Question

If \( \vec{a} \) and \( \vec{b} \) are two non-zero vectors such that \( (\vec{a} + \vec{b}) \perp \vec{a \) and \( (2\vec{a} + \vec{b}) \perp \vec{b} \), then prove that \( |\vec{b}| = \sqrt{2} |\vec{a}| \).}

Hint:

To prove vector relationships, use the perpendicularity condition \( \vec{u} \cdot \vec{v} = 0 \), and simplify dot products by expanding and substituting given constraints.

Answer 1

Given: 1. \( (\vec{a} + \vec{b}) \perp \vec{a} \), which implies: \[ (\vec{a} + \vec{b}) \cdot \vec{a} = 0. \] Expanding the dot product: \[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} = 0. \] Using \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 \), we get: \[ |\vec{a}|^2 + \vec{b} \cdot \vec{a} = 0, \] or \[ \vec{b} \cdot \vec{a} = -|\vec{a}|^2. \tag{1} \] 2. \( (2\vec{a} + \vec{b}) \perp \vec{b} \), which implies: \[ (2\vec{a} + \vec{b}) \cdot \vec{b} = 0. \] Expanding the dot product: \[ 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} = 0. \] Using \( \vec{b} \cdot \vec{b} = |\vec{b}|^2 \), we get: \[ 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 0. \tag{2} \] From equation (1), substitute \( \vec{a} \cdot \vec{b} = -|\vec{a}|^2 \) into equation (2): \[ 2(-|\vec{a}|^2) + |\vec{b}|^2 = 0. \] Simplify: \[ -2|\vec{a}|^2 + |\vec{b}|^2 = 0, \] or \[ |\vec{b}|^2 = 2|\vec{a}|^2. \] Taking the square root on both sides: \[ |\vec{b}| = \sqrt{2} |\vec{a}|. \] Hence, it is proved that \( |\vec{b}| = \sqrt{2} |\vec{a}| \).