The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

Question

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces (A) are equal to each other (B) are equal to each other in magnitude (C) are not equal to each other in magnitude (D) cannot be predicted

Tardigrade Admin · Accepted Answer

Given: ( vecF1+ vecF2) perp( vecF1- vecF2) ∴( vecF1+ vecF2) ⋅( vecF1- vecF2)=0 F12-F22- vecF1 ⋅ vecF2+ vecF2 ⋅ vecF1=0 ⇒ F12=F22 i.e. F1, F2 are equal to each other in magnitude.

Q. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

are equal to each other

are equal to each other in magnitude

are not equal to each other in magnitude

cannot be predicted

Given : $(F_{1} + F_{2}) ⊥ (F_{1} - F_{2})$
$∴ (F_{1} + F_{2}) \cdot (F_{1} - F_{2}) = 0$
$F_{1}^{2} - F_{2}^{2} - F_{1} \cdot F_{2} + F_{2} \cdot F_{1} = 0$
$\Rightarrow F_{1}^{2} = F_{2}^{2}$
i.e. $F_{1}, F_{2}$ are equal to each other in magnitude.

Q. The vector sum of two forces is perpendicular to their vector differences. In that case, the forces

are equal to each other

are equal to each other in magnitude

are not equal to each other in magnitude

cannot be predicted

Given : (F1​+F2​)⊥(F1​−F2​) ∴(F1​+F2​)⋅(F1​−F2​)=0 F12​−F22​−F1​⋅F2​+F2​⋅F1​=0 ⇒F12​=F22​ i.e. F1​,F2​ are equal to each other in magnitude.

Given : $(F_{1} + F_{2}) ⊥ (F_{1} - F_{2})$
$∴ (F_{1} + F_{2}) \cdot (F_{1} - F_{2}) = 0$
$F_{1}^{2} - F_{2}^{2} - F_{1} \cdot F_{2} + F_{2} \cdot F_{1} = 0$
$\Rightarrow F_{1}^{2} = F_{2}^{2}$
i.e. $F_{1}, F_{2}$ are equal to each other in magnitude.