Key Concepts and Formulas

1. Vector Representation: A vector $\vec{v}$ in component form is $\vec{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k}$.
2. Dot Product: For $\vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k}$ and $\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$, $\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z$.
3. Cross Product Property: If $\vec{A} \times \vec{B} = \vec{0}$ and $\vec{A} \neq \vec{0}$, then $\vec{A}$ and $\vec{B}$ are parallel, meaning $\vec{B} = k\vec{A}$ for some scalar $k$.
4. Distributive Property of Cross Product: $\vec{A} \times (\vec{B} - \vec{C}) = \vec{A} \times \vec{B} - \vec{A} \times \vec{C}$.

Step-by-Step Solution

Let the unknown vector $\vec{c}$ be represented as $\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}$. We are given $\vec{a}=6 \hat{i}+9 \hat{j}+12 \hat{k}$ and $\vec{b}=\alpha \hat{i}+11 \hat{j}-2 \hat{k}$.

Step 1: Use the dot product conditions to form linear equations.

The condition $\vec{a} \cdot \vec{c}=-12$ translates to: $(6 \hat{i}+9 \hat{j}+12 \hat{k}) \cdot (c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}) = -12$ $6c_1 + 9c_2 + 12c_3 = -12$ Dividing by 3, we get: $2c_1 + 3c_2 + 4c_3 = -4 \quad \text{...(1)}$

The condition $\vec{c} \cdot(\hat{i}-2 \hat{j}+\hat{k})=5$ translates to: $(c_1 \hat{i}+c_2 \hat{j}+c_3 \hat{k}) \cdot (\hat{i}-2 \hat{j}+\hat{k}) = 5$ $c_1 - 2c_2 + c_3 = 5 \quad \text{...(2)}$

Step 2: Utilize the cross product condition to establish a relationship between $\vec{c}$ and $\vec{b}$.

We are given $\vec{a} \times \vec{c}=\vec{a} \times \vec{b}$. Rearranging this, we get: $\vec{a} \times \vec{c} - \vec{a} \times \vec{b} = \vec{0}$ Using the distributive property of the cross product: $\vec{a} \times (\vec{c} - \vec{b}) = \vec{0}$ Since $\vec{a} \neq \vec{0}$, the vector $(\vec{c} - \vec{b})$ must be parallel to $\vec{a}$. Therefore, we can write: $\vec{c} - \vec{b} = k \vec{a}$ for some scalar $k$. This means $\vec{c} = \vec{b} + k \vec{a}$.

Substituting the components: $c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} = (\alpha \hat{i}+11 \hat{j}-2 \hat{k}) + k(6 \hat{i}+9 \hat{j}+12 \hat{k})$ $c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k} = (\alpha + 6k) \hat{i} + (11 + 9k) \hat{j} + (-2 + 12k) \hat{k}$ Equating the components, we get: $c_1 = \alpha + 6k \quad \text{...(3)}$ $c_2 = 11 + 9k \quad \text{...(4)}$ $c_3 = -2 + 12k \quad \text{...(5)}$

Step 3: Solve the system of equations for $\alpha$ and $k$.

Substitute equations (3), (4), and (5) into equation (2): $(\alpha + 6k) - 2(11 + 9k) + (-2 + 12k) = 5$ $\alpha + 6k - 22 - 18k - 2 + 12k = 5$ $\alpha + (6 - 18 + 12)k - 24 = 5$ $\alpha + 0k - 24 = 5$ $\alpha = 29$

Now substitute equations (3), (4), (5), and the value of $\alpha=29$ into equation (1): $2(\alpha + 6k) + 3(11 + 9k) + 4(-2 + 12k) = -4$ $2(29 + 6k) + 3(11 + 9k) + 4(-2 + 12k) = -4$ $58 + 12k + 33 + 27k - 8 + 48k = -4$ $(12 + 27 + 48)k + (58 + 33 - 8) = -4$ $87k + 83 = -4$ $87k = -87$ $k = -1$

Step 4: Determine the components of $\vec{c}$.

Using $k = -1$ and $\alpha = 29$ in equations (3), (4), and (5): $c_1 = \alpha + 6k = 29 + 6(-1) = 29 - 6 = 23$ $c_2 = 11 + 9k = 11 + 9(-1) = 11 - 9 = 2$ $c_3 = -2 + 12k = -2 + 12(-1) = -2 - 12 = -14$ So, $\vec{c} = 23 \hat{i} + 2 \hat{j} - 14 \hat{k}$.

Step 5: Calculate the final dot product.

We need to find $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k})$: $\vec{c} \cdot(\hat{i}+\hat{j}+\hat{k}) = (23 \hat{i} + 2 \hat{j} - 14 \hat{k}) \cdot (\hat{i}+\hat{j}+\hat{k})$ $= (23)(1) + (2)(1) + (-14)(1)$ $= 23 + 2 - 14$ $= 25 - 14$ $= 11$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs and arithmetic when solving the system of equations.
• Misinterpreting Cross Product: Ensure you correctly apply the property that $\vec{A} \times \vec{B} = \vec{0}$ implies parallelism for non-zero vectors.
• Scalar Representation: When setting up the parallelism, $\vec{c} - \vec{b} = k\vec{a}$ is one way; $\vec{a} = \lambda(\vec{c} - \vec{b})$ is another. Ensure consistency in your chosen notation.

Summary

The problem was solved by first converting the given dot product conditions into a system of linear equations involving the components of $\vec{c}$. The cross product condition was used to establish that $\vec{c} - \vec{b}$ is parallel to $\vec{a}$, leading to expressions for the components of $\vec{c}$ in terms of an unknown scalar $k$ and the component $\alpha$ of $\vec{b}$. Substituting these expressions back into the linear equations allowed us to solve for $\alpha$ and $k$. Finally, the components of $\vec{c}$ were determined, and the required dot product was calculated.

The final answer is $\boxed{11}$.