The angle $\theta$ between two vectors $\mathbf{A} = (1, 1, -1)$ and $\mathbf{B} = (0, 2, 3)$ can be found using the dot product formula:

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

Step 1: Calculate the dot product $\mathbf{A} \cdot \mathbf{B}$

The dot product of two vectors $\mathbf{A} = (A_x, A_y, A_z)$ and $\mathbf{B} = (B_x, B_y, B_z)$ is given by:

$\mathbf{A} \cdot \mathbf{B} = A_x B_x + A_y B_y + A_z B_z$

Substituting the values:

$\mathbf{A} \cdot \mathbf{B} = (1)(0) + (1)(2) + (-1)(3) = 0 + 2 - 3 = -1$

Step 2: Calculate the magnitudes of $\mathbf{A}$ and $\mathbf{B}$

The magnitude of a vector $\mathbf{A} = (A_x, A_y, A_z)$ is given by:

$|\mathbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$

For $\mathbf{A} = (1, 1, -1)$:

$|\mathbf{A}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$

For $\mathbf{B} = (0, 2, 3)$:

$|\mathbf{B}| = \sqrt{0^2 + 2^2 + 3^2} = \sqrt{0 + 4 + 9} = \sqrt{13}$

Step 3: Calculate $\cos \theta$

Now, substitute the dot product and magnitudes into the formula:

$\cos \theta = \frac{-1}{\sqrt{3} \times \sqrt{13}} = \frac{-1}{\sqrt{39}}$

Step 4: Calculate $\theta$

Finally, find the angle $\theta$ by taking the arccosine:

$\theta = \cos^{-1}\left(\frac{-1}{\sqrt{39}}\right)$

This is the exact value of the angle between the two vectors.

If you need the numerical value of the angle, I can calculate it for you as well.

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Would you like me to calculate the numerical value of $\theta$, or do you have any questions about the process? Here are some related questions you might find interesting:

1. How do you find the angle between two 2D vectors?
2. What is the geometric interpretation of the dot product?
3. How can you find the projection of one vector onto another?
4. What is the cross product, and how does it differ from the dot product?
5. Can the angle between two vectors be negative?

Tip: The dot product of two vectors is zero when the vectors are perpendicular (i.e., the angle between them is 90 degrees).