Key Concepts and Formulas

• Vector Cross Product Properties:
  • $(\vec{X} - \vec{Y}) \times \vec{Z} = \vec{X} \times \vec{Z} - \vec{Y} \times \vec{Z}$ (Distributive property)
  • If $\vec{A} \times \vec{B} = \vec{0}$ and $\vec{A}, \vec{B} \neq \vec{0}$, then $\vec{A}$ and $\vec{B}$ are parallel, meaning $\vec{A} = k\vec{B}$ for some scalar $k$.
• Vector Dot Product: For $\vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ and $\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$, $\vec{u} \cdot \vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3$.
• Magnitude of a Vector: For $\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$, $|\vec{v}|^2 = v_1^2 + v_2^2 + v_3^2$.
• Cross Product using Determinant: For $\vec{u} = u_1 \hat{i} + u_2 \hat{j} + u_3 \hat{k}$ and $\vec{v} = v_1 \hat{i} + v_2 \hat{j} + v_3 \hat{k}$, $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix}$

Step-by-Step Solution

Step 1: Analyze the condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ to determine the direction of $\vec{d}$. The given condition is $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$. To proceed, we rearrange the equation by moving all terms to one side: $\vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0}$ Using the distributive property of the cross product, we can factor out $\vec{d}$: $(\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$ This implies that the vector $(\vec{b} - \vec{c})$ and the vector $\vec{d}$ are parallel (or one of them is a zero vector). Since $\vec{a} \cdot \vec{d} = 4$, $\vec{d}$ cannot be the zero vector. Therefore, $\vec{d}$ must be parallel to $(\vec{b} - \vec{c})$. We can express this relationship as: $\vec{d} = \lambda (\vec{b} - \vec{c})$ where $\lambda$ is a scalar constant.

Step 2: Calculate the vector $(\vec{b} - \vec{c})$. We are given $\vec{b} = 3 \hat{i} - 3 \hat{j} + 3 \hat{k}$ and $\vec{c} = 2 \hat{i} - \hat{j} + 2 \hat{k}$. $\vec{b} - \vec{c} = (3 \hat{i} - 3 \hat{j} + 3 \hat{k}) - (2 \hat{i} - \hat{j} + 2 \hat{k})$ $= (3-2) \hat{i} + (-3 - (-1)) \hat{j} + (3-2) \hat{k}$ $= 1 \hat{i} - 2 \hat{j} + 1 \hat{k}$ $= \hat{i} - 2 \hat{j} + \hat{k}$

Step 3: Use the condition $\vec{a} \cdot \vec{d} = 4$ to find the scalar $\lambda$. From Step 1, we have $\vec{d} = \lambda (\hat{i} - 2 \hat{j} + \hat{k})$. We are given $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$. Substitute these into the dot product condition: $(\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\lambda (\hat{i} - 2 \hat{j} + \hat{k})) = 4$ We can take the scalar $\lambda$ out of the dot product: $\lambda [(\hat{i} + 2 \hat{j} + \hat{k}) \cdot (\hat{i} - 2 \hat{j} + \hat{k})] = 4$ Now, perform the dot product: $\lambda [(1)(1) + (2)(-2) + (1)(1)] = 4$ $\lambda [1 - 4 + 1] = 4$ $\lambda [-2] = 4$ Solving for $\lambda$: $\lambda = \frac{4}{-2} = -2$

Step 4: Determine the vector $\vec{d}$. Now that we have found $\lambda = -2$, we can write the complete vector $\vec{d}$: $\vec{d} = -2 (\hat{i} - 2 \hat{j} + \hat{k})$ $\vec{d} = -2 \hat{i} + 4 \hat{j} - 2 \hat{k}$

Step 5: Calculate the cross product $\vec{a} \times \vec{d}$. We have $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{d} = -2 \hat{i} + 4 \hat{j} - 2 \hat{k}$. Using the determinant method for the cross product: $\vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -2 & 4 & -2 \end{vmatrix}$ $= \hat{i} ((2)(-2) - (1)(4)) - \hat{j} ((1)(-2) - (1)(-2)) + \hat{k} ((1)(4) - (2)(-2))$ $= \hat{i} (-4 - 4) - \hat{j} (-2 - (-2)) + \hat{k} (4 - (-4))$ $= \hat{i} (-8) - \hat{j} (0) + \hat{k} (8)$ $= -8 \hat{i} + 0 \hat{j} + 8 \hat{k}$

Step 6: Calculate the magnitude squared of $\vec{a} \times \vec{d}$. The resulting vector is $\vec{a} \times \vec{d} = -8 \hat{i} + 8 \hat{k}$. The magnitude squared is: $|(\vec{a} \times \vec{d})|^2 = (-8)^2 + (0)^2 + (8)^2$ $= 64 + 0 + 64$ $= 128$

Common Mistakes & Tips

• Mistake: Assuming $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ implies $\vec{b} = \vec{c}$. This is incorrect; it implies that $\vec{d}$ is parallel to $(\vec{b} - \vec{c})$.
• Tip: Always check if the given conditions lead to a trivial case (like $\vec{d} = \vec{0}$) before proceeding. Here, $\vec{a} \cdot \vec{d} = 4$ prevents $\vec{d}$ from being zero.
• Tip: When calculating cross products, be careful with the signs and the order of subtraction in the determinant expansion.

Summary

The problem required us to first determine the unknown vector $\vec{d}$ using the given vector conditions. The condition $\vec{b} \times \vec{d} = \vec{c} \times \vec{d}$ was used to establish that $\vec{d}$ is parallel to the vector $(\vec{b} - \vec{c})$. The second condition, $\vec{a} \cdot \vec{d} = 4$, allowed us to find the scalar multiplier that defines $\vec{d}$ completely. Once $\vec{d}$ was determined, we computed the cross product $\vec{a} \times \vec{d}$ and then found the square of its magnitude.

The final answer is $\boxed{128}$.