To find the angle $\theta$ between two vectors $\mathbf{a} = \langle 1, 2, -1 \rangle$ and $\mathbf{b} = \langle 2, 1, -2 \rangle$, we can use the following formula derived from the dot product:

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

Step 1: Compute the dot product $\mathbf{a} \cdot \mathbf{b}$

The dot product is calculated as follows:

$\mathbf{a} \cdot \mathbf{b} = (1)(2) + (2)(1) + (-1)(-2) = 2 + 2 + 2 = 6$

Step 2: Compute the magnitudes $\|\mathbf{a}\|$ and $\|\mathbf{b}\|$

The magnitude of $\mathbf{a}$ is:

$\|\mathbf{a}\| = \sqrt{(1)^2 + (2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

The magnitude of $\mathbf{b}$ is:

$\|\mathbf{b}\| = \sqrt{(2)^2 + (1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 3: Compute $\cos \theta$

Now, using the formula for the cosine of the angle:

$\cos \theta = \frac{6}{\sqrt{6} \times 3} = \frac{6}{3\sqrt{6}} = \frac{2}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}$

Step 4: Find the angle $\theta$

To find $\theta$, take the inverse cosine of $\cos \theta$:

$\theta = \cos^{-1}\left(\frac{\sqrt{6}}{3}\right)$

This gives the angle in radians. Let me calculate this value.

$\theta \approx \cos^{-1}(0.8165) \approx 35.26^\circ$

Thus, the angle between the vectors is approximately $35.26^\circ$.

Would you like further details on any step?

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Related Questions:

1. How is the dot product of two vectors used to determine their angle?
2. What is the geometric interpretation of the dot product?
3. Can the angle between two vectors ever be greater than 90 degrees?
4. How do you calculate the projection of one vector onto another?
5. How do we generalize this to higher-dimensional vectors?

Tip: When dealing with vectors, always ensure that both the magnitudes and the dot product are calculated carefully to avoid errors.