There exists a scalar $c$ so that
\[\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = c \mathbf{v}\]for all vectors $\mathbf{v}.$  Find $c.$
In general, the vector triple product states that for any vectors $\mathbf{a},$ $\mathbf{b},$ and $\mathbf{c},$
\[\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}.\]So
\begin{align*}
\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) &= (\mathbf{i} \cdot \mathbf{i}) \mathbf{v} - (\mathbf{i} \cdot \mathbf{v}) \mathbf{i} = \mathbf{v} - (\mathbf{i} \cdot \mathbf{v}) \mathbf{i}, \\
\mathbf{j} \times (\mathbf{v} \times \mathbf{j}) &= (\mathbf{j} \cdot \mathbf{j}) \mathbf{v} - (\mathbf{j} \cdot \mathbf{v}) \mathbf{j} = \mathbf{v} - (\mathbf{j} \cdot \mathbf{v}) \mathbf{j}, \\
\mathbf{k} \times (\mathbf{v} \times \mathbf{k}) &= (\mathbf{k} \cdot \mathbf{k}) \mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k} = \mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k}.
\end{align*}Hence,
\begin{align*}
&\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) \\
&= 3 \mathbf{v} - ((\mathbf{i} \cdot \mathbf{v}) \mathbf{i} + (\mathbf{j} \cdot \mathbf{v}) \mathbf{j} + (\mathbf{k} \cdot \mathbf{v}) \mathbf{k}) \\
&= 3 \mathbf{v} - \mathbf{v} = 2 \mathbf{v}.
\end{align*}Thus, $c = \boxed{2}.$
2
Level 3
Precalculus
Problem: There exists a scalar $c$ so that
\[\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = c \mathbf{v}\]for all vectors $\mathbf{v}.$  Find $c.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, we need to analyze the expression involving the cross products. We start by applying the vector triple product identity, which states that for any vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, the identity $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}$ holds.

We will apply this identity to each term in the given equation. 

1. For the first term $\mathbf{i} \times (\mathbf{v} \times \mathbf{i})$, we set $\mathbf{a} = \mathbf{i}$, $\mathbf{b} = \mathbf{v}$, and $\mathbf{c} = \mathbf{i}$. Using the identity, we get:
   \[
   \mathbf{i} \times (\mathbf{v} \times \mathbf{i}) = (\mathbf{i} \cdot \mathbf{i}) \mathbf{v} - (\mathbf{i} \cdot \mathbf{v}) \mathbf{i} = 1 \cdot \mathbf{v} - v_1 \mathbf{i} = \mathbf{v} - v_1 \mathbf{i}
   \]

2. For the second term $\mathbf{j} \times (\mathbf{v} \times \mathbf{j})$, we set $\mathbf{a} = \mathbf{j}$, $\mathbf{b} = \mathbf{v}$, and $\mathbf{c} = \mathbf{j}$. Applying the identity gives:
   \[
   \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) = (\mathbf{j} \cdot \mathbf{j}) \mathbf{v} - (\mathbf{j} \cdot \mathbf{v}) \mathbf{j} = 1 \cdot \mathbf{v} - v_2 \mathbf{j} = \mathbf{v} - v_2 \mathbf{j}
   \]

3. For the third term $\mathbf{k} \times (\mathbf{v} \times \mathbf{k})$, we set $\mathbf{a} = \mathbf{k}$, $\mathbf{b} = \mathbf{v}$, and $\mathbf{c} = \mathbf{k}$. Using the identity results in:
   \[
   \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = (\mathbf{k} \cdot \mathbf{k}) \mathbf{v} - (\mathbf{k} \cdot \mathbf{v}) \mathbf{k} = 1 \cdot \mathbf{v} - v_3 \mathbf{k} = \mathbf{v} - v_3 \mathbf{k}
   \]

Now, we combine all three results:
\[
\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = (\mathbf{v} - v_1 \mathbf{i}) + (\mathbf{v} - v_2 \mathbf{j}) + (\mathbf{v} - v_3 \mathbf{k})
\]
This simplifies to:
\[
3\mathbf{v} - (v_1 \mathbf{i} + v_2 \mathbf{j} + v_3 \mathbf{k}) = 3\mathbf{v} - \mathbf{v} = 2\mathbf{v}
\]

Thus, we have:
\[
\mathbf{i} \times (\mathbf{v} \times \mathbf{i}) + \mathbf{j} \times (\mathbf{v} \times \mathbf{j}) + \mathbf{k} \times (\mathbf{v} \times \mathbf{k}) = 2\mathbf{v}
\]

This shows that the scalar $c$ must equal $2$, as the equation holds for all vectors $\mathbf{v}$. Therefore, the final answer is $2$.<|END_OF_LATENT|> The final answer is $2$.