} + \mathbf{b} \times \mathbf{b}.$
Similarly, $\mathbf{b} \times \mathbf{b} = \mathbf{0},$ and I get $\mathbf{v} \times \mathbf{b} = \mathbf{a} \times \mathbf{b},$ as desired.
Therefore, $\mathbf{v} = \mathbf{a} + \mathbf{b}$ is a vector that satisfies the given conditions.
To find its components, I just add the components of $\mathbf{a}$ and $\mathbf{b}$ respectively: $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \\ -1 \end{pmatrix}.$