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

The problem provides two conditions:

1. \( (\vec{a} + \vec{b}) \perp \vec{a} \). This implies: \[ (\vec{a} + \vec{b}) \cdot \vec{a} = 0 \] \[ \vec{a}\cdot\vec{a} + \vec{b}\cdot\vec{a} = 0 \] \[ |\vec{a}|^2 + \vec{a}\cdot\vec{b} = 0 \] \[ \vec{a}\cdot\vec{b} = -|\vec{a}|^2 \quad \text{(Equation 1)} \]

2. \( (2\vec{a} + \vec{b}) \perp \vec{b} \). This gives: \[ (2\vec{a} + \vec{b}) \cdot \vec{b} = 0 \] \[ 2(\vec{a}\cdot\vec{b}) + |\vec{b}|^2 = 0 \quad \text{(Equation 2)} \]

Substitute Equation (1) into Equation (2): \[ 2(-|\vec{a}|^2) + |\vec{b}|^2 = 0 \] \[ |\vec{b}|^2 = 2|\vec{a}|^2 \]

Taking square roots: \[ |\vec{b}| = \sqrt{2}\,|\vec{a}| \]

Hence proved.