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Motion in A Plane question

2004 · Q141
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Motion in A Plane question

2004 · Q141

NEETPhysicsMotion in A PlaneMCQ+4 / −1
If ∣A→×B→∣=3A→.B→\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \overrightarrow A .\overrightarrow B​A×B​=3​A.B then the value of ∣A→+B→∣\left| {\overrightarrow A + \overrightarrow B } \right|​A+B​ is
  1. A
    (A2 + b2 + AB)1/2
  2. B
    (A2+B2+AB3)1/2{\left( {{A^2} + {B^2} + {{AB} \over {\sqrt 3 }}} \right)^{1/2}}(A2+B2+3​AB​)1/2
  3. C
    A + B
  4. D
    (A2 + B2 + 3{\sqrt 3 }3​AB)1/2.
View written solutionFree

Correct answer: A

∣A→×B→∣=ABsin⁡θ\left| {\overrightarrow A \times \overrightarrow B } \right| = AB\sin \theta ​A×B​=ABsinθ

A→B→=ABcos⁡θ\overrightarrow A \overrightarrow B = AB\cos \theta AB=ABcosθ

∣A→×B→∣=3A→B→\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \overrightarrow A \overrightarrow B ​A×B​=3​AB

⇒ABsin⁡θ=3ABcos⁡θ\Rightarrow AB\sin \theta = \sqrt 3 AB\cos \theta⇒ABsinθ=3​ABcosθ

⇒tan⁡θ=3\Rightarrow \tan \theta = \sqrt 3⇒tanθ=3​    ∴θ\therefore \theta∴θ = 60o

∴\therefore∴ ∣A→×B→∣=A2+B2+2ABcos⁡60∘\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} + 2AB\cos 60^\circ } ​A×B​=A2+B2+2ABcos60∘​

=A2+B2+AB= \sqrt {{A^2} + {B^2} + AB}=A2+B2+AB​ = (A2 + b2 + AB)1/2

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