Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

Given that $\vec{a}$ and $\vec{b}$ are unit vectors, we have:

• $|\vec{a}| = 1$
• $|\vec{b}| = 1$

The governing property for the magnitude of a vector is $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$. We use this to expand the given equation: $|\vec{a} + 2\vec{b}| = |2\vec{a} - \vec{b}|$.

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Squaring both sides of the given equation:

$$|\vec{a} + 2\vec{b}|^2 = |2\vec{a} - \vec{b}|^2$$

Expanding using the dot product property $(\vec{u} \pm \vec{v})^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2(\vec{u} \cdot \vec{v})$:

$$|\vec{a}|^2 + |2\vec{b}|^2 + 2(\vec{a} \cdot 2\vec{b}) = |2\vec{a}|^2 + |\vec{b}|^2 - 2(2\vec{a} \cdot \vec{b})$$ $$|\vec{a}|^2 + 4|\vec{b}|^2 + 4(\vec{a} \cdot \vec{b}) = 4|\vec{a}|^2 + |\vec{b}|^2 - 4(\vec{a} \cdot \vec{b})$$

Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$:

$$(1)^2 + 4(1)^2 + 4(\vec{a} \cdot \vec{b}) = 4(1)^2 + (1)^2 - 4(\vec{a} \cdot \vec{b})$$ $$1 + 4 + 4(\vec{a} \cdot \vec{b}) = 4 + 1 - 4(\vec{a} \cdot \vec{b})$$ $$5 + 4(\vec{a} \cdot \vec{b}) = 5 - 4(\vec{a} \cdot \vec{b})$$

Rearranging the terms:

$$8(\vec{a} \cdot \vec{b}) = 0 \implies \vec{a} \cdot \vec{b} = 0$$

Step 3 (Final Concluded Answer): [0.5 Mark]

Since $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = 0$ and $|\vec{a}|, |\vec{b}| \neq 0$, we have:

$$\cos\theta = 0 \implies \theta = \frac{\pi}{2}$$

The angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{2}$ (or 90°).

💡 EL10 Concept Insight: Think of this like a tug-of-war: if the total "pull" of two vectors combined is the same regardless of how you mix them, it means the vectors must be pulling in perfectly independent directions, which only happens when they are at a right angle to each other.