Find the angle between the vectors 2i + 3j and i - j.

To find the angle \( \theta \) between the vectors \(\mathbf{A} = 2\mathbf{i} + 3\mathbf{j}\) and \(\mathbf{B} = \mathbf{i} - \mathbf{j}\), we can use the formula:

\[
\cos \theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}
\]

### Step 1: Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\)

\[
\mathbf{A} \cdot \mathbf{B} = (2)(1) + (3)(-1) = 2 - 3 = -1
\]

### Step 2: Calculate the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\)

\[
|\mathbf{A}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13}
\]

\[
|\mathbf{B}| = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2}
\]

### Step 3: Substitute into the cosine formula

\[
\cos \theta = \frac{-1}{\sqrt{13} \cdot \sqrt{2}} = \frac{-1}{\sqrt{26}}
\]

### Step 4: Find the angle \(\theta\)

\[
\theta = \cos^{-1}\left(\frac{-1}{\sqrt{26}}\right)
\]

Calculating this gives:

\[
\theta \approx 108.21^\circ
\]

Thus, the angle between the vectors \(2\mathbf{i} + 3\mathbf{j}\) and \(\mathbf{i} - \mathbf{j}\) is approximately \(108.21^\circ\).

Do you want details or have any questions? Here are some related questions you might consider:

1. What is the geometric interpretation of the dot product?
2. How do you find the angle between three-dimensional vectors?
3. Can you provide an example of vectors with a right angle?
4. What other methods can be used to find angles between vectors?
5. How does the angle change if the vectors are scaled? 

**Tip:** Always ensure your vectors are expressed in the same coordinate system before performing calculations.

Discover how to find the angle between the vectors 2i + 3j and i - j using vector algebra and trigonometry. The solution includes calculations of the dot product, magnitudes, and the resulting angle of approximately 108.21 degrees.

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