1. Key Concepts and Formulas

• Dot Product: For two vectors $\vec{a}$ and $\vec{b}$, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$, where $\theta$ is the angle between them.
• Magnitude of a Vector: For $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$, $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$, so $|\vec{v}|^2 = v_x^2 + v_y^2 + v_z^2$.
• Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$. This identity is crucial for relating the cross product magnitude to the dot product and vector magnitudes without explicit cross product calculation.

1. Step-by-Step Solution

Step 1: Find the magnitude of vector $\vec{a}$. We are given $\vec{a} \cdot \vec{b} = 3\sqrt{2}$, $|\vec{b}|^2 = 6$, and the angle $\theta = \frac{\pi}{4}$. We use the dot product formula to find $|\vec{a}|$. $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ Substituting the given values: $3\sqrt{2} = |\vec{a}|(\sqrt{6})\cos\left(\frac{\pi}{4}\right)$ Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$ and $\sqrt{6} = \sqrt{3}\sqrt{2}$: $3\sqrt{2} = |\vec{a}|(\sqrt{3}\sqrt{2})\left(\frac{1}{\sqrt{2}}\right)$ $3\sqrt{2} = |\vec{a}|\sqrt{3}$ Solving for $|\vec{a}|$: $|\vec{a}| = \frac{3\sqrt{2}}{\sqrt{3}} = \frac{3\sqrt{6}}{3} = \sqrt{6}$ Therefore, $|\vec{a}|^2 = (\sqrt{6})^2 = 6$.

Step 2: Calculate $\alpha^2 + \beta^2$. The vector $\vec{a}$ is given as $\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$. The square of its magnitude is the sum of the squares of its components. $|\vec{a}|^2 = (1)^2 + \alpha^2 + \beta^2$ Using the result from Step 1, $|\vec{a}|^2 = 6$: $6 = 1 + \alpha^2 + \beta^2$ Rearranging to find $\alpha^2 + \beta^2$: $\alpha^2 + \beta^2 = 6 - 1 = 5$

Step 3: Calculate $|\vec{a} \times \vec{b}|^2$. We use Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$. We have:

• $|\vec{a}|^2 = 6$ (from Step 1)
• $|\vec{b}|^2 = 6$ (given)
• $\vec{a} \cdot \vec{b} = 3\sqrt{2}$ (given), so $(\vec{a} \cdot \vec{b})^2 = (3\sqrt{2})^2 = 9 \times 2 = 18$.

Substitute these values into Lagrange's Identity: $|\vec{a} \times \vec{b}|^2 + 18 = (6)(6)$ $|\vec{a} \times \vec{b}|^2 + 18 = 36$ Solving for $|\vec{a} \times \vec{b}|^2$: $|\vec{a} \times \vec{b}|^2 = 36 - 18 = 18$

Step 4: Compute the final expression $(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2$. We need to multiply the results from Step 2 and Step 3. From Step 2: $\alpha^2 + \beta^2 = 5$. From Step 3: $|\vec{a} \times \vec{b}|^2 = 18$.

$(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2 = (5)(18) = 90$

1. Common Mistakes & Tips

• Lagrange's Identity: This identity is extremely useful. Memorize it and recognize when to apply it, as it bypasses the need to compute the cross product explicitly.
• Magnitude vs. Magnitude Squared: Be careful whether you are working with $|\vec{v}|$ or $|\vec{v}|^2$. The problem provides $|\vec{b}|^2$, and it's often easier to work with squared magnitudes to avoid square roots.
• Trigonometric Values: Ensure accurate recall of $\cos(\pi/4)$ and $\sin(\pi/4)$, which are both $1/\sqrt{2}$.

1. Summary The problem requires us to find the value of $(\alpha^2+\beta^2)|\vec{a} \times \vec{b}|^2$. We first determined the magnitude of $\vec{a}$ using the dot product and the given information. This allowed us to find $\alpha^2 + \beta^2$ from the components of $\vec{a}$. Then, we used Lagrange's Identity, which relates the dot product, cross product magnitude, and vector magnitudes, to efficiently calculate $|\vec{a} \times \vec{b}|^2$. Finally, we multiplied the two calculated values to obtain the answer.

The final answer is $\boxed{90}$.