If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ and $|\vec{a}| = 3, |\vec{b}| = 5, |\vec{c}| = 7$, then the angle between $\vec{a}$ and $\vec{b}$ is

Given three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$ where:

• $\vec{a} + \vec{b} + \vec{c} = \vec{0}$
• $|\vec{a}| = 3$
• $|\vec{b}| = 5$
• $|\vec{c}| = 7$

---

From $\vec{a} + \vec{b} + \vec{c} = \vec{0}$:

$\vec{c} = -(\vec{a} + \vec{b})$

---

Taking magnitude on both sides:

$|\vec{c}|^2 = |\vec{a} + \vec{b}|^2$

Expanding using $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$:

$|\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$

---

Substituting the known values:

$49 = 9 + 25 + 2\vec{a} \cdot \vec{b}$

$49 = 34 + 2\vec{a} \cdot \vec{b}$

$15 = 2\vec{a} \cdot \vec{b}$

$\vec{a} \cdot \vec{b} = \frac{15}{2}$

---

Using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$:

$\frac{15}{2} = 3 \times 5 \times \cos\theta$

$\frac{15}{2} = 15\cos\theta$

$\cos\theta = \frac{1}{2}$

---

When $\cos\theta = \frac{1}{2}$:

$\theta = \frac{\pi}{3}$ (or 60°)

Therefore, the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{3}$.