<p>If \(|\vec{c}|^2 = 60\) and \(\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}\), then a value of \(\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})\) is</p>
Step-by-Step Solution
Key Concept: If \(\vec{c} \times \vec{a} = \vec{0}\), then \(\vec{c}\) is parallel to \(\vec{a}\), so \(\vec{c} = k\vec{a}\) for some scalar k. Use the magnitude condition to find k, then compute the dot product.
Step 1: Since \(\vec{c} \times (\hat{i} + 2\hat{j} + 5\hat{k}) = \vec{0}\), the vector \(\vec{c}\) must be parallel to \((\hat{i} + 2\hat{j} + 5\hat{k})\). Step 2: Write \(\vec{c} = k(\hat{i} + 2\hat{j} + 5\hat{k})\) for some scalar k. Step 3: Use the condition \(|\vec{c}|^2 = 60\):
\(|k(\hat{i} + 2\hat{j} + 5\hat{k})|^2 = 60\)
\(k^2(1 + 4 + 25) = 60\)
\(k^2 \cdot 30 = 60\)
\(k^2 = 2\)
\(k = \pm\sqrt{2}\) Step 4: Calculate \(\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})\):
\(\vec{c} = \pm\sqrt{2}(\hat{i} + 2\hat{j} + 5\hat{k})\)
\(\vec{c} \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) = \pm\sqrt{2}[1(-7) + 2(2) + 5(3)]\)
\(= \pm\sqrt{2}[-7 + 4 + 15]\)
\(= \pm\sqrt{2} \cdot 12\)
\(= \pm 12\sqrt{2}\) ∴ Answer: B
Correct Answer: B