The resultant of two vectors is perpendicular to first vector of magnitude $6 \mathrm{N}$. If the resultant has magnitude $6\sqrt{3}\text{ N}$, then magnitude of second vector is

• A

  $6\sqrt{2}\mathrm{ N}$
• Correct answer
  B

  $12 \mathrm{N}$
• C

  $9\sqrt{3} \mathrm{N}$
• D

  $6\sqrt{3}\mathrm{ N}$

Solution

$\mathrm{B\; Cos \theta }=6$

$\mathrm{B\; Sin} \mathrm{\theta }=6\sqrt{3}$

$\text{B}=\sqrt{6^2+{\left(6\sqrt{3}\right)}^2}=12$

Hence, the correct answer is option: 'B'.