Two forces P and Q have a resultant R. This resultant is doubled, either when Q is doubled, or when Q is reversed. Show that, `P:Q:R=sqrt(2):sqrt(3):sqrt(2)`.

Explanation

Let the angle between P and Q be `alpha`. So,
`R^2=P^2+Q^2+2PQ cos alpha`
`or, 2PQcos alpha=R^2-P^2-Q^2.... (1)`
When Q is doubled, `(2R)^2 =P^2+(2Q)^2+2P*2Q cos alpha`
or, `2PQ cos alpha =2R^2-1/2P^2-2Q^2 ……..(2)`
Again, when Q is reversed,
`(2R)^2=P^2+Q^2+2PQcos (180^@-alpha)`
`or, 2PQ cos alpha=-4R^2+P^2+Q^2 " "......(3)`
From (1) and (3) , `R^2-P^2-Q^2=-4R^2+P^2+Q^2`
`or, 2P^2+2Q^2-5R^2=0 " "......(4)`
From (1) and (2) , `R^2-P^2-Q^2=2R^2-1/2P^2-2Q^2`
`or, P^2-2Q^2+2R^2=0 " ".......(5)`
Adding (4) and (5) ,`3P^2-3R^2=0 or, P=R "Putting this in (5)"`
`P^2-2Q^2+2P^2=0 or,3P^2=2Q^2 or,Q=(sqrt(3))/(sqrt(2))P`
`P:Q:R=P:(sqrt(3))/(sqrt(2)):P=1:(sqrt(3))/(sqrt(2)):1=sqrt(2):sqrt(3):sqrt(2)`