Let \(\vec{a} = 6\hat{i} + 9\hat{j} + 12\hat{k}\), \(\vec{b} = a\hat{i} + 11\hat{j} - 2\hat{k}\), and \(\vec{c}\) be vectors such that \(\vec{a} \times \vec{c} = \vec{a} \times \vec{b}\). If \(\vec{a} \cdot \vec{c} = -12\) and \(\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5\), then \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})\) is equal to:
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Correct Answer: 11
Solution and Explanation
Step 1: Analyze the given condition.
\(\vec{a} \times (\vec{c} - \vec{b}) = 0 \implies \vec{a} || (\vec{c} - \vec{b})\).
Thus, \(\vec{c} - \vec{b} = k \vec{a}\), where \(k\) is a scalar.
Step 2: Solve for \(\vec{c}\).
\[ \vec{c} = \vec{b} + k\vec{a} = (a + 6k)\hat{i} + (11 + 9k)\hat{j} + (-2 + 12k)\hat{k}. \] Step 3: Use the given conditions.
1. \(\vec{a} \cdot \vec{c} = -12\): \[ 6(a + 6k) + 9(11 + 9k) + 12(-2 + 12k) = -12. \] Simplify to solve for \(a\) and \(k\).
2. Substituting \(\vec{c}\) into \(\vec{c} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 5\): \[ (a + 6k) - 2(11 + 9k) + (-2 + 12k) = 5. \] Solving these, we find \(\vec{c} = 23\hat{i} + 2\hat{j} - 14\hat{k}\).
Step 4: Compute \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k})\).
\[ \vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 23 + 2 - 14 = 11. \] Final Answer: \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 11\).
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