Two vectors X → and Y → have equal magnitude. The magnitude of ( X → - Y → ) is n times the magnitude of ( X → + Y → ) . The angle between X → and Y → is :

Given X = Y X 2 + Y 2 - 2 × Y cos θ = n X 2 + Y 2 + 2 × Y cos θ Square both sides 2 X 2 ( 1 - cos θ ) = n 2 · 2 X 2 ( 1 + cos θ ) 1 - cos θ = n 2 + n 2 cos θ cos θ = 1 - n 2 1 + n 2 θ = cos - 1 n 2 - 1 - n 2 - 1

Physics - Vectors Question with Solution | TestHub

PhysicsVectorsMiscellaneousEasy2 minPYQ_2021

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Two vectors $\vec{X}$ and $\vec{Y}$ have equal magnitude. The magnitude of $(\vec{X} - \vec{Y})$ is $n$ times the magnitude of $(\vec{X} + \vec{Y})$ . The angle between $\vec{X}$ and $\vec{Y}$ is :

Options:

Answer:

Solution:

Given $X = Y$

$\sqrt{X^{2} + Y^{2} - 2 \times Y \cos θ}$

$= n \sqrt{X^{2} + Y^{2} + 2 \times Y \cos θ}$

Square both sides

$2 X^{2} (1 - \cos θ) = n^{2} \cdot 2 X^{2} (1 + \cos θ)$

$1 - \cos θ = n^{2} + n^{2} \cos θ$

$\cos θ = \frac{1 - n^{2}}{1 + n^{2}}$

$θ = \cos^{- 1} [\frac{n^{2} - 1}{- n^{2} - 1}]$

Stream:JEESubject:PhysicsTopic:VectorsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2021

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