Question

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

Hint:

When three vectors add to zero, the triangle rule applies: the sum of any two vectors is the negative of the third. You can use the cosine law to find angles between vectors.

Answer 1

Given that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, it follows that $\vec{a} + \vec{b} = -\vec{c}$. Squaring both sides, we get $(\vec{a} + \vec{b})^2 = (-\vec{c})^2 = |\vec{c}|^2 = 49$. To compute $(\vec{a} + \vec{b})^2$, we use the expansion: \[ (\vec{a} + \vec{b})^2 = \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + 2\vec{a} \cdot \vec{b} = |\vec{a}|^2 + |\vec{b}|^2 + 2 |\vec{a}||\vec{b}| \cos\theta \] Substituting the given magnitudes: \[ (3)^2 + (5)^2 + 2(3)(5)\cos\theta = 49 \] This simplifies to: \[ 9 + 25 + 30\cos\theta = 49 \] \[ 34 + 30\cos\theta = 49 \] \[ 30\cos\theta = 15 \] \[ \cos\theta = \frac{1}{2} \] Thus, the angle $\theta$ is: \[ \theta = 60^\circ \]