Key Concepts and Formulas

• Perpendicular Vectors and Cross Product: A vector perpendicular to two non-parallel vectors $\vec{u}$ and $\vec{v}$ is parallel to their cross product, i.e., $\vec{d} = \lambda (\vec{u} \times \vec{v})$ for some scalar $\lambda$.
• Dot Product: The dot product of two vectors $\vec{u} = u_x \hat{i} + u_y \hat{j} + u_z \hat{k}$ and $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$ is $\vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z$. It is used to find unknown scalar multipliers.
• Vector Triple Product Identity: For any three vectors $\vec{A}, \vec{B}, \vec{C}$, the identity is $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$.
• Scalar Triple Product: The scalar triple product $\vec{P} \cdot (\vec{Q} \times \vec{R})$ represents the volume of the parallelepiped formed by $\vec{P}, \vec{Q}, \vec{R}$ and can be computed using a determinant.

Step-by-Step Solution

Step 1: Find a vector perpendicular to $\vec{a}$ and $\vec{b}$

• Why: The problem states that $\vec{d}$ is perpendicular to both $\vec{a}$ and $\vec{b}$. The cross product $\vec{a} \times \vec{b}$ yields a vector that is orthogonal to both $\vec{a}$ and $\vec{b}$. Therefore, $\vec{d}$ must be a scalar multiple of $\vec{a} \times \vec{b}$.
• Calculation: $\vec{a} \times \vec{b} = (2 \hat{i}+7 \hat{j}-\hat{k}) \times (3 \hat{i}+0 \hat{j}+5 \hat{k})$ $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 7 & -1 \\ 3 & 0 & 5 \end{vmatrix}$ $= \hat{i}((7)(5) - (-1)(0)) - \hat{j}((2)(5) - (-1)(3)) + \hat{k}((2)(0) - (7)(3))$ $= \hat{i}(35 - 0) - \hat{j}(10 + 3) + \hat{k}(0 - 21)$ $= 35 \hat{i} - 13 \hat{j} - 21 \hat{k}$
• Result: Thus, $\vec{d} = \lambda (35 \hat{i} - 13 \hat{j} - 21 \hat{k})$ for some scalar $\lambda$.

Step 2: Determine the scalar $\lambda$ using the condition $\vec{c} \cdot \vec{d} = 12$

• Why: This condition allows us to find the specific value of $\lambda$, which uniquely defines the vector $\vec{d}$.
• Calculation: $\vec{c} \cdot \vec{d} = (\hat{i}-\hat{j}+2 \hat{k}) \cdot \lambda (35 \hat{i} - 13 \hat{j} - 21 \hat{k}) = 12$ $\lambda [(1)(35) + (-1)(-13) + (2)(-21)] = 12$ $\lambda [35 + 13 - 42] = 12$ $\lambda [48 - 42] = 12$ $\lambda (6) = 12$ $\lambda = 2$
• Result: So, $\vec{d} = 2 (35 \hat{i} - 13 \hat{j} - 21 \hat{k}) = 70 \hat{i} - 26 \hat{j} - 42 \hat{k}$.

Step 3: Evaluate $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$

• Why: The expression to be evaluated is a scalar triple product. We can simplify this using vector identities to avoid calculating $\vec{c} \times \vec{d}$ directly, which can be tedious.
• Setup: Let $\vec{X} = -\hat{i}+\hat{j}-\hat{k}$. We need to compute $\vec{X} \cdot (\vec{c} \times \vec{d})$. Since $\vec{d} = \lambda (\vec{a} \times \vec{b})$, we have: $\vec{X} \cdot (\vec{c} \times \vec{d}) = \vec{X} \cdot (\vec{c} \times [ \lambda (\vec{a} \times \vec{b}) ])$ $= \lambda [\vec{X} \cdot (\vec{c} \times (\vec{a} \times \vec{b}))]$ Now, we use the vector triple product identity $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. Let $\vec{A} = \vec{c}$, $\vec{B} = \vec{a}$, $\vec{C} = \vec{b}$. $\vec{c} \times (\vec{a} \times \vec{b}) = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$ Substituting this back: $= \lambda [ \vec{X} \cdot ((\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}) ]$ $= \lambda [ (\vec{c} \cdot \vec{b})(\vec{X} \cdot \vec{a}) - (\vec{c} \cdot \vec{a})(\vec{X} \cdot \vec{b}) ]$
• Calculate required dot products:
  • $\lambda = 2$ (from Step 2)
  • $\vec{c} \cdot \vec{b} = (\hat{i}-\hat{j}+2 \hat{k}) \cdot (3 \hat{i}+0 \hat{j}+5 \hat{k}) = (1)(3) + (-1)(0) + (2)(5) = 3 + 0 + 10 = 13$
  • $\vec{c} \cdot \vec{a} = (\hat{i}-\hat{j}+2 \hat{k}) \cdot (2 \hat{i}+7 \hat{j}-\hat{k}) = (1)(2) + (-1)(7) + (2)(-1) = 2 - 7 - 2 = -7$
  • $\vec{X} \cdot \vec{a} = (-\hat{i}+\hat{j}-\hat{k}) \cdot (2 \hat{i}+7 \hat{j}-\hat{k}) = (-1)(2) + (1)(7) + (-1)(-1) = -2 + 7 + 1 = 6$
  • $\vec{X} \cdot \vec{b} = (-\hat{i}+\hat{j}-\hat{k}) \cdot (3 \hat{i}+0 \hat{j}+5 \hat{k}) = (-1)(3) + (1)(0) + (-1)(5) = -3 + 0 - 5 = -8$
• Final Calculation: $= 2 [ (13)(6) - (-7)(-8) ]$ $= 2 [ 78 - 56 ]$ $= 2 [ 22 ]$ $= 44$

Common Mistakes & Tips

• Sign Errors: Be meticulous with signs during cross product and dot product calculations. A single misplaced sign can lead to an incorrect result.
• Vector Triple Product Order: Remember that the order of vectors in the vector triple product identity matters. Ensure you are applying it correctly to $\vec{c} \times (\vec{a} \times \vec{b})$.
• Scalar Multiplication: When a scalar multiplies a vector that is part of a dot or cross product, the scalar can be factored out or distributed as appropriate, simplifying the expression.

Summary

The problem requires finding a vector $\vec{d}$ perpendicular to two given vectors and satisfying a dot product condition. This is achieved by first using the cross product $\vec{a} \times \vec{b}$ to establish the direction of $\vec{d}$, and then using the dot product condition $\vec{c} \cdot \vec{d} = 12$ to determine the scalar multiplier $\lambda$. Finally, the expression $(-\hat{i}+\hat{j}-\hat{k}) \cdot(\vec{c} \times \vec{d})$ is evaluated by substituting $\vec{d} = \lambda (\vec{a} \times \vec{b})$ and employing the vector triple product identity to simplify the calculation into a series of dot products.

The final answer is $\boxed{44}$, which corresponds to option (C).