Question:

Step 1: Apply the distributive property of the cross product. We are asked to find the value of $|\vec{a} \times \vec{b} - \vec{a} \times \vec{c}|$. Using the distributive property $\vec{u} \times \vec{v} - \vec{u} \times \vec{w} = \vec{u} \times (\vec{v} - \vec{w})$, we can simplify the expression: $$|\vec{a} \times \vec{b} - \vec{a} \times \vec{c}| = |\vec{a} \times (\vec{b} - \vec{c})|$$ Step 2: Simplify the magnitude using vector properties. We know that $|\vec{x} \times \vec{y}| = |\vec{x}| |\vec{y}| \sin\theta$, where $\theta$ is the angle between $\vec{x}$ and $\vec{y}$. Given that $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$, it implies that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{a}$ is perpendicular to $\vec{c}$. If a vector is perpendicular to two vectors, it is also perpendicular to their linear combination, specifically their difference. Therefore, $\vec{a}$ is perpendicular to $(\vec{b} - \vec{c})$. So, the angle between $\vec{a}$ and $(\vec{b} - \vec{c})$ is $90^\circ$, which means $\sin(90^\circ) = 1$. Also, $\vec{a}$ is a unit vector, so $|\vec{a}| = 1$. Substituting these values: $$|\vec{a} \times (\vec{b} - \vec{c})| = |\vec{a}| |\vec{b} - \vec{c}| \sin(90^\circ) = (1) |\vec{b} - \vec{c}| (1) = |\vec{b} - \vec{c}|$$ Step 3: Calculate the square of the magnitude of $(\vec{b} - \vec{c})$. We are given that $\vec{b}$ and $\vec{c}$ are unit vectors, so $|\vec{b}| = 1$ and $|\vec{c}| = 1$. The angle between $\vec{b}$ and $\vec{c}$ is $\pi/3$. We can find $|\vec{b} - \vec{c}|^2$ using the dot product property $|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$: $$|\vec{b} - \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 - 2(\vec{b} \cdot \vec{c})$$ Using the definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos\theta$: $$|\vec{b} - \vec{c}|^2 = (1)^2 + (1)^2 - 2(1)(1)\cos(\pi/3)$$ $$|\vec{b} - \vec{c}|^2 = 1 + 1 - 2(1/2)$$ $$|\vec{b} - \vec{c}|^2 = 2 - 1$$ $$|\vec{b} - \vec{c}|^2 = 1$$ Step 4: Find the magnitude of $(\vec{b} - \vec{c})$ and state the final answer. From the previous step, we have $|\vec{b} - \vec{c}|^2 = 1$. Taking the square root: $$|\vec{b} - \vec{c}| = \sqrt{1} = 1$$ Since we established in Step 2 that $|\vec{a} \times \vec{b} - \vec{a} \times \vec{c}| = |\vec{b} - \vec{c}|$, the value of the expression is $1$. The final answer is $\boxed{1}$.