Let vec{a} and vec{b} be two unit vectors. If | vedclass

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(C) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Since $\vec{c}$ and $\vec{d}$ are perpendicular,their

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$\frac{\pi}{3}$

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(C) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,so $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Since $\vec{c}$ and $\vec{d}$ are perpendicular,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$. Substituting the expressions for $\vec{c}$ and $\vec{d}$: $(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$. Expanding the dot product: $5|\vec{a}|^2 - 4(\vec{a} \cdot \vec{b}) + 10(\vec{a} \cdot \vec{b}) - 8|\vec{b}|^2 = 0$. Substituting $|\vec{a}| = 1$ and $|\vec{b}| = 1$: $5(1)^2 + 6(\vec{a} \cdot \vec{b}) - 8(1)^2 = 0$. $5 + 6(\vec{a} \cdot \vec{b}) - 8 = 0 \Rightarrow 6(\vec{a} \cdot \vec{b}) = 3 \Rightarrow \vec{a} \cdot \vec{b} = \frac{1}{2}$. Using the definition of the dot product: $|\vec{a}||\vec{b}| \cos 	heta = \frac{1}{2}$. $(1)(1) \cos 	heta = \frac{1}{2} \Rightarrow \cos 	heta = \frac{1}{2}$. Therefore,$	heta = \frac{\pi}{3}$.

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular to each other,then the angle between $\vec{a}$ and $\vec{b}$ is

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