P→,Q→,R→,S→ are vector of equal magnitude. If P→ + Q→ - R→ = 0 angle between P→ and Q→ is θ1 . If P→ + Q→ - S→ = 0 angle between P→ and S→ is θ2. The ratio of θ1 to θ2 is

P→+Q→=R→;P→+Q→=S→,tan θ2=Qsinθ1P+Qcosθ1 tan θ2=sin θ11+cosθ1 sin θ2cos θ2=sin θ11+cos θ1 sin θ2= sin(θ1-θ2) θ2=θ1-θ2 ⇒θ2=θ12

P→,Q→,R→,S→ are vector of equal magnitude. If P→ + Q→ - R→ = 0 angle between P→ and Q→ is θ1 . If P→ + Q→ - S→ = 0 angle between P→ and S→ is θ2. The ratio of θ1 to θ2 is

P→+Q→=R→;P→+Q→=S→,tan θ2=Qsinθ1P+Qcosθ1 tan θ2=sin θ11+cosθ1 sin θ2cos θ2=sin θ11+cos θ1 sin θ2= sin(θ1-θ2) θ2=θ1-θ2 ⇒θ2=θ12

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$\vec{P}, \vec{Q}, \vec{R}, \vec{S}$ are vector of equal magnitude. If $\vec{P} + \vec{Q} - \vec{R} = 0$ angle between $\vec{P}$ and $\vec{Q}$ is $θ_{1}$ . If $\vec{P} + \vec{Q} - \vec{S} = 0$ angle between $\vec{P}$ and $\vec{S}$ is $θ_{2}$ . The ratio of $θ_{1}$ to $θ_{2}$ is

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1 : 2

2 : 1

1 : 1

1: 3

answer is B.

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Detailed Solution

$\vec{P} + \vec{Q} = \vec{R}; \vec{P} + \vec{Q} = \vec{S},$

$\tan θ_{2} = \frac{Q \sin θ_{1}}{P + Q \cos θ_{1}} \tan θ_{2} = \frac{\sin θ_{1}}{1 + \cos θ_{1}} \frac{\sin θ_{2}}{\cos θ_{2}} = \frac{\sin θ_{1}}{1 + \cos θ_{1}} \sin θ_{2} = \sin (θ_{1} - θ_{2}) θ_{2} = θ_{1} - θ_{2} \Rightarrow θ_{2} = \frac{θ_{1}}{2}$