Given the problem, where vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ have magnitudes of 3, 4, and 5 units respectively, and the vector equation $\vec{A} + \vec{B} = \vec{C}$, we want to find the angle $\theta$ between $\vec{A}$ and $\vec{B}$.

To solve this, we can use the following steps:

1. Vector Equation

Given: $\vec{A} + \vec{B} = \vec{C}$

2. Magnitude Relation

Taking the magnitude on both sides: $|\vec{A} + \vec{B}| = |\vec{C}|$

Using the formula for the magnitude of the sum of two vectors: $|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$

Substituting the magnitudes: $5 = \sqrt{3^2 + 4^2 + 2(3)(4)\cos\theta}$

3. Simplifying the Equation

$5 = \sqrt{9 + 16 + 24\cos\theta}$ $5 = \sqrt{25 + 24\cos\theta}$

Square both sides to remove the square root: $25 = 25 + 24\cos\theta$

4. Solving for $\cos\theta$

$25 - 25 = 24\cos\theta$ $0 = 24\cos\theta$

So: $\cos\theta = 0$

5. Conclusion

If $\cos\theta = 0$, then: $\theta = 90^\circ$

Therefore, the angle between vectors $\vec{A}$ and $\vec{B}$ is $90^\circ$.

Would you like further details on this, or have any questions? Here are some related questions you might find interesting:

1. How do you find the angle between any two vectors in general?
2. What are the applications of the dot product in vector calculations?
3. How does the Pythagorean theorem relate to vector magnitudes?
4. Can the magnitude of a vector be negative? Why or why not?
5. How would the solution change if the vectors did not satisfy $\vec{A} + \vec{B} = \vec{C}$?

Tip: Always double-check the conditions given in vector problems; they often simplify the solution significantly!