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:

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