When working with vectors, remember that the dot product \( \vec{a} \cdot \vec{b} = 0 \) implies that the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular. This property is useful in geometry and physics problems involving right-angled vectors. Also, when two vectors are opposite to each other, the angle between them is \( 180^\circ \), as they point in completely opposite directions.
Provided:
\[\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{c} = -(\vec{a} + \vec{b})\]
Calculate the magnitude of \(\vec{c}\):
\[|\vec{c}|^2 = |\vec{a} + \vec{b}|^2\]
Expand using the vector magnitude formula:
\[|\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}\]
Substitute \(|\vec{a}| = |\vec{b}| = 1\) and \(|\vec{c}| = 2\):
\[2^2 = 1 + 1 + 2(\vec{a} \cdot \vec{b})\]
Simplify the equation:
\[4 = 2 + 2(\vec{a} \cdot \vec{b})\]
Solve for the dot product \(\vec{a} \cdot \vec{b}\):
\[2(\vec{a} \cdot \vec{b}) = 2 \Rightarrow \vec{a} \cdot \vec{b} = 0\]
This implies \(\vec{a}\) and \(\vec{b}\) are orthogonal.
From the relationship \(\vec{c} = -(\vec{a} + \vec{b})\):
\[|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\]
Therefore:
\[\vec{c} = -(\vec{a} + \vec{b})\]
The direction of \(\vec{c}\) is opposite to that of \(\vec{a} + \vec{b}\).
Since \(\vec{c}\) is in the opposite direction to \(\vec{a} + \vec{b}\), and \(\vec{b}\) is a component contributing to \(\vec{c}\), the angle between \(\vec{b}\) and \(\vec{c}\) is:
\[\theta = 180^\circ.\]
The resulting angle is:
\[180^\circ.\]