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

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$1 : \sqrt{3}$

2 : 1

1 : 2

1 : 1

answer is B.

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Detailed Solution

$| \vec{P} | = | \vec{Q} | = | \vec{R} | = | \vec{S} | given$

$\vec{P} + \vec{Q} - \vec{R} = 0; \vec{R} = \vec{P} + \vec{Q}$

$\Rightarrow R^{2} = P^{2} + Q^{2} + 2 PQcos θ_{1} \Rightarrow θ_{1} = + 120^{\circ}$

$Also given \vec{P} + \vec{Q} - \vec{S} = 0$

$\vec{P} - \vec{S} = - \vec{Q} = \vec{S} - \vec{P} = \vec{Q}$

$\Rightarrow Q^{2} = S^{2} + P^{2} + 2 \cdot SPcos (180 - θ_{2})$

$θ_{2} = 60^{\circ}; \frac{θ_{1}}{θ_{2}} = 2$

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