Question

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

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

Answer 1

Answer: (A)

$A\times B=AB\sin \theta$
$A.B=AB\cos \theta$
$|A\times B|=\sqrt{3}A.B$
$|A\times B|=|A||B|\sin \theta$
$=AB\sin \theta$
$A.B=|A||B|\cos \theta =AB\cos \theta$
$AB\sin \theta =\sqrt{2}AB\cos \theta$
$\tan \theta =\sqrt{3},\theta =60{}^{\circ }$
Now ${(A+B)}^2=A^2+B^2+2A.B$
$=A^2+B^2+2AB$
$=A^2+B^2+AB\frac{1}{2}$
$=A^2+B^2+2A{B}^2$
or $|A+B|=(A^2+B^2+A{B}^{1/2})$