Question:
Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k},\ \vec{c} = \hat{j} - \hat{k}\) and a vector \(\vec{b}\) be such that \(\vec{a} \times \vec{b} = \vec{c}\) and \(\vec{a} \cdot \vec{b} = 3\). Then \(|\vec{b}|\) equals :
Let \(\vec{a} = \hat{i} + \hat{j} + \hat{k},\ \vec{c} = \hat{j} - \hat{k}\) and a vector \(\vec{b}\) be such that \(\vec{a} \times \vec{b} = \vec{c}\) and \(\vec{a} \cdot \vec{b} = 3\). Then \(|\vec{b}|\) equals :
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Take the cross product of \(\vec{a}\) with both sides of \(\vec{a}\times\vec{b}=\vec{c}\) and use the vector triple product expansion to find \(\vec{b}\).
Updated On: Nov 11, 2026
- \(\frac{11}{3}\)
- \(\frac{11}{\sqrt{3}}\)
- \(\sqrt{\frac{11}{3}}\)
- \(\frac{\sqrt{11}}{3}\)
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The Correct Option is C
Solution and Explanation
\(\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times \vec{c}\).
Left side: \((\vec{a}\cdot\vec{b})\vec{a} - (\vec{a}\cdot\vec{a})\vec{b} = 3\vec{a} - 3\vec{b}\) (since \(\vec{a}\cdot\vec{a} = 3\)).
Right side: \(\vec{a}\times\vec{c} = (\hat i+\hat j+\hat k)\times(\hat j-\hat k) = -2\hat i + \hat j + \hat k\).
So \(3\vec{b} = 3\vec{a} - (-2\hat i + \hat j + \hat k) = 5\hat i + 2\hat j + 2\hat k\), giving \(\vec{b} = \frac{1}{3}(5\hat i + 2\hat j + 2\hat k)\).
Check: \(\vec{a}\cdot\vec{b} = \frac{5+2+2}{3} = 3\). OK.
\(|\vec{b}|^2 = \frac{25+4+4}{9} = \frac{11}{3}\), so \(|\vec{b}| = \sqrt{\frac{11}{3}}\).
Answer: (3) \(\sqrt{\frac{11}{3}}\)
Left side: \((\vec{a}\cdot\vec{b})\vec{a} - (\vec{a}\cdot\vec{a})\vec{b} = 3\vec{a} - 3\vec{b}\) (since \(\vec{a}\cdot\vec{a} = 3\)).
Right side: \(\vec{a}\times\vec{c} = (\hat i+\hat j+\hat k)\times(\hat j-\hat k) = -2\hat i + \hat j + \hat k\).
So \(3\vec{b} = 3\vec{a} - (-2\hat i + \hat j + \hat k) = 5\hat i + 2\hat j + 2\hat k\), giving \(\vec{b} = \frac{1}{3}(5\hat i + 2\hat j + 2\hat k)\).
Check: \(\vec{a}\cdot\vec{b} = \frac{5+2+2}{3} = 3\). OK.
\(|\vec{b}|^2 = \frac{25+4+4}{9} = \frac{11}{3}\), so \(|\vec{b}| = \sqrt{\frac{11}{3}}\).
Answer: (3) \(\sqrt{\frac{11}{3}}\)
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