Key Concepts and Formulas

1. Vector Cross Product Property: For any vectors $\vec A$, $\vec B$, and $\vec C$, $\vec A \times \vec B = \vec A \times \vec C$ implies $\vec A \times (\vec B - \vec C) = \vec 0$. This further implies that $\vec A$ is parallel to $(\vec B - \vec C)$, meaning $\vec A = k(\vec B - \vec C)$ for some scalar $k$.
2. Scalar (Dot) Product: For vectors $\vec P = P_x\hat i + P_y\hat j + P_z\hat k$ and $\vec Q = Q_x\hat i + Q_y\hat j + Q_z\hat k$, their dot product is $\vec P \cdot \vec Q = P_x Q_x + P_y Q_y + P_z Q_z$.
3. Vector Subtraction: To subtract vectors, subtract their corresponding components: $(A_x - B_x)\hat i + (A_y - B_y)\hat j + (A_z - B_z)\hat k$.

Step-by-Step Solution

Step 1: Simplify the first vector equation to determine the relationship between $\vec r$ and $(\vec a - \vec b)$.

We are given the equation $\vec r \times \vec a = \vec r \times \vec b$.

• Why this step? This equation provides information about the orientation of $\vec r$ relative to $\vec a$ and $\vec b$. We can simplify it using vector cross product properties.

Rearranging the terms, we get: $\vec r \times \vec a - \vec r \times \vec b = \vec 0$ Using the distributive property of the cross product, we have: $\vec r \times (\vec a - \vec b) = \vec 0$

• Why this step? This simplified form tells us that $\vec r$ is parallel to the vector $(\vec a - \vec b)$.

Now, let's calculate the vector $(\vec a - \vec b)$: Given $\vec a = 2\hat i - 3\hat j + 4\hat k$ and $\vec b = 7\hat i + \hat j - 6\hat k$. $\vec a - \vec b = (2 - 7)\hat i + (-3 - 1)\hat j + (4 - (-6))\hat k$ $\vec a - \vec b = -5\hat i - 4\hat j + (4 + 6)\hat k$ $\vec a - \vec b = -5\hat i - 4\hat j + 10\hat k$ Since $\vec r \times (\vec a - \vec b) = \vec 0$, $\vec r$ must be parallel to $(\vec a - \vec b)$. Therefore, we can write $\vec r$ as a scalar multiple of $(\vec a - \vec b)$: $\vec r = \lambda (\vec a - \vec b)$ $\vec r = \lambda (-5\hat i - 4\hat j + 10\hat k)$ where $\lambda$ is a scalar constant.

Step 2: Use the second given condition to find the value of the scalar $\lambda$.

The second condition is $\vec r \cdot (\hat i + 2\hat j + \hat k) = -3$.

• Why this step? We have an expression for $\vec r$ in terms of $\lambda$. Substituting this into the dot product equation will allow us to solve for the unknown scalar $\lambda$.

Substitute $\vec r = \lambda (-5\hat i - 4\hat j + 10\hat k)$ into the equation: $\left( \lambda (-5\hat i - 4\hat j + 10\hat k) \right) \cdot (\hat i + 2\hat j + \hat k) = -3$ Factor out $\lambda$ and compute the dot product: $\lambda \left( (-5)(1) + (-4)(2) + (10)(1) \right) = -3$ $\lambda \left( -5 - 8 + 10 \right) = -3$ $\lambda \left( -13 + 10 \right) = -3$ $\lambda (-3) = -3$ Solving for $\lambda$: $\lambda = \frac{-3}{-3}$ $\lambda = 1$

Step 3: Determine the vector $\vec r$.

• Why this step? Now that we have found the value of $\lambda$, we can determine the exact components of the vector $\vec r$.

Substitute $\lambda = 1$ back into the expression for $\vec r$ from Step 1: $\vec r = 1 \cdot (-5\hat i - 4\hat j + 10\hat k)$ $\vec r = -5\hat i - 4\hat j + 10\hat k$

Step 4: Calculate the required dot product.

We need to find the value of $\vec r \cdot (2\hat i - 3\hat j + \hat k)$.

• Why this step? This is the final quantity we need to compute, and we now have the complete vector $\vec r$.

Substitute the components of $\vec r$ and the target vector into the dot product: $\vec r \cdot (2\hat i - 3\hat j + \hat k) = (-5\hat i - 4\hat j + 10\hat k) \cdot (2\hat i - 3\hat j + \hat k)$ Perform the dot product calculation: $= (-5)(2) + (-4)(-3) + (10)(1)$ $= -10 + 12 + 10$ $= 2 + 10$ $= 12$

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs during vector subtraction and dot product calculations. For example, $4 - (-6)$ is $4+6$.
• Cross Product Order: Remember that the cross product is not commutative ($\vec A \times \vec B \neq \vec B \times \vec A$). Ensure correct order when applying distributive properties.
• Vector vs. Scalar: Differentiate between vector quantities (like $\vec r$) and scalar quantities (like $\lambda$ and the final dot product result).

Summary

The problem starts with a condition on the cross product of an unknown vector $\vec r$ with two given vectors $\vec a$ and $\vec b$. This condition, $\vec r \times \vec a = \vec r \times \vec b$, simplifies to $\vec r \times (\vec a - \vec b) = \vec 0$, implying that $\vec r$ is parallel to $(\vec a - \vec b)$. We then express $\vec r$ as a scalar multiple, $\lambda(\vec a - \vec b)$. The second condition, a dot product involving $\vec r$, is used to solve for the scalar $\lambda$. Once $\lambda$ is found, the vector $\vec r$ is determined, and the final required dot product can be calculated.

The final answer is $\boxed{\text{12}}$.