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(D) Given $|\bar{a}| = |\bar{b}|$. Let $|\bar{a}| = |\bar{b}| = k$. Squaring the given equation $|\bar{a} + 2\bar{b}| = |2\bar{a} - \bar{b}|$, we get:

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$90$

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(D) Given $|\bar{a}| = |\bar{b}|$. Let $|\bar{a}| = |\bar{b}| = k$. Squaring the given equation $|\bar{a} + 2\bar{b}| = |2\bar{a} - \bar{b}|$, we get: $|\bar{a} + 2\bar{b}|^2 = |2\bar{a} - \bar{b}|^2$ $(\bar{a} + 2\bar{b}) \cdot (\bar{a} + 2\bar{b}) = (2\bar{a} - \bar{b}) \cdot (2\bar{a} - \bar{b})$ $|\bar{a}|^2 + 4|\bar{b}|^2 + 4(\bar{a} \cdot \bar{b}) = 4|\bar{a}|^2 + |\bar{b}|^2 - 4(\bar{a} \cdot \bar{b})$ Since $|\bar{a}| = |\bar{b}| = k$, we substitute: $k^2 + 4k^2 + 4(\bar{a} \cdot \bar{b}) = 4k^2 + k^2 - 4(\bar{a} \cdot \bar{b})$ $5k^2 + 4(\bar{a} \cdot \bar{b}) = 5k^2 - 4(\bar{a} \cdot \bar{b})$ $8(\bar{a} \cdot \bar{b}) = 0$ $\bar{a} \cdot \bar{b} = 0$ This implies that $\bar{a}$ is perpendicular to $\bar{b}$. Since $\bar{c}$ is parallel to $\bar{a}$, the angle between $\bar{b}$ and $\bar{c}$ is the same as the angle between $\bar{b}$ and $\bar{a}$, which is $90^{\circ}$.

Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=|\bar{b}|$ and $|\bar{a}+2 \bar{b}|=|2 \bar{a}-\bar{b}|$. If $\bar{c}$ is a vector parallel to $\bar{a}$, then the angle between $\bar{b}$ and $\bar{c}$ is (in $^{\circ}$)

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