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:

Whenever magnitudes of vector sums or differences are involved, squaring is the most common first step. It transforms the absolute magnitude into the dot product, allowing for algebraic manipulation.

Answer 1

Concept:
• For unit vectors, \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\).
• Use the identity \(|\vec{u}|^2 = \vec{u} \cdot \vec{u}\).
• Expansion of dot product: \((\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}\).

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

Step 2: Expand using dot product properties
LHS expansion: \[ |\vec{a}|^2 + |2\vec{b}|^2 + 2(\vec{a} \cdot 2\vec{b}) = |\vec{a}|^2 + 4|\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) \] RHS expansion: \[ |2\vec{a}|^2 + |\vec{b}|^2 - 2(2\vec{a} \cdot \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b}) \]

Step 3: Substitute unit vector magnitudes
Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\): \[ (1)^2 + 4(1)^2 + 4(\vec{a} \cdot \vec{b}) = 4(1)^2 + (1)^2 - 4(\vec{a} \cdot \vec{b}) \] \[ 1 + 4 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 - 4(\vec{a} \cdot \vec{b}) \] \[ 5 + 4(\vec{a} \cdot \vec{b}) = 5 - 4(\vec{a} \cdot \vec{b}) \]

Step 4: Solve for the angle
Cancel the \(5\) from both sides: \[ 4(\vec{a} \cdot \vec{b}) = -4(\vec{a} \cdot \vec{b}) \] \[ 8(\vec{a} \cdot \vec{b}) = 0 \] \[ \vec{a} \cdot \vec{b} = 0 \] Since \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\): \[ (1)(1) \cos \theta = 0 \implies \cos \theta = 0 \] The angle \(\theta\) is \(\frac{\pi}{2}\) or \(90^\circ\).