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

Let $$\theta$$ be the angle between the unit vectors $$\hat{a}$$ and $$\hat{b}$$, so $$\hat{a}\cdot\hat{b} = \cos\theta$$, while $$\hat{a}\cdot\hat{a} = 1$$ and $$\hat{b}\cdot\hat{b} = 1$$ because each is of unit length.

The given vectors are $$\vec{c} = \hat{a} + 2\hat{b}$$ and $$\vec{d} = 5\hat{a} - 4\hat{b}$$. Since they are perpendicular, $$\vec{c}\cdot\vec{d} = 0$$.

Compute the dot product:
$$\vec{c}\cdot\vec{d} = (\hat{a} + 2\hat{b})\cdot(5\hat{a} - 4\hat{b})$$
$$= 5(\hat{a}\cdot\hat{a}) - 4(\hat{a}\cdot\hat{b}) + 10(\hat{b}\cdot\hat{a}) - 8(\hat{b}\cdot\hat{b})$$.

Substitute $$\hat{a}\cdot\hat{a}=1$$, $$\hat{b}\cdot\hat{b}=1$$ and $$\hat{a}\cdot\hat{b} = \cos\theta$$:
$$\vec{c}\cdot\vec{d} = 5(1) - 4\cos\theta + 10\cos\theta - 8(1)$$
$$= -3 + 6\cos\theta$$.

Set this equal to zero for perpendicularity:
$$-3 + 6\cos\theta = 0 \quad \Longrightarrow \quad \cos\theta = \frac{1}{2}$$.

The angle whose cosine is $$\frac{1}{2}$$ is $$\theta = \frac{\pi}{3}$$.

Hence the angle between $$\hat{a}$$ and $$\hat{b}$$ is $$\frac{\pi}{3}$$.

Option C which is: $$\frac{\pi}{3}$$