Question

Let $\vec{a} = 3\hat{i} + 2\hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$, and $\vec{c}$ be a vector such that

$(\vec{a} + \vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k}$ and $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3.$Then $|\vec{c}|^2$ is equal to \\_\\_\\_\\_\\_.

Answer 1

Calculate $(\vec{a} + \vec{b}) \times \vec{c}$:

$\vec{a} + \vec{b} = (3 + 5) \hat{i} + (2 - 1) \hat{j} + (1 + 3) \hat{k} = 8 \hat{i} + \hat{j} + 4 \hat{k}.$

Then,

$(\vec{a} + \vec{b}) \times \vec{c} = 2 (\vec{a} \times \vec{b}) + 24 \hat{j} - 6 \hat{k}.$

Solving for $\vec{c}$ using the vector equation and substituting values, we get:

$|\vec{c}|^2 = 25 + 9 + 4 = 38.$

Therefore, the answer is: 38.