To find the value of $\left\| (\vec{a} - \vec{b}) \times (\vec{a} + 5\vec{b}) \right\|$, we will use the properties of the cross product and the magnitudes of vectors.

Given:

• $\|\vec{a}\| = 1$
• $\|\vec{b}\| = 3$
• The angle between $\vec{a}$ and $\vec{b}$ is $\theta = \frac{\pi}{3}$ radians.

Steps to solve the problem:

1. Express the vectors and use the distributive property of the cross product:

   $(\vec{a} - \vec{b}) \times (\vec{a} + 5\vec{b}) = \vec{a} \times \vec{a} + 5\vec{a} \times \vec{b} - \vec{b} \times \vec{a} - 5\vec{b} \times \vec{b}$

2. Simplify each term:

   • $\vec{a} \times \vec{a} = \vec{0}$ (cross product of any vector with itself is zero)
   • $\vec{b} \times \vec{b} = \vec{0}$ (same reason as above)
   • $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ (property of cross product)

   Therefore, the expression simplifies to:

   $5\vec{a} \times \vec{b} - \vec{b} \times \vec{a} = 6\vec{a} \times \vec{b}$

3. Calculate the magnitude $\|6\vec{a} \times \vec{b}\|$:

   The magnitude of a cross product $\vec{u} \times \vec{v}$ is given by $\|\vec{u}\| \|\vec{v}\| \sin \theta$.

   Here, $\|\vec{a}\| = 1$, $\|\vec{b}\| = 3$, and $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$.

   $\|\vec{a} \times \vec{b}\| = \|\vec{a}\| \|\vec{b}\| \sin\left(\frac{\pi}{3}\right) = 1 \times 3 \times \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}$

   Therefore,

   $\|6\vec{a} \times \vec{b}\| = 6 \times \frac{3\sqrt{3}}{2} = 9\sqrt{3}$

Hence, the value of $\left\| (\vec{a} - \vec{b}) \times (\vec{a} + 5\vec{b}) \right\|$ is $\boxed{9\sqrt{3}}$.

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Would you like more details or have any questions?

Here are some related questions:

1. What is the geometrical interpretation of the cross product result?
2. How would the result change if the angle between $\vec{a}$ and $\vec{b}$ was different?
3. Can the cross product be used to determine the area of a parallelogram formed by two vectors?
4. What is the difference between the dot product and the cross product in vector analysis?
5. How do vector magnitudes affect the cross product?

Tip: The cross product results in a vector that is perpendicular to the plane formed by the original vectors, and its magnitude represents the area of the parallelogram defined by those vectors.