Magnitude of Vector Combination Given Pairwise Vector Distances

To find the magnitude $|2\vec{a} + 5\vec{b} + 5\vec{c}|$, we start by using the properties of the given unit vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Since $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, we have: $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$

We expand the given condition: $|\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9$

Expanding each term using the scalar product: $|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 1 + 1 - 2(\vec{a} \cdot \vec{b}) = 2 - 2(\vec{a} \cdot \vec{b})$ $|\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c}) = 1 + 1 - 2(\vec{b} \cdot \vec{c}) = 2 - 2(\vec{b} \cdot \vec{c})$ $|\vec{c} - \vec{a}|^2 = |\vec{c}|^2 + |\vec{a}|^2 - 2(\vec{c} \cdot \vec{a}) = 1 + 1 - 2(\vec{c} \cdot \vec{a}) = 2 - 2(\vec{c} \cdot \vec{a})$

Summing these equations: $(2 - 2(\vec{a} \cdot \vec{b})) + (2 - 2(\vec{b} \cdot \vec{c})) + (2 - 2(\vec{c} \cdot \vec{a})) = 9$ $6 - 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 9$ $2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 6 - 9 = -3$

Now, consider the square of the magnitude of $\vec{a} + \vec{b} + \vec{c}$: $|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a})$ Substituting the values: $|\vec{a} + \vec{b} + \vec{c}|^2 = 1 + 1 + 1 + (-3) = 0$

Since the squared magnitude of a real vector is zero if and only if the vector itself is the zero vector: $\vec{a} + \vec{b} + \vec{c} = \vec{0}$

This implies: $\vec{b} + \vec{c} = -\vec{a}$

Now, substituting $\vec{b} + \vec{c} = -\vec{a}$ into the required expression: $2\vec{a} + 5\vec{b} + 5\vec{c} = 2\vec{a} + 5(\vec{b} + \vec{c}) = 2\vec{a} + 5(-\vec{a}) = -3\vec{a}$

Taking the magnitude: $|2\vec{a} + 5\vec{b} + 5\vec{c}| = |-3\vec{a}| = 3|\vec{a}| = 3(1) = 3$