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:

Use dot product properties and given perpendicularity conditions to derive relationships between the magnitudes of vectors.

Answer 1

1. Condition 1: \( (\vec{a} + \vec{b}) \perp \vec{a} \): \[ (\vec{a} + \vec{b}) \cdot \vec{a} = 0 \quad \Rightarrow \quad \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} = 0. \] \[ |\vec{a}|^2 + \vec{a} \cdot \vec{b} = 0 \quad \Rightarrow \quad \vec{a} \cdot \vec{b} = -|\vec{a}|^2. \] 2. Condition 2: \( (2\vec{a} + \vec{b}) \perp \vec{b} \): \[ (2\vec{a} + \vec{b}) \cdot \vec{b} = 0 \quad \Rightarrow \quad 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 0. \] Substitute \( \vec{a} \cdot \vec{b} = -|\vec{a}|^2 \): \[ 2(-|\vec{a}|^2) + |\vec{b}|^2 = 0 \quad \Rightarrow \quad -2|\vec{a}|^2 + |\vec{b}|^2 = 0. \] Rearrange: \[ |\vec{b}|^2 = 2|\vec{a}|^2 \quad \Rightarrow \quad |\vec{b}| = \sqrt{2} |\vec{a}|. \] Final Answer: \[ \boxed{|\vec{b}| = \sqrt{2} |\vec{a}|.} \]