Two vectors P → and Q → have equal magnitudes. If the magnitude of P → + Q → is n times the magnitude of P → - Q → , then angle between P → and Q → is

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PhysicsVectorsAddition/SubtractionMedium2 minPYQ_2021

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Two vectors $\vec{P}$ and $\vec{Q}$ have equal magnitudes. If the magnitude of $\vec{P} + \vec{Q}$ is $n$ times the magnitude of $\vec{P} - \vec{Q},$ then angle between $\vec{P}$ and $\vec{Q}$ is

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Answer:

Solution:

$| \vec{P} | = | \vec{Q} | = x \dots (i)$

$| \vec{P} + \vec{Q} | = n | \vec{P} - \vec{Q} |$

$P^{2} + Q^{2} + 2 P Q \cos θ = n^{2} (P^{2} + Q^{2} - 2 P Q \cos θ)$

Using $(i)$ in above equation,

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

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

Stream:JEESubject:PhysicsTopic:VectorsSubtopic:Addition/Subtraction

⏱ 2mℹ️ Source: PYQ_2021

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