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JEE Mains · Maths · STD 12 - 10. vector algebra

Let \(\vec{a}\) and \(\vec{b}\) be two vectors such that \(|\vec{b}|=1\) and \(|\vec{b} \times \vec{a}|=2\). Then \(|(\vec{b} \times \vec{a})-\vec{b}|^2\) is equal to

  1. A \(3\)
  2. B \(5\)
  3. C \(1\)
  4. D \(4\)
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Answer & Solution

Correct Answer

(B) \(5\)

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Detailed explanation

\(|\vec{b}|=1 \&|\vec{b} \times \vec{a}|=2 \) \((\vec{b} \times \vec{a}) \cdot \vec{b}=\vec{b} \cdot(\vec{b} \times \vec{a})=0 \) \(|(\vec{b} \times \vec{a})-\vec{b}|^2=|\vec{b} \times \vec{a}|^2+|\vec{b}|^2 \) \(=4+1=5\)
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