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

Correct Answer: 38

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.