A) 1
B) \[\frac{3}{2}\]
C) 2
D) 4
Correct Answer: C
Solution :
\[{{R}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\,\,\cos \theta \] Þ \[{{(\sqrt{7}Q)}^{2}}={{P}^{2}}+{{Q}^{2}}+2PQ\cos 60{}^\circ \] Þ \[7{{Q}^{2}}={{P}^{2}}+Q+PQ\] Þ \[{{P}^{2}}+PQ-6{{Q}^{2}}=0\] Þ \[{{P}^{2}}+3PQ-2PQ-6{{Q}^{2}}=0\] Þ \[P(P+3Q)-2Q(P+3Q)=0\] Þ \[(P-2Q)(P+3Q)=0\] Þ \[P-2Q=0\] or \[P+3Q=0\] From \[P-2Q=0\] Þ \[\frac{P}{Q}=2\].You need to login to perform this action.
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