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Vector Algebra question

2021 · 25 Jul · Shift 2 · Q63
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  5. /2021 · 25 Jul · Shift 2 · Q63

Vector Algebra question

2021 · 25 Jul · Shift 2 · Q63

JEE MainPhysicsVector AlgebraMCQ+4 / −1
Two vectors X→\overrightarrow XX and Y→\overrightarrow YY have equal magnitude. The magnitude of (X→−Y→\overrightarrow X -\overrightarrow YX−Y) is n times the magnitude of (X→\overrightarrow XX+Y→\overrightarrow YY). The angle between X→\overrightarrow XX and Y→\overrightarrow YY is :
  1. A
    cos⁡−1(−n2−1n2−1){\cos ^{ - 1}}\left( {{{ - {n^2} - 1} \over {{n^2} - 1}}} \right)cos−1(n2−1−n2−1​)
  2. B
    cos⁡−1(n2−1−n2−1){\cos ^{ - 1}}\left( {{{{n^2} - 1} \over { - {n^2} - 1}}} \right)cos−1(−n2−1n2−1​)
  3. C
    cos⁡−1(n2+1−n2−1){\cos ^{ - 1}}\left( {{{{n^2} + 1} \over { - {n^2} - 1}}} \right)cos−1(−n2−1n2+1​)
  4. D
    cos⁡−1(n2+1n2−1){\cos ^{ - 1}}\left( {{{{n^2} + 1} \over {{n^2} - 1}}} \right)cos−1(n2−1n2+1​)
View written solutionFree

Correct answer: B

  1. Let the magnitudes be equal

Since ∣X⃗∣=∣Y⃗∣|\vec X|=|\vec Y|∣X∣=∣Y∣, let

∣X⃗∣=∣Y⃗∣=a|\vec X|=|\vec Y|=a∣X∣=∣Y∣=a

and let the angle between them be θ\thetaθ.

  1. Use magnitude formulas

We know:

∣X⃗−Y⃗∣=n∣X⃗+Y⃗∣|\vec X-\vec Y|=n|\vec X+\vec Y|∣X−Y∣=n∣X+Y∣

Squaring both sides,

∣X⃗−Y⃗∣2=n2∣X⃗+Y⃗∣2|\vec X-\vec Y|^2=n^2|\vec X+\vec Y|^2∣X−Y∣2=n2∣X+Y∣2

Now,

∣X⃗−Y⃗∣2=∣X⃗∣2+∣Y⃗∣2−2X⃗⋅Y⃗|\vec X-\vec Y|^2=|\vec X|^2+|\vec Y|^2-2\vec X\cdot\vec Y∣X−Y∣2=∣X∣2+∣Y∣2−2X⋅Y

Since X⃗⋅Y⃗=a2cos⁡θ\vec X\cdot\vec Y=a^2\cos\thetaX⋅Y=a2cosθ,

∣X⃗−Y⃗∣2=a2+a2−2a2cos⁡θ=2a2(1−cos⁡θ)|\vec X-\vec Y|^2=a^2+a^2-2a^2\cos\theta=2a^2(1-\cos\theta)∣X−Y∣2=a2+a2−2a2cosθ=2a2(1−cosθ)

Similarly,

∣X⃗+Y⃗∣2=∣X⃗∣2+∣Y⃗∣2+2X⃗⋅Y⃗|\vec X+\vec Y|^2=|\vec X|^2+|\vec Y|^2+2\vec X\cdot\vec Y∣X+Y∣2=∣X∣2+∣Y∣2+2X⋅Y

so

∣X⃗+Y⃗∣2=2a2(1+cos⁡θ)|\vec X+\vec Y|^2=2a^2(1+\cos\theta)∣X+Y∣2=2a2(1+cosθ)
  1. Substitute into the given relation
2a2(1−cos⁡θ)=n2 2a2(1+cos⁡θ)2a^2(1-\cos\theta)=n^2\,2a^2(1+\cos\theta)2a2(1−cosθ)=n22a2(1+cosθ)

Cancel 2a22a^22a2:

1−cos⁡θ=n2(1+cos⁡θ)1-\cos\theta=n^2(1+\cos\theta)1−cosθ=n2(1+cosθ)

Expand:

1−cos⁡θ=n2+n2cos⁡θ1-\cos\theta=n^2+n^2\cos\theta1−cosθ=n2+n2cosθ

Rearrange:

1−n2=cos⁡θ(1+n2)1-n^2=\cos\theta(1+n^2)1−n2=cosθ(1+n2)

Hence,

cos⁡θ=1−n21+n2\cos\theta=\frac{1-n^2}{1+n^2}cosθ=1+n21−n2​

Therefore,

θ=cos⁡−1(1−n21+n2)\theta=\cos^{-1}\left(\frac{1-n^2}{1+n^2}\right)θ=cos−1(1+n21−n2​)
  1. Match with the options

Option B is

cos⁡−1(n2−1−n2−1)\cos^{-1}\left(\frac{n^2-1}{-n^2-1}\right)cos−1(−n2−1n2−1​)

which simplifies to

cos⁡−1(−n2−1n2+1)=cos⁡−1(1−n2n2+1)\cos^{-1}\left(-\frac{n^2-1}{n^2+1}\right)=\cos^{-1}\left(\frac{1-n^2}{n^2+1}\right)cos−1(−n2+1n2−1​)=cos−1(n2+11−n2​)

This matches our result.

  1. Check other options briefly
  • A: −n2−1n2−1=−n2+1n2−1\dfrac{-n^2-1}{n^2-1}= -\dfrac{n^2+1}{n^2-1}n2−1−n2−1​=−n2−1n2+1​, not equal to our expression.
  • B: matches exactly.
  • C: n2+1−n2−1=−1\dfrac{n^2+1}{-n^2-1}=-1−n2−1n2+1​=−1, gives constant angle π\piπ, incorrect.
  • D: n2+1n2−1\dfrac{n^2+1}{n^2-1}n2−1n2+1​, generally outside valid cosine range, incorrect.

So the correct option is B.

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