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Motion in A Plane question

2003 · Q153
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Motion in A Plane question

2003 · Q153

NEETPhysicsMotion in A PlaneMCQ+4 / −1
The vector sum of two forces is perpendicular to their vector differences. In that case, the forces
  1. A
    are equal to each other
  2. B
    are equal to each other in magnitude
  3. C
    are not equal to each other in magnitude
  4. D
    cannot be predicted.
View written solutionFree

Correct answer: B

Given : (F1→+F2→)⊥(F1→−F2→)\left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right) \bot \left( {\overrightarrow {{F_1}} - \overrightarrow {{F_2}} } \right)(F1​​+F2​​)⊥(F1​​−F2​​)

(F1→+F2→).(F1→−F2→)=0\left( {\overrightarrow {{F_1}} + \overrightarrow {{F_2}} } \right).\left( {\overrightarrow {{F_1}} - \overrightarrow {{F_2}} } \right) = 0(F1​​+F2​​).(F1​​−F2​​)=0

F12−F22−F1→.F2→+F2→.F1→=0F_1^2 - F_2^2 - \overrightarrow {{F_1}} .\overrightarrow {{F_2}} + \overrightarrow {{F_2}} .\overrightarrow {{F_1}} = 0F12​−F22​−F1​​.F2​​+F2​​.F1​​=0

⇒F12=F22 \Rightarrow F_1^2 = F_2^2⇒F12​=F22​   (i.e. F1, F2 are equal to each other in magnitude.)

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