Question

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:

• A. \(4\sqrt{2}\)
• B. \(12\)
• C. \(24\)
• D. \(12\sqrt{2}\)

Hint:

Zero cross product means \(\vec{c}\) is parallel to \(\hat{i}+2\hat{j}+5\hat{k}\); use \(|\vec{c}|^2 = 60\) to fix the scalar.

Answer 1

The Correct Option is D

\(\vec{c}\times(\hat{i}+2\hat{j}+5\hat{k}) = \vec{0}\) means \(\vec{c} = \lambda(\hat{i}+2\hat{j}+5\hat{k})\).
\(|\vec{c}|^2 = \lambda^2(1+4+25) = 30\lambda^2 = 60\), so \(\lambda = \pm\sqrt{2}\).
\(\vec{c}\cdot(-7\hat{i}+2\hat{j}+3\hat{k}) = \lambda(-7 + 4 + 15) = 12\lambda = \pm12\sqrt{2}\).
A value is \(12\sqrt{2}\).
Answer: (4) \(12\sqrt{2}\)