Given the vectors $\vec{u}=7 i+10 j$ and $\vec{v}=3 i+2 j$, find the angle (in degrees) between them.

Step 1 ：Given the vectors \(\vec{u}=7 i+10 j\) and \(\vec{v}=3 i+2 j\), we are to find the angle (in degrees) between them.

Step 2 ：The angle between two vectors can be found using the dot product formula: \[\vec{u} \cdot \vec{v} = ||\vec{u}|| ||\vec{v}|| \cos(\theta)\] where \(\vec{u} \cdot \vec{v}\) is the dot product of \(\vec{u}\) and \(\vec{v}\), \(||\vec{u}||\) and \(||\vec{v}||\) are the magnitudes of \(\vec{u}\) and \(\vec{v}\) respectively, and \(\theta\) is the angle between \(\vec{u}\) and \(\vec{v}\).

Step 3 ：We can rearrange this formula to solve for \(\theta\): \[\theta = \cos^{-1}\left(\frac{\vec{u} \cdot \vec{v}}{||\vec{u}|| ||\vec{v}||}\right)\]

Step 4 ：We can calculate the dot product \(\vec{u} \cdot \vec{v}\) as \(u_i \cdot v_i + u_j \cdot v_j\), and the magnitudes \(||\vec{u}||\) and \(||\vec{v}||\) as \(\sqrt{u_i^2 + u_j^2}\) and \(\sqrt{v_i^2 + v_j^2}\) respectively.

Step 5 ：Let's calculate these values and find the angle.

Step 6 ：For the given vectors, the dot product is 41, the magnitude of \(\vec{u}\) is approximately 12.21, and the magnitude of \(\vec{v}\) is approximately 3.61.

Step 7 ：Substituting these values into the formula for \(\theta\), we find that the angle in radians is approximately 0.37.

Step 8 ：Converting this angle to degrees, we find that the angle between the vectors \(\vec{u}=7 i+10 j\) and \(\vec{v}=3 i+2 j\) is approximately \(\boxed{21.32}\) degrees.