Let $\mathbf{a}$ and $\mathbf{b}$ be vectors, and let $\mathbf{m}$ be the midpoint of $\mathbf{a}$ and $\mathbf{b}.$  Given $\mathbf{m} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ and $\mathbf{a} \cdot \mathbf{b} = 6,$ find $\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2.$
Since $\mathbf{m}$ is the midpoint of $\mathbf{a}$ and $\mathbf{b},$
\[\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}.\]Hence, $\mathbf{a} + \mathbf{b} = 2 \mathbf{m} = \begin{pmatrix} 6 \\ 14 \end{pmatrix}.$  Then
\[\|\mathbf{a} + \mathbf{b}\|^2 = \left\| \begin{pmatrix} 6 \\ 14 \end{pmatrix} \right\|^2 = 6^2  + 14^2 = 232.\]But
\begin{align*}
\|\mathbf{a} + \mathbf{b}\|^2 &= (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b}) \\
&= \mathbf{a} \cdot \mathbf{a} + 2 \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{b} \\
&= \|\mathbf{a}\|^2 + 2 \mathbf{a} \cdot \mathbf{b} + \|\mathbf{b}\|^2,
\end{align*}so
\[\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 = \|\mathbf{a} + \mathbf{b}\|^2 - 2 \mathbf{a} \cdot \mathbf{b} = 232 - 2 \cdot 6 = \boxed{220}.\]
220
Level 4
Precalculus
Problem: Let $\mathbf{a}$ and $\mathbf{b}$ be vectors, and let $\mathbf{m}$ be the midpoint of $\mathbf{a}$ and $\mathbf{b}.$  Given $\mathbf{m} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}$ and $\mathbf{a} \cdot \mathbf{b} = 6,$ find $\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2, we start by using the definition of the midpoint. The midpoint \mathbf{m} of vectors \mathbf{a} and \mathbf{b} is given by:

\mathbf{m} = \frac{\mathbf{a} + \mathbf{b}}{2}.

Given \mathbf{m} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}, we can express this as:

\frac{\mathbf{a} + \mathbf{b}}{2} = \begin{pmatrix} 3 \\ 7 \end{pmatrix}.

Multiplying both sides by 2 yields:

\mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ 14 \end{pmatrix}.

Next, we denote \|\mathbf{a}\|^2 and \|\mathbf{b}\|^2 using the formula for the dot product:

\|\mathbf{a}\|^2 = \mathbf{a} \cdot \mathbf{a}, \quad \|\mathbf{b}\|^2 = \mathbf{b} \cdot \mathbf{b}.

We also know that:

\|\mathbf{a} + \mathbf{b}\|^2 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + 2(\mathbf{a} \cdot \mathbf{b}).

Substituting \mathbf{a} + \mathbf{b} = \begin{pmatrix} 6 \\ 14 \end{pmatrix}, we calculate:

\|\mathbf{a} + \mathbf{b}\|^2 = \begin{pmatrix} 6 \\ 14 \end{pmatrix} \cdot \begin{pmatrix} 6 \\ 14 \end{pmatrix} = 6^2 + 14^2 = 36 + 196 = 232.

Now, we substitute \mathbf{a} \cdot \mathbf{b} = 6$ into the equation:

232 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + 2(6).

This simplifies to:

232 = \|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 + 12.

Rearranging gives:

\|\mathbf{a}\|^2 + \|\mathbf{b}\|^2 = 232 - 12 = 220.

Thus, the final answer is 220.<|END_OF_LATENT|> The final answer is $220$.