If bar{a} and bar{b} are unit vectors such th | vedclass

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(B) Given that $(\bar{a} + 2\bar{b})$ and $(5\bar{a} - 4\bar{b})$ are perpendicular,their dot product is zero:$(\bar{a} + 2\bar{b}) \cdot (5\bar{a} -

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$60$

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(B) Given that $(\bar{a} + 2\bar{b})$ and $(5\bar{a} - 4\bar{b})$ are perpendicular,their dot product is zero:$(\bar{a} + 2\bar{b}) \cdot (5\bar{a} - 4\bar{b}) = 0$Expanding the dot product:$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{a} \cdot \bar{b}) - 8|\bar{b}|^2 = 0$Since $\bar{a}$ and $\bar{b}$ are unit vectors,$|\bar{a}| = 1$ and $|\bar{b}| = 1$:$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}$Let $	heta$ be the angle between $\bar{a}$ and $\bar{b}$. Then $\cos 	heta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}||\bar{b}|} = \frac{1/2}{1 	imes 1} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}\left(\frac{1}{2}ight) = 60^o$.

If $\bar{a}$ and $\bar{b}$ are 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 .....$^o$.

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