1. Key Concepts and Formulas

• Properties of the Cross Product:
  • Distributive Property: $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \times \vec{c}$ and $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$.
  • Scalar Multiplication: $(k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b})$ and $\vec{a} \times (k\vec{b}) = k(\vec{a} \times \vec{b})$.
  • Anti-commutativity: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$.
  • Cross Product of Parallel Vectors: $\vec{a} \times \vec{a} = \vec{0}$.
• Magnitude of the Cross Product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
• Dot Product of Perpendicular Vectors: If $\vec{u}$ and $\vec{v}$ are perpendicular, then $\vec{u} \cdot \vec{v} = 0$.
• Magnitude of a Vector: $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.

2. Step-by-Step Solution

Step 1: Expand the cross product. We are asked to find the magnitude of the cross product $(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})$. We will use the distributive property of the cross product to expand this expression. $(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b}) = \vec{a} \times (2 \vec{a}) + \vec{a} \times (-3 \vec{b}) + (2 \vec{b}) \times (2 \vec{a}) + (2 \vec{b}) \times (-3 \vec{b})$

Step 2: Apply scalar multiplication and anti-commutativity. Now we simplify the expanded terms using the properties of scalar multiplication and the anti-commutative property of the cross product. $= 2(\vec{a} \times \vec{a}) - 3(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{a}) - 6(\vec{b} \times \vec{b})$ Using $\vec{a} \times \vec{a} = \vec{0}$, $\vec{b} \times \vec{b} = \vec{0}$, and $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$: $= 2(\vec{0}) - 3(\vec{a} \times \vec{b}) + 4(-(\vec{a} \times \vec{b})) - 6(\vec{0})$ $= \vec{0} - 3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{b}) - \vec{0}$ $= -7(\vec{a} \times \vec{b})$

Step 3: Calculate the magnitude of the resulting vector. We need to find the magnitude of $-7(\vec{a} \times \vec{b})$. $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})| = |-7(\vec{a} \times \vec{b})|$ Using the property $|k\vec{v}| = |k| |\vec{v}|$: $= |-7| |\vec{a} \times \vec{b}|$ $= 7 |\vec{a} \times \vec{b}|$

Step 4: Calculate the magnitude of $\vec{a} \times \vec{b}$. We are given $|\vec{a}|=2$, $|\vec{b}|=3$, and the angle between them is $\theta = \frac{\pi}{4}$. We use the formula for the magnitude of the cross product: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$. $|\vec{a} \times \vec{b}| = (2)(3) \sin\left(\frac{\pi}{4}\right)$ $= 6 \times \frac{\sqrt{2}}{2}$ $= 3\sqrt{2}$

Step 5: Substitute the value back into the magnitude expression. Now we substitute the value of $|\vec{a} \times \vec{b}|$ back into the expression from Step 3. $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})| = 7 |\vec{a} \times \vec{b}|$ $= 7 (3\sqrt{2})$ $= 21\sqrt{2}$

Step 6: Calculate the square of the magnitude. The question asks for the square of the magnitude: $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2$. $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^2 = (21\sqrt{2})^2$ $= (21)^2 \times (\sqrt{2})^2$ $= 441 \times 2$ $= 882$

3. Common Mistakes & Tips

• Forgetting the Anti-commutative Property: Incorrectly assuming $\vec{a} \times \vec{b} = \vec{b} \times \vec{a}$ will lead to an incorrect cancellation of terms. Always remember $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$.
• Errors in Scalar Multiplication: Be careful when distributing scalars. For example, $(2\vec{b}) \times (2\vec{a}) = 4(\vec{b} \times \vec{a})$.
• Squaring Errors: Ensure you correctly square the entire result, including any square roots. $(21\sqrt{2})^2 = 21^2 \times (\sqrt{2})^2$, not $21 \times 2$.

4. Summary

The problem requires us to compute the magnitude of the cross product of two linear combinations of vectors $\vec{a}$ and $\vec{b}$. We first expanded the cross product using its distributive and anti-commutative properties, simplifying it to a scalar multiple of $(\vec{a} \times \vec{b})$. We then calculated the magnitude of this resulting vector by finding the magnitude of $(\vec{a} \times \vec{b})$ using the given magnitudes of $\vec{a}$ and $\vec{b}$ and the angle between them. Finally, we squared this magnitude to obtain the required answer.

The final answer is \boxed{882}, which corresponds to option (D).