Question:

## Step 1: Given Vectors and Initial Equation We are given vectors $\vec{a}=\hat{i}-3\hat{j}+7\hat{k}$ and $\vec{b}=2\hat{i}-\hat{j}+\hat{k}$, and we know that $(\vec{a}+2\vec{b})\times\vec{c}=3(\vec{c}\times\vec{a})$. To start solving this problem, let's first compute $\vec{a}+2\vec{b}$. $$ \begin{aligned} \vec{a}+2\vec{b} &= (\hat{i}-3\hat{j}+7\hat{k}) + 2(2\hat{i}-\hat{j}+\hat{k}) \\ &= \hat{i}-3\hat{j}+7\hat{k} + 4\hat{i}-2\hat{j}+2\hat{k} \\ &= 5\hat{i}-5\hat{j}+9\hat{k} \end{aligned} $$ ## Step 2: Simplify the Given Equation Now, let's use the given equation $(\vec{a}+2\vec{b})\times\vec{c}=3(\vec{c}\times\vec{a})$ and apply properties of the cross product to simplify it. $$ \begin{aligned} (\vec{a}+2\vec{b})\times\vec{c} &= 3(\vec{c}\times\vec{a}) \\ (\vec{a}+2\vec{b})\times\vec{c} + 3(\vec{c}\times\vec{a}) &= 0 \\ (\vec{a}+2\vec{b})\times\vec{c} + 3(\vec{a}\times\vec{c}) &= 0 \\ (\vec{a}+2\vec{b} + 3\vec{a})\times\vec{c} &= 0 \\ (4\vec{a}+2\vec{b})\times\vec{c} &= 0 \end{aligned} $$ This implies that $4\vec{a}+2\vec{b}$ is parallel to $\vec{c}$, or $4\vec{a}+2\vec{b} = k\vec{c}$ for some scalar $k$. ## Step 3: Express $\vec{c}$ as a Linear Combination Given that $\vec{a}\cdot\vec{c}=130$, we need to find a way to express $\vec{c}$ in terms of $\vec{a}$ and $\vec{b}$ to utilize this information. ## Step 4: Utilize the Dot Product Since we know that $(\vec{a}+2\vec{b})\times\vec{c}=3(\vec{c}\times\vec{a})$, let's take the dot product of both sides with $\vec{b}$. $$ \begin{aligned} \vec{b}\cdot((\vec{a}+2\vec{b})\times\vec{c}) &= \vec{b}\cdot(3(\vec{c}\times\vec{a})) \\ \vec{b}\cdot((\vec{a}+2\vec{b})\times\vec{c}) &= 3\vec{b}\cdot(\vec{c}\times\vec{a}) \end{aligned} $$ Using the scalar triple product, $\vec{b}\cdot(\vec{c}\times\vec{a}) = \vec{c}\cdot(\vec{a}\times\vec{b})$, we have $$ \begin{aligned} \vec{b}\cdot((\vec{a}+2\vec{b})\times\vec{c}) &= 3\vec{c}\cdot(\vec{a}\times\vec{b}) \end{aligned} $$ ## Step 5: Calculate $\vec{a}\times\vec{b}$ To simplify this, let's calculate $\vec{a}\times\vec{b}$. $$ \begin{aligned} \vec{a}\times\vec{b} &= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 7 \\ 2 & -1 & 1 \end{vmatrix} \\ &= \hat{i}(-3+7) - \hat{j}(1-14) + \hat{k}(-1+6) \\ &= 4\hat{i} + 13\hat{j} + 5\hat{k} \end{aligned} $$ ## Step 6: Find $\vec{c}\cdot(\vec{a}\times\vec{b})$ However, we realize that to directly solve for $\vec{b}\cdot\vec{c}$, we should utilize the relationship between $\vec{a}$, $\vec{b}$, and $\vec{c}$ more effectively. ## Step 7: Rearrange and Simplify Since we have $(\vec{a}+2\vec{b})\times\vec{c}=3(\vec{c}\times\vec{a})$, let's rearrange and use properties of cross products to simplify our approach. $$ \begin{aligned} (\vec{a}+2\vec{b})\times\vec{c} + 3\vec{c}\times\vec{a} &= 0 \\ (\vec{a}+2\vec{b} + 3\vec{a})\times\vec{c} &= 0 \\ (4\vec{a}+2\vec{b})\times\vec{c} &= 0 \end{aligned} $$ This implies that $\vec{c}$ is parallel to $4\vec{a}+2\vec{b}$. ## Step 8: Express $\vec{c}$ in Terms of $\vec{a}$ and $\vec{b}$ Assume $\vec{c} = k(4\vec{a}+2\vec{b})$ for some scalar $k$. ## Step 9: Calculate $\vec{c}$ Using $\vec{a}\cdot\vec{c}=130$ $$ \begin{aligned} \vec{a}\cdot\vec{c} &= 130 \\ \vec{a}\cdot(k(4\vec{a}+2\vec{b})) &= 130 \\ k(4\vec{a}\cdot\vec{a} + 2\vec{a}\cdot\vec{b}) &= 130 \end{aligned} $$ First, calculate $\vec{a}\cdot\vec{a}$ and $\vec{a}\cdot\vec{b}$. $$ \begin{aligned} \vec{a}\cdot\vec{a} &= (1)^2 + (-3)^2 + (7)^2 = 1 + 9 + 49 = 59 \\ \vec{a}\cdot\vec{b} &= (1)(2) + (-3)(-1) + (7)(1) = 2 + 3 + 7 = 12 \end{aligned} $$ Then, $$ \begin{aligned} k(4(59) + 2(12)) &= 130 \\ k(236 + 24) &= 130 \\ k(260) &= 130 \\ k &= \frac{130}{260} = \frac{1}{2} \end{aligned} $$ ## Step 10: Find $\vec{c}$ Using $k$ $$ \begin{aligned} \vec{c} &= \frac{1}{2}(4\vec{a}+2\vec{b}) \\ &= \frac{1}{2}(4(\hat{i}-3\hat{j}+7\hat{k}) + 2(2\hat{i}-\hat{j}+\hat{k})) \\ &= \frac{1}{2}(4\hat{i}-12\hat{j}+28\hat{k} + 4\hat{i}-2\hat{j}+2\hat{k}) \\ &= \frac{1}{2}(8\hat{i}-14\hat{j}+30\hat{k}) \\ &= 4\hat{i}-7\hat{j}+15\hat{k} \end{aligned} $$ ## Step 11: Calculate $\vec{b}\cdot\vec{c}$ $$ \begin{aligned} \vec{b}\cdot\vec{c} &= (2\hat{i}-\hat{j}+\hat{k})\cdot(4\hat{i}-7\hat{j}+15\hat{k}) \\ &= (2)(4) + (-1)(-7) + (1)(15) \\ &= 8 + 7 + 15 \\ &= 30 \end{aligned} $$ Therefore: $\boxed{30}$ <div class="key-concept"><strong>Key Concept:</strong> $(\vec{a}+2\vec{b})\times\vec{c}=3(\vec{c}\times\vec{a})=-3(\vec{a}\times\vec{c})$. Rearranging: $(\vec{a}+2\vec{b})\times\vec{c}+3(\vec{a}\times\vec{c})=0\Rightarrow(2\vec{b}+4\vec{a})\times\vec{c}=0\Rightarrow\vec{c}=\lambda(4\vec{a}+2\vec{b})$.</div> <div class="trap-box"><strong>Trap:</strong> $4\vec{a}+2\vec{b}=8\hat{i}-14\hat{j}+30\hat{k}$. $\vec{a}\cdot\vec{c}=\lambda(8+42+210)=260\lambda=130\Rightarrow\lambda=1/2$. $\vec{c}=4\hat{i}-7\hat{j}+15\hat{k}$. $\vec{b}\cdot\vec{c}=8+7+15=30$.</div>