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,we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Since $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d}

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$\frac{\pi}{3}$

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(C) Given that $\vec{a}$ and $\vec{b}$ are unit vectors,we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$. Since $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ are perpendicular,their dot product is zero: $\vec{c} \cdot \vec{d} = 0$. $(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$. Expanding the dot product: $5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$. Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$,and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$: $5(1) + 6(\vec{a} \cdot \vec{b}) - 8(1) = 0$. $6(\vec{a} \cdot \vec{b}) - 3 = 0$. $6(\vec{a} \cdot \vec{b}) = 3$. $\vec{a} \cdot \vec{b} = \frac{3}{6} = \frac{1}{2}$. We know $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos 	heta$,where $	heta$ is the angle between $\vec{a}$ and $\vec{b}$. $1 \cdot 1 \cdot \cos 	heta = \frac{1}{2}$. $\cos 	heta = \frac{1}{2} \implies 	heta = \frac{\pi}{3}$.

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the vectors $\vec{c} = \vec{a} + 2\vec{b}$ 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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