Let $\mathbf{u},$ $\mathbf{v},$ and $\mathbf{w}$ be vectors such that $\|\mathbf{u}\| = 3,$ $\|\mathbf{v}\| = 4,$ and $\|\mathbf{w}\| = 5,$ and
\[\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}.\]Compute $\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}.$
From $\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0},$ we have $(\mathbf{u} + \mathbf{v} + \mathbf{w}) \cdot (\mathbf{u} + \mathbf{v} + \mathbf{w}) = 0.$  Expanding, we get
\[\mathbf{u} \cdot \mathbf{u} + \mathbf{v} \cdot \mathbf{v} + \mathbf{w} \cdot \mathbf{w} + 2 (\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}) = 0.\]Note that $\mathbf{u} \cdot \mathbf{u} = \|\mathbf{u}\|^2 = 9,$ $\mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^2 = 16,$ and $\mathbf{w} \cdot \mathbf{w} = \|\mathbf{w}\|^2 = 25,$ so
\[2 (\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}) + 50 = 0.\]Therefore, $\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w} = \boxed{-25}.$
-25
Level 4
Precalculus
Problem: Let $\mathbf{u},$ $\mathbf{v},$ and $\mathbf{w}$ be vectors such that $\|\mathbf{u}\| = 3,$ $\|\mathbf{v}\| = 4,$ and $\|\mathbf{w}\| = 5,$ and
\[\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}.\]Compute $\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w}$, we start from the equation $\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}$. This implies that $\mathbf{w} = -(\mathbf{u} + \mathbf{v})$. 

Next, we compute the square of the norm of $\mathbf{w}$:
\[
\|\mathbf{w}\|^2 = \|-\mathbf{u} - \mathbf{v}\|^2 = \|\mathbf{u} + \mathbf{v}\|^2.
\]
Using the property of norms, we have:
\[
\|\mathbf{u} + \mathbf{v}\|^2 = \|\mathbf{u}\|^2 + \|\mathbf{v}\|^2 + 2 \mathbf{u} \cdot \mathbf{v}.
\]
Substituting the known norms:
\[
\|\mathbf{w}\|^2 = 5^2 = 25,
\]
\[
\|\mathbf{u}\|^2 = 3^2 = 9, \quad \|\mathbf{v}\|^2 = 4^2 = 16.
\]
Thus, we can write:
\[
25 = 9 + 16 + 2 \mathbf{u} \cdot \mathbf{v}.
\]
This simplifies to:
\[
25 = 25 + 2 \mathbf{u} \cdot \mathbf{v}.
\]
From this, we deduce:
\[
0 = 2 \mathbf{u} \cdot \mathbf{v} \implies \mathbf{u} \cdot \mathbf{v} = 0.
\]
This indicates that $\mathbf{u}$ and $\mathbf{v}$ are orthogonal.

Next, we compute $\mathbf{u} \cdot \mathbf{w}$:
\[
\mathbf{u} \cdot \mathbf{w} = \mathbf{u} \cdot (-\mathbf{u} - \mathbf{v}) = -\|\mathbf{u}\|^2 - \mathbf{u} \cdot \mathbf{v} = -9 - 0 = -9.
\]

Now, we compute $\mathbf{v} \cdot \mathbf{w}$:
\[
\mathbf{v} \cdot \mathbf{w} = \mathbf{v} \cdot (-\mathbf{u} - \mathbf{v}) = -\mathbf{v} \cdot \mathbf{u} - \|\mathbf{v}\|^2 = -0 - 16 = -16.
\]

Finally, we combine these results:
\[
\mathbf{u} \cdot \mathbf{v} + \mathbf{u} \cdot \mathbf{w} + \mathbf{v} \cdot \mathbf{w} = 0 - 9 - 16 = -25.
\]
Thus, the final answer is $-25$.<|END_OF_LATENT|> The final answer is $-25$.