Key Concepts and Formulas

• Properties of the Cross Product:
  • Distributive Property: $\vec{X} \times (\vec{Y} + \vec{Z}) = \vec{X} \times \vec{Y} + \vec{X} \times \vec{Z}$ and $(\vec{X} + \vec{Y}) \times \vec{Z} = \vec{X} \times \vec{Z} + \vec{Y} \times \vec{Z}$.
  • Anti-commutativity: $\vec{X} \times \vec{Y} = -(\vec{Y} \times \vec{X})$.
  • If $\vec{X} \times \vec{Y} = \vec{0}$ and $\vec{X}, \vec{Y}$ are non-zero, then $\vec{X}$ and $\vec{Y}$ are parallel, meaning $\vec{X} = k\vec{Y}$ for some scalar $k$.
• Dot Product: For vectors $\vec{P} = P_x\hat{i} + P_y\hat{j} + P_z\hat{k}$ and $\vec{Q} = Q_x\hat{i} + Q_y\hat{j} + Q_z\hat{k}$, $\vec{P} \cdot \vec{Q} = P_x Q_x + P_y Q_y + P_z Q_z$.

Step-by-Step Solution

Step 1: Simplify the given vector equation We are given the equation $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. Our goal is to rearrange this equation to find a relationship between $\vec{c}$ and a combination of $\vec{a}$ and $\vec{b}$. Using the distributive property of the cross product on the left side: $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} + 2 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = 3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$ Applying the anti-commutativity property $\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}} = -(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})$: $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} + 2 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = 3(-(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}))$ $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} + 2 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = -3(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})$ Now, we rearrange the terms to group the $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}$ terms: $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} + 3(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) + 2 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = \vec{0}$ $4(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) + 2 \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = \vec{0}$ Divide by 2: $2(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) + \overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}} = \vec{0}$ Factor out $\overrightarrow{\mathrm{c}}$ using the distributive property in reverse: $(2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}} = \vec{0}$

Step 2: Deduce the relationship between $\vec{c}$ and other vectors The equation $(2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}} = \vec{0}$ implies that the vector $(2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$ and the vector $\overrightarrow{\mathrm{c}}$ are parallel. This is because their cross product is the zero vector, and we are given $\vec{a} \cdot \vec{c}=130$, which means $\vec{c}$ is not the zero vector. Therefore, $\overrightarrow{\mathrm{c}}$ must be a scalar multiple of $(2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$. Let this scalar be $\lambda$. $\overrightarrow{\mathrm{c}} = \lambda (2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$ Now, we compute the vector $(2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}})$. Given $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}$ and $\overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$: $2\overrightarrow{\mathrm{a}} = 2(\hat{i}-3 \hat{j}+7 \hat{k}) = 2\hat{i}-6 \hat{j}+14 \hat{k}$ $2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} = (2\hat{i}-6 \hat{j}+14 \hat{k}) + (2 \hat{i}-\hat{j}+\hat{k})$ $2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} = (2+2)\hat{i} + (-6-1)\hat{j} + (14+1)\hat{k}$ $2\overrightarrow{\mathrm{a}} + \overrightarrow{\mathrm{b}} = 4\hat{i}-7 \hat{j}+15 \hat{k}$ So, we have: $\overrightarrow{\mathrm{c}} = \lambda (4\hat{i}-7 \hat{j}+15 \hat{k})$

Step 3: Determine the scalar $\lambda$ using the dot product We are given $\vec{a} \cdot \vec{c}=130$. Substitute the expression for $\overrightarrow{\mathrm{c}}$ into this equation: $\overrightarrow{\mathrm{a}} \cdot [\lambda (4\hat{i}-7 \hat{j}+15 \hat{k})] = 130$ Pull the scalar $\lambda$ out of the dot product: $\lambda [\overrightarrow{\mathrm{a}} \cdot (4\hat{i}-7 \hat{j}+15 \hat{k})] = 130$ Substitute the components of $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}$: $\lambda [(\hat{i}-3 \hat{j}+7 \hat{k}) \cdot (4\hat{i}-7 \hat{j}+15 \hat{k})] = 130$ Calculate the dot product: $\lambda [(1)(4) + (-3)(-7) + (7)(15)] = 130$ $\lambda [4 + 21 + 105] = 130$ $\lambda [130] = 130$ Solving for $\lambda$: $\lambda = \frac{130}{130} = 1$

Step 4: Find the vector $\vec{c}$ Now that we have found $\lambda = 1$, we can determine the vector $\vec{c}$: $\overrightarrow{\mathrm{c}} = 1 \cdot (4\hat{i}-7 \hat{j}+15 \hat{k})$ $\overrightarrow{\mathrm{c}} = 4\hat{i}-7 \hat{j}+15 \hat{k}$

Step 5: Calculate $\vec{b} \cdot \vec{c}$ Finally, we need to find the dot product of $\vec{b}$ and $\vec{c}$. Given $\overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and we found $\overrightarrow{\mathrm{c}}=4\hat{i}-7 \hat{j}+15 \hat{k}$: $\vec{b} \cdot \vec{c} = (2 \hat{i}-\hat{j}+\hat{k}) \cdot (4\hat{i}-7 \hat{j}+15 \hat{k})$ $\vec{b} \cdot \vec{c} = (2)(4) + (-1)(-7) + (1)(15)$ $\vec{b} \cdot \vec{c} = 8 + 7 + 15$ $\vec{b} \cdot \vec{c} = 30$

Common Mistakes & Tips

• Sign errors with cross products: Be careful when using the anti-commutativity property ($\vec{X} \times \vec{Y} = -\vec{Y} \times \vec{X}$), as sign mistakes are common.
• Algebraic manipulation of vector equations: Ensure all steps in rearranging vector equations are logically sound and correctly apply vector properties.
• Calculation of dot products: Double-check the multiplication and addition when computing dot products to avoid arithmetic errors.

Summary The problem was solved by first simplifying the given vector cross product equation to establish that $\vec{c}$ is parallel to a linear combination of $\vec{a}$ and $\vec{b}$. This parallelism allowed us to express $\vec{c}$ as a scalar multiple of $(2\vec{a} + \vec{b})$. Using the given dot product condition $\vec{a} \cdot \vec{c} = 130$, we determined the scalar multiplier. Finally, with the explicit form of $\vec{c}$, we computed the required dot product $\vec{b} \cdot \vec{c}$.

The final answer is $\boxed{30}$.