3-1.Vectors
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यदि $\overrightarrow{ A }$ और $\overrightarrow{ B }$ ऐसे दो सदिश हैं जो संबंध $\overrightarrow{ A } \cdot \overrightarrow{ B }=|\overrightarrow{ A } \times \overrightarrow{ B }|$ की पुष्टि करते है तब $|\overrightarrow{ A }-\overrightarrow{ B }|$ का मान होगा।

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

$\sqrt{A^{2}+B^{2}+\sqrt{2} A B}$

(JEE MAIN-2021)

Solution

$\vec{A} \vec{B}=|\vec{A} \times \vec{B}|$

$A B \cos \theta=A B \sin \theta \Rightarrow \theta=45^{\circ}$

$|\vec{A}-\vec{B}|=\sqrt{A^{2}+B^{2}-2 A B \cos 45^{\circ}}$

$=\sqrt{A^{2}+B^{2}-\sqrt{2} A B}$

Standard 11
Physics

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