Key Concepts and Formulas

1. Distributive Property of Cross Product: $(\vec{u} - \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} - \vec{v} \times \vec{w}$.
2. Anti-commutative Property of Cross Product: $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$.
3. Scalar Triple Product: $[\vec{u} \ \vec{v} \ \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$. A key property is that $[\vec{u} \ \vec{v} \ \vec{w}] = [\vec{v} \ \vec{w} \ \vec{u}] = [\vec{w} \ \vec{u} \ \vec{v}]$ and if any two vectors are identical, the scalar triple product is zero, e.g., $[\vec{u} \ \vec{u} \ \vec{v}] = 0$.
4. Vector Operations: Given $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ and $\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$:
   • $\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$.
   • $\vec{a} \times \vec{b} = (a_2 b_3 - a_3 b_2) \hat{i} - (a_1 b_3 - a_3 b_1) \hat{j} + (a_1 b_2 - a_2 b_1) \hat{k}$.

Step-by-Step Solution

We are given the vectors $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$. We are also given:

1. $(\vec{a}-\vec{c}) \times \vec{b}=-18 \hat{i}-3 \hat{j}+12 \hat{k}$
2. $\vec{a} \cdot \vec{c}=3$ And the definition $\vec{d} = \vec{b} \times \vec{c}$. Our objective is to find $|\vec{a} \cdot \vec{d}|$.

Step 1: Expand the given cross product equation.

We use the distributive property of the cross product on the first given condition: $(\vec{a}-\vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$ So, we have: $\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18 \hat{i}-3 \hat{j}+12 \hat{k} \quad (*)$ Reasoning: This step breaks down the complex cross product into simpler terms involving $\vec{a}$, $\vec{c}$, and $\vec{b}$.

Step 2: Introduce $\vec{d}$ into the equation using the anti-commutative property.

We are given $\vec{d} = \vec{b} \times \vec{c}$. From the anti-commutative property, we know that $\vec{c} \times \vec{b} = -(\vec{b} \times \vec{c})$. Substituting this into equation $(*)$: $\vec{a} \times \vec{b} - (-\vec{b} \times \vec{c}) = -18 \hat{i}-3 \hat{j}+12 \hat{k}$ $\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = -18 \hat{i}-3 \hat{j}+12 \hat{k}$ Now, substitute $\vec{d} = \vec{b} \times \vec{c}$: $\vec{a} \times \vec{b} + \vec{d} = -18 \hat{i}-3 \hat{j}+12 \hat{k} \quad (**)$ Reasoning: This manipulation is done to incorporate the vector $\vec{d}$ into the equation, which is part of what we need to calculate.

Step 3: Calculate $\vec{a} \times \vec{b}$.

We are given $\vec{a}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+2 \hat{j}+5 \hat{k}$. Using the formula for the cross product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$ $= \hat{i}((-3)(5) - (1)(2)) - \hat{j}((2)(5) - (1)(3)) + \hat{k}((2)(2) - (-3)(3))$ $= \hat{i}(-15 - 2) - \hat{j}(10 - 3) + \hat{k}(4 - (-9))$ $= -17 \hat{i} - 7 \hat{j} + 13 \hat{k}$ Reasoning: We need the explicit value of $\vec{a} \times \vec{b}$ to isolate $\vec{d}$ from equation $(**)$.

Step 4: Calculate $\vec{d}$.

Substitute the calculated value of $\vec{a} \times \vec{b}$ into equation $(**)$: $(-17 \hat{i} - 7 \hat{j} + 13 \hat{k}) + \vec{d} = -18 \hat{i}-3 \hat{j}+12 \hat{k}$ Now, solve for $\vec{d}$: $\vec{d} = (-18 \hat{i}-3 \hat{j}+12 \hat{k}) - (-17 \hat{i} - 7 \hat{j} + 13 \hat{k})$ $\vec{d} = (-18 - (-17)) \hat{i} + (-3 - (-7)) \hat{j} + (12 - 13) \hat{k}$ $\vec{d} = (-18 + 17) \hat{i} + (-3 + 7) \hat{j} + (-1) \hat{k}$ $\vec{d} = -1 \hat{i} + 4 \hat{j} - 1 \hat{k}$ Reasoning: By isolating $\vec{d}$, we obtain its explicit vector form, which is necessary for the final calculation.

Step 5: Calculate $\vec{a} \cdot \vec{d}$.

We need to find the dot product of $\vec{a}$ and the calculated $\vec{d}$: $\vec{a} = 2 \hat{i}-3 \hat{j}+\hat{k}$ $\vec{d} = -1 \hat{i} + 4 \hat{j} - 1 \hat{k}$ Using the dot product formula: $\vec{a} \cdot \vec{d} = (2)(-1) + (-3)(4) + (1)(-1)$ $= -2 - 12 - 1$ $= -15$ Reasoning: This step directly calculates the scalar value of $\vec{a} \cdot \vec{d}$.

Step 6: Find the magnitude $|\vec{a} \cdot \vec{d}|$.

The problem asks for the absolute value of the dot product: $|\vec{a} \cdot \vec{d}| = |-15|$ $|\vec{a} \cdot \vec{d}| = 15$ Reasoning: The absolute value is taken to ensure the magnitude is a non-negative quantity.

Common Mistakes & Tips

• Sign Errors in Cross Product: Be extremely careful with the signs when calculating the cross product, especially with the $\hat{j}$ component and when applying the anti-commutative property.
• Misinterpreting Scalar Triple Product: While not directly used for calculation here, remembering that $\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ can sometimes simplify problems or act as a check.
• Order of Operations: Ensure that vector operations are performed in the correct order as specified by the problem and algebraic manipulations.

Summary

The problem was solved by first expanding the given cross product equation using the distributive and anti-commutative properties to introduce the vector $\vec{d}$. We then calculated the explicit vector forms of $\vec{a} \times \vec{b}$ and subsequently $\vec{d}$. Finally, the dot product $\vec{a} \cdot \vec{d}$ was computed, and its absolute value was taken to find the required magnitude. The condition $\vec{a} \cdot \vec{c}=3$ was not directly needed for the calculation of $|\vec{a} \cdot \vec{d}|$, which is a common characteristic of some vector algebra problems where extraneous information might be provided.

The final answer is 15, which corresponds to option (A).

The final answer is \boxed{15}.