Question

If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are three vectors such that \( \vec{a} + \vec{b} + \vec{c} = 0 \), 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:

• A. \( 60^\circ \)
• B. \( 90^\circ \)
• C. \( 120^\circ \)
• D. \( 180^\circ \)

Answer 1

The Correct Option is D

Given:

\[\vec{a} + \vec{b} + \vec{c} = \vec{0} \Rightarrow \vec{c} = -(\vec{a} + \vec{b})\]

The magnitude of \(\vec{c}\) is:

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

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:

\[4 = 2 + 2(\vec{a} \cdot \vec{b})\]

Solve for \(\vec{a} \cdot \vec{b}\):

\[2(\vec{a} \cdot \vec{b}) = 2 \Rightarrow \vec{a} \cdot \vec{b} = 0\]

This means \(\vec{a}\) and \(\vec{b}\) are perpendicular.

From the equation \(\vec{c} = -(\vec{a} + \vec{b})\):

• Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, their resultant \(\vec{a} + \vec{b}\) forms a diagonal of a square with side length 1.
• The magnitude of \(\vec{a} + \vec{b}\) is:

\[|\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \sqrt{1^2 + 1^2} = \sqrt{2}\]

Thus:

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

and its direction is opposite to \(\vec{a} + \vec{b}\).

Since \(\vec{c}\) is opposite to \(\vec{a} + \vec{b}\), and \(\vec{b}\) contributes to \(\vec{c}\), the angle between \(\vec{b}\) and \(\vec{c}\) is

\[\theta = 180^\circ.\]

Thus:

\[180^\circ.\]