### Instruction: Find the vector $\mathbf{v}$ such that
\[\mathbf{i} \times [(\mathbf{v} - \mathbf{j}) \times \mathbf{i}] + \mathbf{j} \times [(\mathbf{v} - \mathbf{k}) \times \mathbf{j}] + \mathbf{k} \times [(\mathbf{v} - \mathbf{i}) \times \mathbf{k}] = \mathbf{0}.\].### Response: I notice that this equation involves a lot of cross products, which are anti-commutative and distributive over addition and subtraction.
I also notice that each term has a unit vector $\mathbf{i}, \mathbf{j}, \mathbf{k}$ crossed with a vector of the form $(\mathbf{v} - \mathbf{u}) \times \mathbf{u}$, where $\mathbf{u}$ is another unit vector.
I wonder if I can simplify each term by using some properties of cross products.
I recall that $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$ for any vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$.
I also recall that $\mathbf{u} \cdot \mathbf{u} = 1$ and $\mathbf{u} \cdot \mathbf{v} = 0$ for any orthogonal unit vectors $\mathbf{u}$ and $\mathbf{v}$.
I apply these properties to the first term and get \[\mathbf{i} \times [(\mathbf{v} - \mathbf{j}) \times \mathbf{i}] = (\mathbf{i} \cdot \mathbf{i}) (\mathbf{v} - \mathbf{j}) - (\mathbf{i} \cdot (\mathbf{v} - \mathbf{j})) \mathbf{i} = \mathbf{v} - \mathbf{j} - v_i \mathbf{i},\] where $v_i$ is the $i$-component of $\mathbf{v}$.
I apply the same properties to the second and third terms and get \[\mathbf{j} \times [(\mathbf{v} - \mathbf{k}) \times \mathbf{j}] = \mathbf{v} - \mathbf{k} - v_j \mathbf{j}\] and \[\mathbf{k} \times [(\mathbf{v} - \mathbf{i}) \times \mathbf{k}] = \mathbf{v} - \mathbf{i} - v_k \mathbf{k},\] where $v_j$ and $v_k$ are the $j$ and $k$-components of $\mathbf{v}$, respectively.
I add these three terms and set them equal to $\mathbf{0}$ and get \[3\mathbf{v} - (\mathbf{i} + \mathbf{j} + \mathbf{k}) - (v_i \mathbf{i} + v_j \mathbf{j} + v_k \mathbf{k}) = \mathbf{0}.\]
I rearrange this equation and get \[\mathbf{v} = \frac{1}{3} (\mathbf{i} + \mathbf{j} + \mathbf{k}) + \frac{1}{3} (v_i \mathbf{i} + v_j \mathbf{j} + v_k \mathbf{k}).\]
I realize that this is a general solution for any vector $\mathbf{v}$ that satisfies the given equation.
I wonder if there is a more specific solution that depends on the values of $v_i, v_j, v_k$.
I notice that the second term in the right-hand side of the equation is just $\frac{1}{3} \mathbf{v}$.
I substitute this into the equation and get \[\mathbf{v} = \frac{1}{3} (\mathbf{i} + \mathbf{j} + \mathbf{k}) + \frac{1}{3} \mathbf{v}.\]
I subtract $\frac{1}{3} \mathbf{v}$ from both sides and get \[\frac{2}{3} \mathbf{v} = \frac{1}{3} (\mathbf{i} + \mathbf{j} + \mathbf{k}).\]
I multiply both sides by $\frac{3}{2}$ and get \[\mathbf{v} = \frac{1}{2} (\mathbf{i} + \mathbf{j} + \mathbf{k}).\]