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. \( 180^\circ \)

Hint:

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.

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:

\[180^\circ.\]

Answer 2

Given:

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

Step 1: Find the magnitude of \( \vec{c} \):

The magnitude of \( \vec{c} \) is given by: \[ |\vec{c}|^2 = |\vec{a} + \vec{b}|^2 \] Expanding using the vector magnitude formula: \[ |\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2 \vec{a} \cdot \vec{b} \]

Step 2: Substitute the known values:

We are given that \( |\vec{a}| = |\vec{b}| = 1 \) and \( |\vec{c}| = 2 \). Substituting these values into the equation: \[ 2^2 = 1^2 + 1^2 + 2 (\vec{a} \cdot \vec{b}) \] Simplifying: \[ 4 = 2 + 2 (\vec{a} \cdot \vec{b}) \]

Step 3: Solve for \( \vec{a} \cdot \vec{b} \):

Rearranging the equation: \[ 2 (\vec{a} \cdot \vec{b}) = 2 \Rightarrow \vec{a} \cdot \vec{b} = 0 \] This means that \( \vec{a} \) and \( \vec{b} \) are perpendicular to each other.

Step 4: Analyze the geometry of the vectors:

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

Step 5: Find the magnitude of \( \vec{a} + \vec{b} \):

The magnitude of \( \vec{a} + \vec{b} \) is: \[ |\vec{a} + \vec{b}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \]

Step 6: Magnitude of \( \vec{c} \):

Since \( \vec{c} = -(\vec{a} + \vec{b}) \), its magnitude is: \[ |\vec{c}| = \sqrt{2} \] and its direction is opposite to \( \vec{a} + \vec{b} \).

Step 7: Find the angle between \( \vec{b} \) and \( \vec{c} \):

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

Conclusion:

Thus, the angle between \( \vec{b} \) and \( \vec{c} \) is: \[ 180^\circ. \]