To calculate the length (or magnitude) of each vector $\vec{a}$ and $\vec{b}$, let's go through both methods.

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

1. Using the Pythagorean Theorem

For a vector $\vec{a} = (x, y)$, the length (magnitude) $|\vec{a}|$ is given by: $|\vec{a}| = \sqrt{x^2 + y^2}$

Applying this to $\vec{a} = (5, 4)$: $|\vec{a}| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.4$

2. Using the Dot Product

The length of a vector $\vec{a}$ can also be found by taking the dot product of the vector with itself: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ Thus, $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$

Calculating $\vec{a} \cdot \vec{a}$ for $(5, 4)$: $\vec{a} \cdot \vec{a} = 5 \cdot 5 + 4 \cdot 4 = 25 + 16 = 41$ So, $|\vec{a}| = \sqrt{41} \approx 6.4$

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

1. Using the Pythagorean Theorem

For a vector $\vec{b} = (x, y, z)$, the length $|\vec{b}|$ is given by: $|\vec{b}| = \sqrt{x^2 + y^2 + z^2}$

Applying this to $\vec{b} = (-1, 3, 2)$: $|\vec{b}| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \approx 3.74$

2. Using the Dot Product

Similarly, we can find $|\vec{b}|$ using the dot product: $|\vec{b}| = \sqrt{\vec{b} \cdot \vec{b}}$

Calculating $\vec{b} \cdot \vec{b}$ for $(-1, 3, 2)$: $\vec{b} \cdot \vec{b} = (-1) \cdot (-1) + 3 \cdot 3 + 2 \cdot 2 = 1 + 9 + 4 = 14$ So, $|\vec{b}| = \sqrt{14} \approx 3.74$

---

Summary

• Length of $\vec{a} = (5, 4)$: $\sqrt{41} \approx 6.4$
• Length of $\vec{b} = (-1, 3, 2)$: $\sqrt{14} \approx 3.74$

Would you like more details on these calculations, or do you have any questions?

---

Here are some related questions to explore:

1. How can we calculate the angle between two different vectors?
2. What is the geometric interpretation of the dot product?
3. Can we use the cross product to find vector lengths?
4. What are some applications of vector magnitude in physics?
5. How does the dot product formula generalize to higher dimensions?

Tip: The dot product is especially useful for finding the projection of one vector onto another, making it a powerful tool in both geometry and physics.