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Vector Addition by Analytical Method Physics Motion in Plane Physics Class 11th

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

Option 1 -

$s i n^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

Option 2 -

$c o s^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

Option 3 -

$s i n^{- 1} (\frac{n - 1}{n + 1})$

Option 4 -

$c o s^{- 1} (\frac{n - 1}{n + 1})$

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Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

$| \vec{P} | = | \vec{Q} | = x$

let the angle between $\vec{P} a n d \vec{Q}$ is , so according to question, we can write $^{}^{}^{}^{}$

^{$| \vec{P} + \vec{Q} | = n \times | \vec{P} - \vec{Q} | \Rightarrow \sqrt[]{P^{2} + Q^{2} + 2 P Q c o s θ} = n \times \sqrt[]{P^{2} + Q^{2} - 2 P Q c o s θ}$}

$\Rightarrow \sqrt[]{x^{2} + x^{2} + 2 x^{2} c o s θ} = n \times \sqrt[]{x^{2} + x^{2} - 2 x^{2} c o s θ}$

$\Rightarrow θ = c o s^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

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

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Two vectors P → a n d Q → have equal magnitudes. If the magnitude of P → + Q → is n times the magnitude of P → − Q → , then angle between P → a n d Q → is:

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