To calculate the lengths of the given vectors $\vec{a} = (5, 4)$ and $\vec{b} = (-1, 3, 2)$ in two different ways, we'll use:

1. The Pythagorean Theorem (or magnitude formula) for each vector.
2. The dot product approach.

Vector $\vec{a} = (5, 4)$

1. Using the Pythagorean Theorem:

The length (or magnitude) of $\vec{a}$ is given by: $|\vec{a}| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41}$ Thus, $|\vec{a}| = \sqrt{41} \approx 6.4$.

2. Using the Dot Product:

The dot product of $\vec{a}$ with itself is: $\vec{a} \cdot \vec{a} = 5 \times 5 + 4 \times 4 = 25 + 16 = 41$ Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, we have: $|\vec{a}| = \sqrt{41} \approx 6.4$

Both methods yield the same result: $|\vec{a}| = \sqrt{41} \approx 6.4$.

---

Vector $\vec{b} = (-1, 3, 2)$

1. Using the Pythagorean Theorem:

The length of $\vec{b}$ is: $|\vec{b}| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$ Thus, $|\vec{b}| = \sqrt{14} \approx 3.74$.

2. Using the Dot Product:

The dot product of $\vec{b}$ with itself is: $\vec{b} \cdot \vec{b} = (-1) \times (-1) + 3 \times 3 + 2 \times 2 = 1 + 9 + 4 = 14$ Since $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, we have: $|\vec{b}| = \sqrt{14} \approx 3.74$

Both methods yield the same result: $|\vec{b}| = \sqrt{14} \approx 3.74$.

---

Would you like further details on any step?

Here are some related questions:

1. What are the properties of dot products and how are they useful in vector calculations?
2. How do you calculate the angle between two different vectors using the dot product?
3. How does the Pythagorean Theorem generalize for vectors in higher dimensions?
4. What are unit vectors, and how can you find the unit vector of $\vec{a}$ and $\vec{b}$?
5. How does the concept of orthogonality relate to the dot product?

Tip: The dot product is a versatile tool in vector analysis, allowing you to determine angles, lengths, and even projections of vectors onto one another.