The resultant of forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$. If $\overrightarrow{\mathrm{Q}}$ is doubled then $\overrightarrow{\mathrm{R}}$ is doubled. If the direction of $\overrightarrow{\mathrm{Q}}$ is reversed, then $\overrightarrow{\mathrm{R}}$ is again doubled. Then $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2$ is

• A

  $2: 3: 1$
• B

  $3: 1: 1$
• Correct answer
  C

  $2: 3: 2$
• D

  $1: 2: 3$

Solution

$\mathrm{R}^2=\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta$
$4 \mathrm{R}^2=\mathrm{P}^2+4 \mathrm{Q}^2+4 \mathrm{PQ} \cos \theta$
$4 \mathrm{R}^2=\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{PQ} \cos \theta$
$\mathrm{On}(1)+(2), 5 \mathrm{R}^2=2 \mathrm{P}^2+2 \mathrm{Q}^2$
$\mathrm{On}(3) \times 2+(2), 12 \mathrm{R}^2=3 \mathrm{P}^2+6 \mathrm{Q}^2$
$2 \mathrm{P}^2+2 \mathrm{Q}^2-5 \mathrm{R}^2=0$
$3 \mathrm{P}^2+6 \mathrm{Q}^2-12 \mathrm{R}^2=0$
$\frac{\mathrm{P}^2}{-24+30}=\frac{\mathrm{Q}^2}{24-15}=\frac{\mathrm{R}^2}{12-6}$ $\frac{\mathrm{P}^2}{6}=\frac{\mathrm{Q}^2}{9}=\frac{\mathrm{R}^2}{6}$ or $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2=2: 3: 2$

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