Key Concepts and Formulas

• Cross Product Properties:
  • Distributive Property: $\vec{A} \times (\vec{B} + \vec{C}) = \vec{A} \times \vec{B} + \vec{A} \times \vec{C}$ and $(\vec{A} + \vec{B}) \times \vec{C} = \vec{A} \times \vec{C} + \vec{B} \times \vec{C}$.
  • Anticommutative Property: $\vec{A} \times \vec{B} = -(\vec{B} \times \vec{A})$.
• Parallel Vectors: If $\vec{P} \times \vec{Q} = \vec{0}$ and $\vec{P}, \vec{Q}$ are non-zero vectors, then $\vec{P} = \lambda \vec{Q}$ for some scalar $\lambda$.
• Dot Product: For $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.

Step-by-Step Solution

Given vectors: $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}$ $\vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$

Given vector equation: $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$

Given dot product condition: $\vec{a} \cdot \vec{c}=-29$

We need to find the value of $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$.

Step 1: Simplify the given vector equation. Our goal is to manipulate the equation to isolate a cross product that equals the zero vector, which will imply parallelism. $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$ Move all terms to one side: $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}) - \vec{a} \times(\vec{c}+\hat{i}) = \vec{0}$ Using the anticommutative property, $\vec{a} \times(\vec{c}+\hat{i}) = -(\vec{c}+\hat{i}) \times \vec{a}$. Substituting this: $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}) + (\vec{c}+\hat{i}) \times \vec{a} = \vec{0}$ Now, factor out the common vector $(\vec{c}+\hat{i})$ using the distributive property: $(\vec{c}+\hat{i}) \times [(\vec{a}+\vec{b}+\hat{i}) + \vec{a}] = \vec{0}$ Simplify the expression inside the square brackets: $(\vec{c}+\hat{i}) \times (2\vec{a}+\vec{b}+\hat{i}) = \vec{0}$

Step 2: Deduce the relationship between vectors. The equation $(\vec{c}+\hat{i}) \times (2\vec{a}+\vec{b}+\hat{i}) = \vec{0}$ implies that the vector $(\vec{c}+\hat{i})$ is parallel to the vector $(2\vec{a}+\vec{b}+\hat{i})$. Therefore, we can write: $\vec{c}+\hat{i} = \lambda (2\vec{a}+\vec{b}+\hat{i})$ for some scalar $\lambda$.

Step 3: Calculate the vector $(2\vec{a}+\vec{b}+\hat{i})$. Let's compute the components of this vector: $2\vec{a} = 2(2 \hat{i}+5 \hat{j}-\hat{k}) = 4 \hat{i}+10 \hat{j}-2 \hat{k}$ $2\vec{a}+\vec{b}+\hat{i} = (4 \hat{i}+10 \hat{j}-2 \hat{k}) + (2 \hat{i}-2 \hat{j}+2 \hat{k}) + (1 \hat{i}+0 \hat{j}+0 \hat{k})$ $= (4+2+1)\hat{i} + (10-2+0)\hat{j} + (-2+2+0)\hat{k}$ $= 7\hat{i} + 8\hat{j}$

Step 4: Express $\vec{c}$ in terms of $\lambda$. Substitute the calculated vector back into the parallelism equation: $\vec{c}+\hat{i} = \lambda (7\hat{i} + 8\hat{j})$ Isolate $\vec{c}$: $\vec{c} = \lambda (7\hat{i} + 8\hat{j}) - \hat{i}$ $\vec{c} = (7\lambda - 1)\hat{i} + 8\lambda \hat{j}$

Step 5: Use the dot product condition to find $\lambda$. We are given $\vec{a} \cdot \vec{c}=-29$. Substitute the expressions for $\vec{a}$ and $\vec{c}$: $\vec{a} \cdot \vec{c} = (2 \hat{i}+5 \hat{j}-\hat{k}) \cdot ((7\lambda - 1)\hat{i} + 8\lambda \hat{j} + 0\hat{k})$ $= (2)(7\lambda - 1) + (5)(8\lambda) + (-1)(0)$ $= 14\lambda - 2 + 40\lambda$ $= 54\lambda - 2$ Equate this to the given value: $54\lambda - 2 = -29$ $54\lambda = -27$ $\lambda = -\frac{27}{54} = -\frac{1}{2}$

Step 6: Calculate the explicit vector $\vec{c}$. Substitute $\lambda = -\frac{1}{2}$ back into the expression for $\vec{c}$: $\vec{c} = (7(-\frac{1}{2}) - 1)\hat{i} + 8(-\frac{1}{2})\hat{j}$ $\vec{c} = (-\frac{7}{2} - \frac{2}{2})\hat{i} - 4\hat{j}$ $\vec{c} = -\frac{9}{2}\hat{i} - 4\hat{j}$

Step 7: Compute the final dot product. We need to find $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$. $\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k}) = (-\frac{9}{2}\hat{i} - 4\hat{j} + 0\hat{k}) \cdot (-2 \hat{i}+1 \hat{j}+1 \hat{k})$ $= (-\frac{9}{2})(-2) + (-4)(1) + (0)(1)$ $= 9 - 4 + 0$ $= 5$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when using the anticommutative property of the cross product or when distributing scalar multiples.
• Algebraic Simplification: Ensure all algebraic manipulations, especially when combining terms and solving for $\lambda$, are accurate.
• Component-wise Operations: Remember that dot products and cross products involve specific rules for combining vector components.

Summary The problem was solved by first simplifying the vector equation using properties of the cross product to establish parallelism between $(\vec{c}+\hat{i})$ and $(2\vec{a}+\vec{b}+\hat{i})$. This allowed us to express $\vec{c}$ in terms of a scalar parameter $\lambda$. The given dot product condition was then used to solve for $\lambda$. Finally, with $\vec{c}$ fully determined, the required dot product was computed.

The final answer is $\boxed{5}$, which corresponds to option (C).