If bar{a} and bar{b} are two unit vectors suc | vedclass

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(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Since $(\bar{a}+2\bar{b})$ and $(5\bar{a}-4\bar{b})$ a

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$\frac{\pi}{3}$

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(B) Given that $\bar{a}$ and $\bar{b}$ are unit vectors,so $|\bar{a}| = 1$ and $|\bar{b}| = 1$. Since $(\bar{a}+2\bar{b})$ and $(5\bar{a}-4\bar{b})$ are perpendicular,their dot product is zero: $(\bar{a}+2\bar{b}) \cdot (5\bar{a}-4\bar{b}) = 0$ $5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{a} \cdot \bar{b}) - 8|\bar{b}|^2 = 0$ $5(1)^2 + 6(\bar{a} \cdot \bar{b}) - 8(1)^2 = 0$ $5 + 6(\bar{a} \cdot \bar{b}) - 8 = 0$ $6(\bar{a} \cdot \bar{b}) = 3$ $\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$ Since $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos 	heta$,we have: $(1)(1) \cos 	heta = \frac{1}{2}$ $\cos 	heta = \frac{1}{2}$ $	heta = \frac{\pi}{3}$

If $\bar{a}$ and $\bar{b}$ are two unit vectors such that $\bar{a}+2 \bar{b}$ and $5 \bar{a}-4 \bar{b}$ are perpendicular to each other,then the angle between $\bar{a}$ and $\bar{b}$ is

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