Question

The resultant of $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$. If $\overrightarrow{Q}$ isdoubled, $\overrightarrow{R}$ is doubled; when $\overrightarrow{Q}$ is reversed, $\overrightarrow{R}$ is again doubled. Find $P:Q:R$.

Answer 1

Let $\theta$ be the angle between $\overrightarrow{P}$ and $\overrightarrow{Q}$. Then

$R^2=|\overrightarrow{P}+\overrightarrow{P}|=P^2+Q^2+2PQ\cos \theta$ ...(i)

If $\overrightarrow{Q}$ is doubled, $\overrightarrow{R}$ is doubled. That means, the magnitude of resultant of $2\overrightarrow{Q}$ and $\overrightarrow{P}$ is

$(2R{)}^2=P^2+(2Q{)}^2+2P\left(PQ\right)\cos \theta$

This yields $4{\overrightarrow{R}}^2=P^2+4Q^2+4PQ\cos \theta$ ...(ii)

When $\overrightarrow{Q}$ is reversed, $\overrightarrow{R}$ is doubled. Hence, the magnitude of resultant of $\overrightarrow{P}$ and $(-\overrightarrow{Q})$ is $2R$.

Then $(2R{)}^2=P^2+Q^2+2PQ\cos (180-\theta )$ ...(iii)

This yields $4R^2=P^2+Q^2-2PQ\cos \theta$ ...(iii)

(i) - (ii) yields $PQ\cos \theta =\frac{-3R^2}{4}$ ...(iv)

(i) + (iii) yields $P^2+Q^2=\frac{5R^2}{2}$ ...(v)

(ii) + (iv) yields $P^2+4Q^2=7R^2$ ...(vi)

Solving (v) and (vi), we obtain $Q=\sqrt{\frac{3}{2}}R$ and $P=R$.

Hence, $P:Q:R=\sqrt{2}:\sqrt{3}:\sqrt{2}$