Given, force vectors P⊥Q, i.e. θ=90° Let resultant of P + Q = x and resultant of P − Q = y ∴x2=P2+Q2+2PQcos90°=P2+Q2 ...(i) y2=P2+Q2−2PQcos90°=P2+Q2 ...(ii) When θ1 is the angle between (P + Q) and (P − Q), then, their resultant is given by |x+y|=√3(P2+Q2) ∴√x2+y2+2xycosθ1=√3P2+3Q2 ...(iii) Substituting the values of x2 and y2 in Eq. (iii), we get
P2+Q2+P2+Q2+2√P2+Q2√P2+Q2cosθ1=3P2+3Q2
2P2+2Q2+(2P2+2Q2)cosθ1=3P2+3Q2 ⇒cosθ1=
P2+Q2
2(P2+Q2)
=
1
2
∴θ1=60° When θ2 is the angle between (P + Q) and (P − Q), then the magnitude of their resultant is given by |x+y|=√2(P2+Q2) x2+y2+2xycosθ2=2P2+2Q2 ...(iv) Substituting the values, we get P2+Q2+P2+Q2+2(P2+Q2)cosθ2=2P2+2Q2 cos θ2=0 ⇒θ2=90° So, θ1=60° and θ2=90° and θ1<θ2 Hence, both statement I and statement II are true.