To find the magnitude of vector $\vec{C}$, where $\vec{C} = \vec{A} - \vec{B}$, we can apply the law of cosines. Since the angle between vectors $\vec{A}$ and $\vec{B}$ is 60 degrees, the formula for the magnitude of $\vec{C}$ is:

$|\vec{C}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2 \cdot |\vec{A}| \cdot |\vec{B}| \cdot \cos(\theta)}$

where:

• $|\vec{A}|$ is the magnitude of vector $\vec{A}$ (40 units),
• $|\vec{B}|$ is the magnitude of vector $\vec{B}$ (50 units),
• $\theta$ is the angle between $\vec{A}$ and $\vec{B}$, which is 60 degrees.

Let's calculate step-by-step:

1. First, convert the angle to radians if needed, but since the cosine function works with degrees, we can directly calculate: $\cos(60^\circ) = 0.5$

2. Now, substitute the values into the formula: $|\vec{C}| = \sqrt{40^2 + 50^2 - 2 \cdot 40 \cdot 50 \cdot 0.5}$ $|\vec{C}| = \sqrt{1600 + 2500 - 2000}$ $|\vec{C}| = \sqrt{4100 - 2000} = \sqrt{2100}$

3. Calculate the square root of 2100: $|\vec{C}| \approx 45.83 \, \text{units}$

Thus, the magnitude of vector $\vec{C}$ is approximately 45.83 units.

Would you like more details on this calculation or have any questions?

Here are 5 related questions:

1. How would the magnitude change if the angle between $\vec{A}$ and $\vec{B}$ were different?
2. How can you represent vector $\vec{C}$ graphically?
3. Can you solve the problem using vector components?
4. What would the magnitude of $\vec{C}$ be if the angle between $\vec{A}$ and $\vec{B}$ were 90 degrees?
5. How would the magnitude of $\vec{C}$ change if both $\vec{A}$ and $\vec{B}$ had the same magnitude?

Tip: In vector subtraction, understanding the relative angle between vectors is key to determining their resultant magnitude.