MCQ
If $|\vec A \times \vec B| = \sqrt 3 \vec A.\vec B,$ then the value of$|\vec A + \vec B|$ is
- A${\left( {{A^2} + {B^2} + \frac{{AB}}{{\sqrt 3 }}} \right)^{1/2}}$
- B$A + B$
- C${({A^2} + {B^2} + \sqrt 3 AB)^{1/2}}$
- ✓${({A^2} + {B^2} + AB)^{1/2}}$
✓
Answer
Correct option: D.
${({A^2} + {B^2} + AB)^{1/2}}$
d
(d) $|\,\overrightarrow A \times \overrightarrow B |\, = \sqrt 3 (\overrightarrow A .\overrightarrow B )$
(d) $|\,\overrightarrow A \times \overrightarrow B |\, = \sqrt 3 (\overrightarrow A .\overrightarrow B )$
$AB\sin \theta = \sqrt 3 AB\cos \theta $$ \Rightarrow $$\tan \theta = \sqrt 3 $ $⇒$ $\theta = 60^\circ $
Now $|\overrightarrow R |\, = \,|\overrightarrow A + \overrightarrow B |\, = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $
$ = \sqrt {{A^2} + {B^2} + 2AB\left( {\frac{1}{2}} \right)} $
$ = {({A^2} + {B^2} + AB)^{1/2}}$
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