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

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

Method 1: Using the Pythagorean Theorem

The length $|\vec{a}|$ of a vector $\vec{a} = (x, y)$ in two dimensions is given by: $|\vec{a}| = \sqrt{x^2 + y^2}$ For $\vec{a} = (5, 4)$: $|\vec{a}| = \sqrt{5^2 + 4^2} = \sqrt{25 + 16} = \sqrt{41} \approx 6.40$

Method 2: Using the Dot Product

The dot product of a vector $\vec{a}$ with itself is $\vec{a} \cdot \vec{a} = x^2 + y^2$, which equals $|\vec{a}|^2$. Therefore: $|\vec{a}| = \sqrt{\vec{a} \cdot \vec{a}}$ For $\vec{a} = (5, 4)$: $\vec{a} \cdot \vec{a} = 5^2 + 4^2 = 25 + 16 = 41$ So, $|\vec{a}| = \sqrt{41} \approx 6.40$

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

Method 1: Using the Pythagorean Theorem

The length $|\vec{b}|$ of a vector $\vec{b} = (x, y, z)$ in three dimensions is given by: $|\vec{b}| = \sqrt{x^2 + y^2 + z^2}$ For $\vec{b} = (-1, 3, 2)$: $|\vec{b}| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \approx 3.74$

Method 2: Using the Dot Product

Similarly, the dot product of $\vec{b}$ with itself is $\vec{b} \cdot \vec{b} = x^2 + y^2 + z^2$, which equals $|\vec{b}|^2$. Therefore: $|\vec{b}| = \sqrt{\vec{b} \cdot \vec{b}}$ For $\vec{b} = (-1, 3, 2)$: $\vec{b} \cdot \vec{b} = (-1)^2 + 3^2 + 2^2 = 1 + 9 + 4 = 14$ So, $|\vec{b}| = \sqrt{14} \approx 3.74$

Summary of Results

• $|\vec{a}| = \sqrt{41} \approx 6.40$
• $|\vec{b}| = \sqrt{14} \approx 3.74$

Would you like further details or have any questions?

Here are some related questions for further exploration:

1. How would you calculate the angle between two different vectors?
2. How does the dot product change if the vectors are perpendicular?
3. What is the significance of the dot product being zero?
4. How can the cross product be used to find vector lengths in 3D space?
5. What are some applications of vector magnitude in physics?

Tip: The length (magnitude) of a vector gives an intuitive sense of its "size" or "strength" in space.