Question

If $|\overrightarrow{ A } \times \overrightarrow{ B }|=\sqrt{3} \overrightarrow{ A } .\overrightarrow{ B }$, then the value of $|\overrightarrow{ A } \times \overrightarrow{ B }|$ is :

• A. $\left(A^{2}+B^{2}+A B\right)^{1 / 2}$
• B. $ {{\left( {{A}^{2}}+{{B}^{2}}+\frac{AB}{\sqrt{3}} \right)}^{1/2}} $
• C. $ A+B $
• D. $\left(A^{2}+B^{2}+\sqrt{3} A B\right)^{1 / 2}$

Answer 1

Answer: (A)

$\vec{A} \times \vec{B}=A B \sin \theta$
and $\vec{A} . \vec{B}=A B \cos \theta$.
Given, $|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} . \vec{B} \ldots$ (i)
but $|\vec{A} \times \vec{B}|=|\vec{A}||\vec{B}| \sin \theta=A B \sin \theta$
and $\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta=A B \cos \theta$ Putting these values in Eq. (i), we get
$A B \sin \theta=\sqrt{3} A B \cos \theta$
or $\tan \theta=\sqrt{3} $
$\therefore \theta=60^{\circ}$
The addition of vector $\vec{A}$ and $\vec{B}$ can be given by the law of parallelogram.
$\therefore |\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos 60^{\circ}}$
$=\sqrt{A^{2}+B^{2}+2 A B \times \frac{1}{2}} $
$=\left(A^{2}+B^{2}+A B\right)^{1 / 2}$