Key Concepts and Formulas

• Vector Cross Product Property: If $\vec{X} \times \vec{Y} = \vec{0}$, then $\vec{X}$ and $\vec{Y}$ are parallel, meaning $\vec{X} = \lambda \vec{Y}$ for some scalar $\lambda$.
• Distributive Property of Dot Product: $\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}$.
• Scalar Multiplication in Dot Product: $(\lambda \vec{A}) \cdot \vec{B} = \lambda (\vec{A} \cdot \vec{B})$.
• Magnitude of a Vector: For a vector $\vec{V} = V_x \hat{i} + V_y \hat{j} + V_z \hat{k}$, its squared magnitude is $|\vec{V}|^2 = V_x^2 + V_y^2 + V_z^2$.

Step-by-Step Solution

We are given the vectors: $\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}$ $\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}$ $\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}$

And two conditions for vector $\vec{d}$:

1. $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$
2. $\vec{d} \cdot \vec{a}=24$

Our goal is to find $|\vec{d}|^2$.

Step 1: Simplify the Cross Product Condition

The first condition is $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$. We can rearrange this equation by moving the term from the right side to the left side: $\vec{d} \times \vec{b} - \vec{c} \times \vec{b} = \vec{0}$ Using the distributive property of the cross product, we can factor out $\vec{b}$: $(\vec{d} - \vec{c}) \times \vec{b} = \vec{0}$ This implies that the vector $(\vec{d} - \vec{c})$ is parallel to the vector $\vec{b}$. Therefore, $(\vec{d} - \vec{c})$ must be a scalar multiple of $\vec{b}$. Let this scalar be $\lambda$. $\vec{d} - \vec{c} = \lambda \vec{b}$ From this, we can express $\vec{d}$ as: $\vec{d} = \vec{c} + \lambda \vec{b}$ This expression relates $\vec{d}$ to the known vectors $\vec{b}$ and $\vec{c}$, and an unknown scalar $\lambda$.

Step 2: Use the Dot Product Condition to Find $\lambda$

Now we use the second condition, $\vec{d} \cdot \vec{a} = 24$. Substitute the expression for $\vec{d}$ from Step 1 into this equation: $(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$ Using the distributive property of the dot product: $\vec{c} \cdot \vec{a} + (\lambda \vec{b}) \cdot \vec{a} = 24$ Using the property that scalar multiples can be factored out of a dot product: $\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 24$ Now, we calculate the dot products $\vec{c} \cdot \vec{a}$ and $\vec{b} \cdot \vec{a}$: $\vec{a} \cdot \vec{c} = (\hat{i}+4 \hat{j}+2 \hat{k}) \cdot (2 \hat{i}-\hat{j}+4 \hat{k}) = (1)(2) + (4)(-1) + (2)(4) = 2 - 4 + 8 = 6$ $\vec{b} \cdot \vec{a} = (3 \hat{i}-2 \hat{j}+7 \hat{k}) \cdot (\hat{i}+4 \hat{j}+2 \hat{k}) = (3)(1) + (-2)(4) + (7)(2) = 3 - 8 + 14 = 9$ Substitute these values back into the equation: $6 + \lambda (9) = 24$ Now, solve for $\lambda$: $9\lambda = 24 - 6$ $9\lambda = 18$ $\lambda = \frac{18}{9} = 2$

Step 3: Determine the Vector $\vec{d}$

With the value of $\lambda=2$, we can now find the explicit form of vector $\vec{d}$ by substituting $\lambda$ back into the expression from Step 1: $\vec{d} = \vec{c} + \lambda \vec{b}$ $\vec{d} = (2 \hat{i}-\hat{j}+4 \hat{k}) + 2(3 \hat{i}-2 \hat{j}+7 \hat{k})$ First, perform the scalar multiplication: $\vec{d} = (2 \hat{i}-\hat{j}+4 \hat{k}) + (6 \hat{i}-4 \hat{j}+14 \hat{k})$ Next, perform the vector addition by summing the corresponding components: $\vec{d} = (2+6)\hat{i} + (-1-4)\hat{j} + (4+14)\hat{k}$ $\vec{d} = 8\hat{i} - 5\hat{j} + 18\hat{k}$

Step 4: Calculate the Squared Magnitude of $\vec{d}$

Finally, we need to find $|\vec{d}|^2$. For the vector $\vec{d} = 8\hat{i} - 5\hat{j} + 18\hat{k}$, the components are $d_x=8$, $d_y=-5$, and $d_z=18$. The squared magnitude is calculated as: $|\vec{d}|^2 = d_x^2 + d_y^2 + d_z^2$ $|\vec{d}|^2 = (8)^2 + (-5)^2 + (18)^2$ $|\vec{d}|^2 = 64 + 25 + 324$ $|\vec{d}|^2 = 89 + 324$ $|\vec{d}|^2 = 413$

Common Mistakes & Tips

• Arithmetic Errors: Be meticulous when performing dot product calculations and squaring components. Small errors can propagate and lead to an incorrect final answer.
• Misinterpreting Cross Product: Remember that $\vec{X} \times \vec{Y} = \vec{0}$ implies parallelism, not equality or perpendicularity.
• Algebraic Manipulation: Ensure correct application of vector properties, especially distributivity of dot and cross products.

Summary

The problem was solved by first using the cross product condition $\vec{d} \times \vec{b}=\vec{c} \times \vec{b}$ to establish that $\vec{d} = \vec{c} + \lambda \vec{b}$. Subsequently, the dot product condition $\vec{d} \cdot \vec{a}=24$ was used to form an equation involving only the scalar $\lambda$, which was solved to be $\lambda=2$. With $\lambda$ determined, the vector $\vec{d}$ was explicitly found. Finally, the squared magnitude of $\vec{d}$ was computed using its components.

The final answer is \boxed{413}.