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Motion in A Plane question

2006 · Q156
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Motion in A Plane question

2006 · Q156

NEETPhysicsMotion in A PlaneMCQ+4 / −1
The vectors A→\overrightarrow AA and B→\overrightarrow BB are such that ∣A→+B→∣=∣A→−B→∣.\left| {\overrightarrow A + \overrightarrow B } \right| = \left| {\overrightarrow A - \overrightarrow B } \right|.​A+B​=​A−B​. The angle between the two vectors is
  1. A
    45o
  2. B
    90o
  3. C
    60o
  4. D
    75o
View written solutionFree

Correct answer: B

∣A→+B→∣2=∣A→−B→∣2{\left| {\overrightarrow A + \overrightarrow B } \right|^2} = {\left| {\overrightarrow A - \overrightarrow B } \right|^2}​A+B​2=​A−B​2

=∣A→∣2+∣B→∣2+2A→B→=A2+B2+2ABcos⁡θ= {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} + 2\overrightarrow A \overrightarrow B = {A^2} + {B^2} + 2AB\cos \theta=​A​2+​B​2+2AB=A2+B2+2ABcosθ

=∣A→−B→∣2=∣A→∣2+∣B→∣2−2A→.B→= {\left| {\overrightarrow A - \overrightarrow B } \right|^2} = {\left| {\overrightarrow A } \right|^2} + {\left| {\overrightarrow B } \right|^2} - 2\overrightarrow A .\overrightarrow B=​A−B​2=​A​2+​B​2−2A.B

=A2+B2−2ABcos⁡θ= {A^2} + {B^2} - 2AB\cos \theta=A2+B2−2ABcosθ

So, A2 + B2 + 2AB cos θ\theta θ

= A2 + B2 – 2AB cos θ\theta θ

4AB cosθ\theta θ = 0 ⇒\Rightarrow⇒ cosθ\theta θ = 0

θ\theta θ = 90º

So, angle between A & B is 90º.

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