To determine if the given set of vectors $\mathbf{u}_1$ and $\mathbf{u}_2$ is orthonormal or orthogonal, we follow these steps:

Step 1: Check for orthogonality

Two vectors are orthogonal if their dot product is zero: $\mathbf{u}_1 \cdot \mathbf{u}_2 = \sum_{i=1}^{n} u_{1i} \cdot u_{2i}$

Vectors:

$\mathbf{u}_1 = \begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{bmatrix}, \quad \mathbf{u}_2 = \begin{bmatrix} \frac{1}{10} \\ 0 \\ \frac{1}{10} \end{bmatrix}$

Dot Product:

$\mathbf{u}_1 \cdot \mathbf{u}_2 = \left(-\frac{1}{3}\right)\left(\frac{1}{10}\right) + \left(\frac{2}{3}\right)(0) + \left(\frac{1}{3}\right)\left(\frac{1}{10}\right)$ $\mathbf{u}_1 \cdot \mathbf{u}_2 = -\frac{1}{30} + 0 + \frac{1}{30} = 0$

Since the dot product is zero, the vectors are orthogonal.

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Step 2: Check for orthonormality

Vectors are orthonormal if they are orthogonal and each has a norm of 1. To check this, calculate the norm of each vector.

Norm of $\mathbf{u}_1$:

$\|\mathbf{u}_1\| = \sqrt{\left(-\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2}$ $\|\mathbf{u}_1\| = \sqrt{\frac{1}{9} + \frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{6}{9}} = \sqrt{\frac{2}{3}}$

Norm of $\mathbf{u}_2$:

$\|\mathbf{u}_2\| = \sqrt{\left(\frac{1}{10}\right)^2 + 0^2 + \left(\frac{1}{10}\right)^2}$ $\|\mathbf{u}_2\| = \sqrt{\frac{1}{100} + \frac{1}{100}} = \sqrt{\frac{2}{100}} = \frac{\sqrt{2}}{10}$

Neither vector has a norm of 1, so the set is not orthonormal.

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Step 3: Normalize the vectors

To normalize a vector $\mathbf{u}$, divide each component by its norm.

Normalizing $\mathbf{u}_1$:

$\mathbf{u}_1^{\text{normalized}} = \frac{\mathbf{u}_1}{\|\mathbf{u}_1\|} = \frac{\begin{bmatrix} -\frac{1}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{bmatrix}}{\sqrt{\frac{2}{3}}}$ $\mathbf{u}_1^{\text{normalized}} = \begin{bmatrix} -\frac{1}{3\sqrt{\frac{2}{3}}} \\ \frac{2}{3\sqrt{\frac{2}{3}}} \\ \frac{1}{3\sqrt{\frac{2}{3}}} \end{bmatrix} = \begin{bmatrix} -\frac{\sqrt{3}}{3\sqrt{2}} \\ \frac{\sqrt{3}}{\sqrt{2}} \\ \frac{\sqrt{3}}{3\sqrt{2}} \end{bmatrix}$

Normalizing $\mathbf{u}_2$:

$\mathbf{u}_2^{\text{normalized}} = \frac{\mathbf{u}_2}{\|\mathbf{u}_2\|} = \frac{\begin{bmatrix} \frac{1}{10} \\ 0 \\ \frac{1}{10} \end{bmatrix}}{\frac{\sqrt{2}}{10}}$ $\mathbf{u}_2^{\text{normalized}} = \begin{bmatrix} \frac{1}{10} \cdot \frac{10}{\sqrt{2}} \\ 0 \\ \frac{1}{10} \cdot \frac{10}{\sqrt{2}} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}$

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Final Orthonormal Set:

$\mathbf{u}_1^{\text{normalized}} = \begin{bmatrix} -\frac{\sqrt{3}}{3\sqrt{2}} \\ \frac{\sqrt{3}}{\sqrt{2}} \\ \frac{\sqrt{3}}{3\sqrt{2}} \end{bmatrix}, \quad \mathbf{u}_2^{\text{normalized}} = \begin{bmatrix} \frac{1}{\sqrt{2}} \\ 0 \\ \frac{1}{\sqrt{2}} \end{bmatrix}$

Let me know if you'd like detailed calculations or further questions addressed!

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Related Questions:

1. What is the geometric meaning of orthonormal vectors?
2. How does normalization affect the length of a vector?
3. Can any set of orthogonal vectors be normalized to form an orthonormal set?
4. How does the dot product verify orthogonality?
5. Why is orthonormality important in applications like computer graphics?

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Tip:

When normalizing a vector, always divide by its norm to ensure its length becomes 1!