If A 2 + B 2 represents the magnitude of resultant of two vectors A → + B → and A → - B → , then the angle between two vectors is

A 2 + B 2 = A + B 2 + A - B 2 + 2 A + B A - B . c o s θ ⇒ A 2 + B 2 = 2 A 2 + B 2 + 2 A 2 - B 2 . c o s θ ⇒ c o s θ = - A 2 + B 2 2 A 2 - B 2 ⇒ θ = c o s - 1 - A 2 + B 2 2 A 2 - B 2

Physics - Vectors Question with Solution | TestHub

PhysicsVectorsAddition/SubtractionMedium2 minPYQ_2019

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If $\sqrt{A^{2} + B^{2}}$ represents the magnitude of resultant of two vectors $(\vec{A} + \vec{B})$ and $(\vec{A} - \vec{B}),$ then the angle between two vectors is

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Answer:

Solution:

$\sqrt{A^{2} + B^{2}} = \sqrt{{(A + B)}^{2} + {(A - B)}^{2} + 2 (A + B) (A - B) . c o s θ}$
$\Rightarrow A^{2} + B^{2} = 2 (A^{2} + B^{2}) + 2 (A^{2} - B^{2}) . c o s θ$
$\Rightarrow c o s θ = \frac{- (A^{2} + B^{2})}{2 (A^{2} - B^{2})}$
$\Rightarrow θ = {c o s}^{- 1} [\frac{- (A^{2} + B^{2})}{2 (A^{2} - B^{2})}]$

Stream:MHT_CETSubject:PhysicsTopic:VectorsSubtopic:Addition/Subtraction

⏱ 2mℹ️ Source: PYQ_2019

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