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

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:

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