The angle between $\overrightarrow{x} = [1, -\sqrt(2), -\sqrt(2)]^T $ and $\overrightarrow{y} = [2\sqrt(2), 1, 1]^T $ 

To find the angle between two vectors $\overrightarrow{x}$ and $\overrightarrow{y}$, we can use the dot product formula. The formula for the angle $\theta$ between two vectors is:

\[
\cos \theta = \frac{\overrightarrow{x} \cdot \overrightarrow{y}}{|\overrightarrow{x}| |\overrightarrow{y}|}
\]

Where:
- $\overrightarrow{x} \cdot \overrightarrow{y}$ is the dot product of the vectors.
- $ |\overrightarrow{x}| $ and $ |\overrightarrow{y}| $ are the magnitudes (lengths) of the vectors.

Given:
\[
\overrightarrow{x} = [1, -\sqrt{2}, -\sqrt{2}]^T, \quad \overrightarrow{y} = [2\sqrt{2}, 1, 1]^T
\]

First, we calculate the dot product $\overrightarrow{x} \cdot \overrightarrow{y}$:

\[
\overrightarrow{x} \cdot \overrightarrow{y} = 1 \cdot 2\sqrt{2} + (-\sqrt{2}) \cdot 1 + (-\sqrt{2}) \cdot 1
\]

\[
\overrightarrow{x} \cdot \overrightarrow{y} = 2\sqrt{2} - \sqrt{2} - \sqrt{2}
\]

\[
\overrightarrow{x} \cdot \overrightarrow{y} = 2\sqrt{2} - 2\sqrt{2} = 0
\]

Since the dot product is 0, the vectors $\overrightarrow{x}$ and $\overrightarrow{y}$ are orthogonal, meaning the angle $\theta$ between them is 90 degrees.

Thus, the angle between $\overrightarrow{x}$ and $\overrightarrow{y}$ is $\theta = 90^\circ$ or $\theta = \frac{\pi}{2}$ radians.

### Do you want more details or have any questions?
Here are some questions you might consider:
1. How to compute the magnitude of a vector?
2. Why does a dot product of zero indicate orthogonality?
3. What are some applications of finding angles between vectors?
4. Can you explain the geometric interpretation of the dot product?
5. How to find the cross product of two vectors?

**Tip:** Always check your vector components and operations carefully to ensure accuracy in calculations.

Learn how to calculate the angle between vectors x = [1, -√2, -√2]^T and y = [2√2, 1, 1]^T using the dot product formula. Understand the concept of orthogonality and how it applies to vectors. Suitable for advanced high school students and beyond.

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