If $\vec{a}, \vec{b}$ and $\vec{c}$ are three vectors such that $\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}$, where $\vec{a}$ and $\vec{b}$ are unit vectors and $|\vec{c}|=2$, then the angle between the vectors $\vec{b}$ and $\vec{c}$ is :

Given:

• $\vec{a} + \vec{b} + \vec{c} = \vec{0}$
  • $|\vec{a}| = |\vec{b}| = 1$ (unit vectors)
  • $|\vec{c}| = 2$

Rearranging the equation: $\vec{c} = -\vec{a} - \vec{b}$

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The angle between vectors can be found using the dot product formula:

$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$

Where $\theta$ is the angle between the vectors.

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Calculating $\vec{b} \cdot \vec{c}$:

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

Since $\vec{b}$ is a unit vector: $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$

So: $\vec{b} \cdot \vec{c} = -\vec{b} \cdot \vec{a} - 1$

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Using the magnitude of $\vec{c}$:

$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (-\vec{a} - \vec{b}) \cdot (-\vec{a} - \vec{b})$

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

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

$= 1 + 2\vec{a} \cdot \vec{b} + 1$

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

Since $|\vec{c}|^2 = 4$:

$4 = 2 + 2\vec{a} \cdot \vec{b}$

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

$\vec{a} \cdot \vec{b} = 1$

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Calculating final dot product:

$\vec{b} \cdot \vec{c} = -\vec{a} \cdot \vec{b} - 1 = -1 - 1 = -2$

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Finding the angle:

$\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta$

$-2 = 1 \times 2 \times \cos\theta$

$-2 = 2\cos\theta$

$\cos\theta = -1$

When $\cos\theta = -1$, $\theta = 180°$ (or $\pi$ radians)

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Therefore, the angle between vectors $\vec{b}$ and $\vec{c}$ is $180°$.