996Let a be the angle between the forces P and Q. Now from the given conditions, we have
R2 + P2 + Q2 + 2PQ cos α …(1)
and (2R)2 + P2 + (2Q)2 + 2P(2Q)cos α
4R2 = P2 + 4Q2 + 4PQ cosα …(2)
and (2R)2 = P2 + (–Q)2 + 2P (–Q) cosα
=> 4R2 = P2+Q2 - 2PQ cosα …(3)
Adding (1) and (3), we get 2P2 + 2Q2 – 5R2 = 0 …(4)
Eliminating a from (2) and (3), we get
P2 + 2Q2 – 4R2 = 0 …(5)
From (4) and (5), we have
P2/–8+10 = Q2/–5+8 = R2/4–2 => P2/2 = Q2/3 = R2/2 => P:Q:R::√2: √3: √2.
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Akash Anand Excellent Work Upvote·0· Reply ·2013-06-19 23:55:10