To find the angle between the vectors $\mathbf{a} = 2\mathbf{i} + 3\mathbf{j}$ and $\mathbf{b} = \mathbf{i} - \mathbf{j}$, use the formula:

$\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|}$

1. Calculate the dot product $\mathbf{a} \cdot \mathbf{b}$: $\mathbf{a} \cdot \mathbf{b} = (2)(1) + (3)(-1) = 2 - 3 = -1$

2. Calculate the magnitudes: $|\mathbf{a}| = \sqrt{(2^2 + 3^2)} = \sqrt{4 + 9} = \sqrt{13}$ $|\mathbf{b}| = \sqrt{(1^2 + (-1)^2)} = \sqrt{1 + 1} = \sqrt{2}$

3. Substitute these values into the cosine formula: $\cos \theta = \frac{-1}{\sqrt{13} \cdot \sqrt{2}} = \frac{-1}{\sqrt{26}}$

4. Finally, find the angle: $\theta = \cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)$

You can plot the vectors in a coordinate system, showing their directions and the angle between them.

Would you like any additional details or help with something else? Here are some related questions you might consider:

1. How do you derive the formula for the angle between two vectors?
2. Can you explain how to find the unit vector of a given vector?
3. What are some applications of vector angles in physics?
4. How do you visualize vectors in a 3D space?
5. What are other methods to find the angle between two vectors?

Tip: Always sketch the vectors to better understand their orientation and the resulting angle.