Question

If for two unit vectors \( \vec{a} \) and \( \vec{b} \), \( |\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}| \), then find the angle between \( \vec{a} \) and \( \vec{b} \).

Hint:

Squaring vector magnitude equations is the single most common technique used to break them down into basic dot product and scalar value expansions.

Answer 1

Concept: For unit vectors, their magnitudes are equal to 1 (\( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \)). To eliminate the vector magnitude bars, we square both sides of the equation and expand using the vector identity \( |\vec{x}|^2 = \vec{x} \cdot \vec{x} \).

Step 1: Square both sides of the given equation.
Given: \[ |\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}| \] Squaring both sides: \[ |\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2 \]

Step 2: Expand using the dot product formula.
Expanding both sides: \[ |\vec{a}|^2 + 4(\vec{a} \cdot \vec{b}) + 4|\vec{b}|^2 = 4|\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \]

Step 3: Substitute the magnitudes of the unit vectors.
Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, substitute \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \): \[ 1 + 4(\vec{a} \cdot \vec{b}) + 4(1) = 4(1) - 4(\vec{a} \cdot \vec{b}) + 1 \] \[ 5 + 4(\vec{a} \cdot \vec{b}) = 5 - 4(\vec{a} \cdot \vec{b}) \]

Step 4: Solve for the dot product and the angle.
Subtract 5 from both sides: \[ 4(\vec{a} \cdot \vec{b}) = -4(\vec{a} \cdot \vec{b}) \quad \Rightarrow \quad 8(\vec{a} \cdot \vec{b}) = 0 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = 0 \] The dot product definition states \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0 \). Since \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \): \[ \cos \theta = 0 \quad \Rightarrow \quad \theta = \frac{\pi}{2} \text{ or } 90^\circ \]