Question

Let \(\vec{a} = \hat{i} - 3\hat{j} + 7\hat{k}, \quad \vec{b} = 2\hat{i} - \hat{j} + \hat{k}, \quad \text{and} \quad \vec{c} \text{ be a vector such that}\) \((\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a}).\)
If \(\vec{a} \cdot \vec{c} = 130\), then \(\vec{b} \cdot \vec{c}\) is equal to \(\_\_\_\_\_\_\_\_ .\)

Answer 1

Correct Answer: 30

Given:

\[ (\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a}) \]

This implies:

\[ 2\vec{b} \times \vec{c} = 0 \quad (\text{since cross product is zero}) \]

Also:

\[ \vec{c} = \lambda (\vec{a} + 2\vec{b}) = \lambda (8\hat{i} - 14\hat{j} + 30\hat{k}) \]

Given:

\[ \vec{a} \cdot \vec{c} = 130 \]

Substitute values:

\[ 8\lambda + 42\lambda + 210\lambda = 130 \quad \implies \quad \lambda = \frac{1}{2} \]

Therefore:

\[ \vec{c} = 4\hat{i} - 7\hat{j} + 15\hat{k} \]

Now:

\[ \vec{b} \cdot \vec{c} = 8 + 7 + 15 = 30 \]

Answer 2

Step 1: Rewrite the vector equation
Given \((\vec{a} + 2\vec{b}) \times \vec{c} = 3(\vec{c} \times \vec{a})\). Since \(\vec{c} \times \vec{a} = -(\vec{a} \times \vec{c})\), we have:
\[ (\vec{a} + 2\vec{b}) \times \vec{c} \;=\; -3(\vec{a} \times \vec{c}) \;\;\Longrightarrow\;\; \vec{a}\times\vec{c} + 2\,\vec{b}\times\vec{c} \;=\; -3\,\vec{a}\times\vec{c} \;\;\Longrightarrow\;\; \vec{b}\times\vec{c} \;=\; -2\,\vec{a}\times\vec{c}. \]

Step 2: Cross both sides with \(\vec{c}\)
Use \((\vec{u}\times\vec{v})\times\vec{w}=\vec{v}(\vec{u}\cdot\vec{w})-\vec{u}(\vec{v}\cdot\vec{w})\) with \(\vec{v}=\vec{w}=\vec{c}\):
\[ (\vec{b}\times\vec{c})\times\vec{c} = \vec{c}(\vec{b}\cdot\vec{c}) - \vec{b}\,\lVert\vec{c}\rVert^2, \qquad (\vec{a}\times\vec{c})\times\vec{c} = \vec{c}(\vec{a}\cdot\vec{c}) - \vec{a}\,\lVert\vec{c}\rVert^2. \] Thus \[ \vec{c}(\vec{b}\cdot\vec{c}) - \vec{b}\,\lVert\vec{c}\rVert^2 = -2\big[\vec{c}(\vec{a}\cdot\vec{c}) - \vec{a}\,\lVert\vec{c}\rVert^2\big]. \] Rearrange the \(\vec{c}\)-terms and the others: \[ \vec{c}\big(\vec{b}\cdot\vec{c} + 2\,\vec{a}\cdot\vec{c}\big) = \lVert\vec{c}\rVert^2(\vec{b}+2\vec{a}). \] Therefore \(\vec{c}\) is parallel to \(\vec{b}+2\vec{a}\). Hence for some scalar \(k\), \[ \vec{c} = k(\vec{b}+2\vec{a}). \]

Step 3: Use the given dot product \(\vec{a}\cdot\vec{c}=130\) to find \(k\)
Compute \(\vec{b}+2\vec{a}\): with \(\vec{a}=\langle 1,-3,7\rangle\) and \(\vec{b}=\langle 2,-1,1\rangle\), \[ \vec{b}+2\vec{a} = \langle 2, -1, 1\rangle + \langle 2, -6, 14\rangle = \langle 4, -7, 15\rangle. \] Then \[ \vec{a}\cdot(\vec{b}+2\vec{a}) = \langle 1,-3,7\rangle\cdot\langle 4,-7,15\rangle = 4 + 21 + 105 = 130. \] Since \(\vec{c}=k(\vec{b}+2\vec{a})\) and \(\vec{a}\cdot\vec{c}=130\), we get \[ 130 = \vec{a}\cdot\vec{c} = k\,\vec{a}\cdot(\vec{b}+2\vec{a}) = k\cdot 130 \;\;\Longrightarrow\;\; k=1. \] Thus \(\vec{c}=\langle 4,-7,15\rangle\).

Step 4: Compute \(\vec{b}\cdot\vec{c}\)
\[ \vec{b}\cdot\vec{c} = \langle 2,-1,1\rangle\cdot\langle 4,-7,15\rangle = 8 + 7 + 15 = 30. \]

Final answer

30