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

Answer: (D)

$180^\circ$

Answer 2

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:

$180^\circ.$