If |vec A times vec B| = sqrt 3 vec A.vec B, | vedclass

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Question

(d) $|\,\overrightarrow A 	imes \overrightarrow B |\, = \sqrt 3 (\overrightarrow A .\overrightarrow B )$

$AB\sin 	heta = \sqrt 3 AB\cos 	heta $$

Vedclass · Accepted Answer

${({A^2} + {B^2} + AB)^{1/2}}$

Vedclass · Answer

(d) $|\,\overrightarrow A 	imes \overrightarrow B |\, = \sqrt 3 (\overrightarrow A .\overrightarrow B )$

$AB\sin 	heta = \sqrt 3 AB\cos 	heta $$ \Rightarrow $$	an 	heta = \sqrt 3 $ $&rArr;$ $	heta = 60^\circ $

Now $|\overrightarrow R |\, = \,|\overrightarrow A + \overrightarrow B |\, = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $

$ = \sqrt {{A^2} + {B^2} + 2AB\left( {\frac{1}{2}} ight)} $

$ = {({A^2} + {B^2} + AB)^{1/2}}$

If $|\vec A \times \vec B| = \sqrt 3 \vec A.\vec B,$ then the value of$|\vec A + \vec B|$ is

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