Two forces $F_{1}$ and $F_{2}$ acting on a body have a resultant $R_{1}$ if $F_{2}$ is doubled the $R_{2}$ new resultant forms are a right angle with $F_{1}$ Prove that $R_{1}$ and $F_{2}$ have the same magnitude.

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Solution

$R_{1} = F_{1} + F_{2} . . . e q 1$
$R_{2} = F_{1} + 2 * F_{2} . . . e q 2$
$F_{1} . R_{2} = 0$ (vector dot product for perpendicular/right angle vectors)... eq3
Prove $| F_{2} | = | R_{1} |$
Sub eq2 into eq3
$F_{1} . (F_{1} + 2 * F_{2}) = 0$
$F_{1} . F_{1} + 2 * F_{1} . F_{2} = 0$
$| F_{1} |^{2} = - 2 * F_{1} . F_{2}$
Get the magnitude of R1 from eq1
$| R_{1} |^{2} = (F_{1} + F_{2}) . (F_{1} + F_{2})$
$| R_{1} |^{2} = F_{1} . F_{1} + F_{2} . F_{2} + 2 * F_{1} . F_{2}$
$| R_{1} |^{2} = | F_{1} |^{2} + | F_{2} |^{2} + 2. F_{1} . F_{2}$
Sub in |F_1|^2
$| R_{1} |^{2} = | F_{1} |^{2} + | F_{2} |^{2} - | F_{1} |^{2}$
$| R_{1} |^{2} = | F_{2} |^{2}$
$| R_{1} | = | F_{2} |$

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