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

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

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$60$

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(B) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$.Since $(\vec{a} + 2\vec{b})$ and $(5\vec{a} - 4\vec{b})$ are perpendicular,their dot product is $0$.$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$$5|\vec{a}|^2 - 4\vec{a} \cdot \vec{b} + 10\vec{a} \cdot \vec{b} - 8|\vec{b}|^2 = 0$$5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$$6(\vec{a} \cdot \vec{b}) - 3 = 0$$6(\vec{a} \cdot \vec{b}) = 3$$\vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$We know that $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$.$1 \cdot 1 \cdot \cos 	heta = \frac{1}{2}$$\cos 	heta = \frac{1}{2}$$	heta = 60^\circ$.

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

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