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Vector Algebra question

2021 · 20 Jul · Shift 2 · Q61

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JEE Main
Physics
Vector Algebra
2021 · 20 Jul · Shift 2 · Q61

Vector Algebra question

2021 · 20 Jul · Shift 2 · Q61

JEE MainPhysicsVector AlgebraMCQ+4 / −1

Two vectors

{\overrightarrow P }

and

{\overrightarrow Q }

have equal magnitudes. If the magnitude of

{\overrightarrow P + \overrightarrow Q }

is n times the magnitude of

{\overrightarrow P - \overrightarrow Q }

, then angle between

{\overrightarrow P }

and

{\overrightarrow Q }

is :

A
${\sin ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$
B
${\cos ^{ - 1}}\left( {{{n - 1} \over {n + 1}}} \right)$
C
${\sin ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$
D
${\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$

View written solutionFree

Correct answer: D

Let the magnitudes be equal

Suppose $|\vec P|=|\vec Q|=a$ and the angle between them is $\theta$ .

We are given: $|\vec P+\vec Q|=n\,|\vec P-\vec Q|$

Find the magnitudes using dot product

For two vectors $\vec P$ and $\vec Q$ , $|\vec P+\vec Q|^2=|\vec P|^2+|\vec Q|^2+2\vec P\cdot \vec Q$

Since $|\vec P|=|\vec Q|=a$ and $\vec P\cdot \vec Q=a^2\cos\theta$ , $|\vec P+\vec Q|^2=a^2+a^2+2a^2\cos\theta=2a^2(1+\cos\theta)$

Similarly, $|\vec P-\vec Q|^2=|\vec P|^2+|\vec Q|^2-2\vec P\cdot \vec Q$ $|\vec P-\vec Q|^2=2a^2(1-\cos\theta)$

Use the given condition

Given $|\vec P+\vec Q|=n|\vec P-\vec Q|$

Squaring both sides, $|\vec P+\vec Q|^2=n^2|\vec P-\vec Q|^2$

So, $2a^2(1+\cos\theta)=n^2\cdot 2a^2(1-\cos\theta)$

Cancel $2a^2$ : $1+\cos\theta=n^2(1-\cos\theta)$

Expand: $1+\cos\theta=n^2-n^2\cos\theta$

Bring terms together: $\cos\theta+n^2\cos\theta=n^2-1$ $\cos\theta(1+n^2)=n^2-1$

Hence, $\cos\theta=\frac{n^2-1}{n^2+1}$

Therefore, $\theta=\cos^{-1}\left(\frac{n^2-1}{n^2+1}\right)$

Match with the options

This corresponds to:

$\boxed{\cos^{-1}\left(\frac{n^2-1}{n^2+1}\right)}$

So the correct option is D.

Comparison with stored answer

Stored correct answer: D

My derived answer: D

They match.

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