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

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(A) 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 $(5 \bar{a}-4 \bar{b})$

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$\frac{\pi}{3}$

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(A) 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$$5|\bar{a}|^2 - 4(\bar{a} \cdot \bar{b}) + 10(\bar{b} \cdot \bar{a}) - 8|\bar{b}|^2 = 0$Substituting $|\bar{a}|^2 = 1$ and $|\bar{b}|^2 = 1$:$5(1) + 6(\bar{a} \cdot \bar{b}) - 8(1) = 0$$6(\bar{a} \cdot \bar{b}) - 3 = 0$$\bar{a} \cdot \bar{b} = \frac{3}{6} = \frac{1}{2}$Since $\bar{a} \cdot \bar{b} = |\bar{a}||\bar{b}| \cos 	heta$,we have:$1 \cdot 1 \cdot \cos 	heta = \frac{1}{2}$$\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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