The non-zero vectors $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ are related by $$\vec{a} = 8\vec{b}$$ and $$\vec{c} = -7\vec{b}$$. Then the angle between $$\vec{a}$$ and $$\vec{c}$$ is

Given $$\vec{a} = 8\vec{b}$$ and $$\vec{c} = -7\vec{b}$$, both $$\vec{a}$$ and $$\vec{c}$$ are scalar multiples of the same non-zero vector $$\vec{b}$$.

We use the dot-product formula for the angle $$\theta$$ between two vectors:
$$\vec{a}\cdot\vec{c} = |\vec{a}|\,|\vec{c}| \cos\theta$$

Substituting $$\vec{a} = 8\vec{b}$$ and $$\vec{c} = -7\vec{b}$$:
$$\vec{a}\cdot\vec{c} = (8\vec{b})\cdot(-7\vec{b}) = -56\,(\vec{b}\cdot\vec{b}) = -56\,|\vec{b}|^{2}$$

The magnitudes are
$$|\vec{a}| = |8\vec{b}| = 8|\vec{b}|,\qquad |\vec{c}| = |-7\vec{b}| = 7|\vec{b}|$$

Thus
$$|\vec{a}|\,|\vec{c}| \cos\theta = (8|\vec{b}|)(7|\vec{b}|)\cos\theta = 56\,|\vec{b}|^{2}\cos\theta$$

Equating the two expressions for $$\vec{a}\cdot\vec{c}$$:
$$-56\,|\vec{b}|^{2} = 56\,|\vec{b}|^{2}\cos\theta$$
$$\Rightarrow \cos\theta = -1$$

The only angle in $$0 \le \theta \le \pi$$ with $$\cos\theta = -1$$ is $$\theta = \pi$$ (180°). Hence, $$\vec{a}$$ and $$\vec{c}$$ are collinear but point in opposite directions.

Option D which is: $$\pi$$