JEE Mains · Physics · STD 11 - 3.1 vectors
If \(\vec{A}\) and \(\vec{B}\) are two vectors satisfying the relation \(\vec{A} . \vec{B}=[\vec{A} \times \vec{B}]\). Then the value of \([\vec{A}-\vec{B}]\). will be :
- A \(\sqrt{A^{2}+B^{2}-\sqrt{2} A B}\)
- B \(\sqrt{A^{2}+B^{2}}\)
- C \(\sqrt{A^{2}+B^{2}+\sqrt{2} A B}\)
- D \(\sqrt{A^{2}+B^{2}+\sqrt{2} A B}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{A^{2}+B^{2}-\sqrt{2} A B}\)
Step-by-step Solution
Detailed explanation
\(\vec{A} \vec{B}=|\vec{A} \times \vec{B}|\) \(A B \cos \theta=A B \sin \theta \Rightarrow \theta=45^{\circ}\) \(|\vec{A}-\vec{B}|=\sqrt{A^{2}+B^{2}-2 A B \cos 45^{\circ}}\) \(=\sqrt{A^{2}+B^{2}-\sqrt{2} A B}\)
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