Given vectors \( \mathbf{A} = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{B} = \mathbf{i} + 2\mathbf{j} - 2\mathbf{k} \), find the angle between these two vectors.

Step 1 ：First, we calculate the dot product of vectors \( \mathbf{A} \) and \( \mathbf{B} \). The dot product of two vectors \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}||\mathbf{B}|cos\theta \), where \(|\mathbf{A}|\) and \(|\mathbf{B}|\) are the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \) respectively, and \(\theta\) is the angle between them. Thus, \( \mathbf{A} \cdot \mathbf{B} = (3)(1) + (-2)(2) + (1)(-2) = -3 \).

Step 2 ：Next, calculate the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \). The magnitude of a vector \( \mathbf{A} = \sqrt{(3)^2 + (-2)^2 + (1)^2} = \sqrt{14} \), and the magnitude of vector \( \mathbf{B} = \sqrt{(1)^2 + (2)^2 + (-2)^2} = \sqrt{9} = 3 \).

Step 3 ：Substitute the dot product and the magnitudes of \( \mathbf{A} \) and \( \mathbf{B} \) into the formula for the dot product: \( cos\theta = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}||\mathbf{B}|} = \frac{-3}{\sqrt{14} \cdot 3} = -\frac{1}{\sqrt{14}} \).

Step 4 ：Finally, find the angle \(\theta\) by taking the inverse cosine (or arccos) of \(-\frac{1}{\sqrt{14}}\).