Key Concepts and Formulas

• Vector Perpendicularity and Cross Product: A vector perpendicular to two given vectors is parallel to their cross product. If $\vec{d}$ is perpendicular to $\vec{b}$ and $\vec{c}$, then $\vec{d} = \lambda(\vec{b} \times \vec{c})$ for some scalar $\lambda$.
• Dot Product: The dot product of two vectors $\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ and $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + u_3v_3$.
• Magnitude of a Vector: The magnitude of a vector $\vec{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ is $|\vec{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}$.
• Lagrange's Identity: $|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$. This identity is derived from $|\vec{a} \times \vec{d}| = |\vec{a}||\vec{d}|\sin\theta$ and $\vec{a} \cdot \vec{d} = |\vec{a}||\vec{d}|\cos\theta$.

Step-by-Step Solution

Step 1: Determine the direction of $\vec{d}$ We are given that $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$. This implies that $\vec{d}$ must be parallel to the cross product of $\vec{b}$ and $\vec{c}$. Therefore, we can write $\vec{d}$ as: $\vec{d} = \lambda(\vec{b} \times \vec{c})$ where $\lambda$ is a scalar.

First, we compute the cross product $\vec{b} \times \vec{c}$:

$$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3 \end{vmatrix}$$

Expanding the determinant:

$$\begin{aligned} \vec{b} \times \vec{c} &= \hat{i}((-2)(3) - (-2)(4)) - \hat{j}((1)(3) - (-2)(-1)) + \hat{k}((1)(4) - (-2)(-1)) \\ &= \hat{i}(-6 + 8) - \hat{j}(3 - 2) + \hat{k}(4 - 2) \\ &= 2\hat{i} - \hat{j} + 2\hat{k} \end{aligned}$$

So, $\vec{d} = \lambda(2\hat{i} - \hat{j} + 2\hat{k})$.

Step 2: Find the scalar $\lambda$ using the dot product condition We are given that $\vec{a} \cdot \vec{d} = 18$. Substituting the expression for $\vec{d}$: $(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot \lambda(2\hat{i} - \hat{j} + 2\hat{k}) = 18$ $\lambda[(2)(2) + (3)(-1) + (4)(2)] = 18$ $\lambda[4 - 3 + 8] = 18$ $\lambda[9] = 18$ Solving for $\lambda$: $\lambda = \frac{18}{9} = 2$ Now we have the complete vector $\vec{d}$: $\vec{d} = 2(2\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} - 2\hat{j} + 4\hat{k}$

Step 3: Calculate $|\vec{a} \times \vec{d}|^2$ using Lagrange's Identity Lagrange's identity provides a direct way to compute $|\vec{a} \times \vec{d}|^2$: $|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$ We are given $\vec{a} \cdot \vec{d} = 18$. We need to find $|\vec{a}|^2$ and $|\vec{d}|^2$.

Calculate $|\vec{a}|^2$: $|\vec{a}|^2 = (2)^2 + (3)^2 + (4)^2 = 4 + 9 + 16 = 29$

Calculate $|\vec{d}|^2$: $|\vec{d}|^2 = (4)^2 + (-2)^2 + (4)^2 = 16 + 4 + 16 = 36$

Now, substitute these values into Lagrange's identity: $|\vec{a} \times \vec{d}|^2 = (29)(36) - (18)^2$ $|\vec{a} \times \vec{d}|^2 = 1044 - 324$ $|\vec{a} \times \vec{d}|^2 = 720$

Common Mistakes & Tips

• Sign Errors in Cross Product: Be meticulous with signs when calculating the determinant for the cross product, especially for the $\hat{j}$ component.
• Direct Calculation vs. Identity: While calculating $\vec{a} \times \vec{d}$ and then its magnitude is possible, it is more computationally intensive and prone to errors than using Lagrange's identity, which directly utilizes the given dot product and calculated magnitudes.
• Magnitude Calculation: Ensure you are squaring the magnitude of the vectors and not the vectors themselves, and that you are taking the square root correctly when calculating magnitudes if needed.

Summary

The problem requires finding the magnitude squared of the cross product of vectors $\vec{a}$ and $\vec{d}$. We first established that $\vec{d}$ is parallel to $\vec{b} \times \vec{c}$ and used the given dot product condition $\vec{a} \cdot \vec{d} = 18$ to find the scalar multiplier for $\vec{d}$. Finally, we efficiently calculated $|\vec{a} \times \vec{d}|^2$ using Lagrange's identity, which relates the squared magnitude of a cross product to the product of the squared magnitudes of the vectors and the square of their dot product.

The final answer is $\boxed{720}$.