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Vector Algebra question

2019 · 10 Jan · Shift 2 · Q70
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Vector Algebra question

2019 · 10 Jan · Shift 2 · Q70

JEE MainPhysicsVector AlgebraMCQ+4 / −1
Two vectors A→\overrightarrow AA and B→\overrightarrow BB have equal magnitudes. The magnitude of (A→+B→)\left( {\overrightarrow A + \overrightarrow B } \right)(A+B) is 'n' times the magnitude of (A→−B→)\left( {\overrightarrow A - \overrightarrow B } \right)(A−B). The angle between A→{\overrightarrow A }A and B→{\overrightarrow B }B is -
  1. A
    sin⁡−1[n−1n+1]{\sin ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]sin−1[n+1n−1​]
  2. B
    sin⁡−1[n2−1n2+1]{\sin ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]sin−1[n2+1n2−1​]
  3. C
    cos⁡−1[n2−1n2+1]{\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over {{n^2} + 1}}} \right]cos−1[n2+1n2−1​]
  4. D
    cos⁡−1[n−1n+1]{\cos ^{ - 1}}\left[ {{{n - 1} \over {n + 1}}} \right]cos−1[n+1n−1​]
View written solutionFree

Correct answer: C

  1. Let the magnitudes be equal

Since ∣A⃗∣=∣B⃗∣|\vec A|=|\vec B|∣A∣=∣B∣, let

∣A⃗∣=∣B⃗∣=a|\vec A|=|\vec B|=a∣A∣=∣B∣=a

and let the angle between them be θ\thetaθ.

We are given that

∣A⃗+B⃗∣=n ∣A⃗−B⃗∣.|\vec A+\vec B|=n\,|\vec A-\vec B|.∣A+B∣=n∣A−B∣.
  1. Use the formula for magnitude of sum and difference

For two vectors with angle θ\thetaθ between them,

∣A⃗+B⃗∣2=∣A⃗∣2+∣B⃗∣2+2∣A⃗∣∣B⃗∣cos⁡θ|\vec A+\vec B|^2=|\vec A|^2+|\vec B|^2+2|\vec A||\vec B|\cos\theta∣A+B∣2=∣A∣2+∣B∣2+2∣A∣∣B∣cosθ ∣A⃗−B⃗∣2=∣A⃗∣2+∣B⃗∣2−2∣A⃗∣∣B⃗∣cos⁡θ.|\vec A-\vec B|^2=|\vec A|^2+|\vec B|^2-2|\vec A||\vec B|\cos\theta.∣A−B∣2=∣A∣2+∣B∣2−2∣A∣∣B∣cosθ.

Since ∣A⃗∣=∣B⃗∣=a|\vec A|=|\vec B|=a∣A∣=∣B∣=a,

∣A⃗+B⃗∣2=2a2(1+cos⁡θ),|\vec A+\vec B|^2=2a^2(1+\cos\theta),∣A+B∣2=2a2(1+cosθ), ∣A⃗−B⃗∣2=2a2(1−cos⁡θ).|\vec A-\vec B|^2=2a^2(1-\cos\theta).∣A−B∣2=2a2(1−cosθ).
  1. Apply the given condition

Given

∣A⃗+B⃗∣=n∣A⃗−B⃗∣.|\vec A+\vec B|=n|\vec A-\vec B|.∣A+B∣=n∣A−B∣.

Squaring both sides,

∣A⃗+B⃗∣2=n2∣A⃗−B⃗∣2.|\vec A+\vec B|^2=n^2|\vec A-\vec B|^2.∣A+B∣2=n2∣A−B∣2.

So,

2a2(1+cos⁡θ)=n2⋅2a2(1−cos⁡θ).2a^2(1+\cos\theta)=n^2\cdot 2a^2(1-\cos\theta).2a2(1+cosθ)=n2⋅2a2(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⁡θ.1+\cos\theta=n^2-n^2\cos\theta.1+cosθ=n2−n2cosθ.

Bring cosine terms together:

cos⁡θ+n2cos⁡θ=n2−1.\cos\theta+n^2\cos\theta=n^2-1.cosθ+n2cosθ=n2−1. cos⁡θ(1+n2)=n2−1.\cos\theta(1+n^2)=n^2-1.cosθ(1+n2)=n2−1.

Hence,

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

Therefore,

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

This corresponds to:

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

which is Option C.

  1. Verification with stored answer

Stored correct answer: C

Our derived answer: C

So, they agree.

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