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3D Geometry question

2004 · Shift 0 · Q96
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3D Geometry question

2004 · Shift 0 · Q96

JEE MainMathematics3D GeometryMCQ+4 / −1
A line makes the same angle θ\thetaθ, with each of the xxx and zzz axis. If the angle β \beta \,β, which it makes with y-axis, is such that  sin⁡2β=3sin⁡2θ,\,{\sin ^2}\beta = 3{\sin ^2}\theta ,sin2β=3sin2θ, then cos⁡2θ{\cos ^2}\thetacos2θ equals :
  1. A
    25{2 \over 5}52​
  2. B
    15{1 \over 5}51​
  3. C
    35{3 \over 5}53​
  4. D
    23{2 \over 3}32​
View written solutionFree

Correct answer: C

  1. Use direction cosines

If a line makes angles α,β,γ\alpha,\beta,\gammaα,β,γ with the x,y,zx,y,zx,y,z axes respectively, then its direction cosines satisfy

cos⁡2α+cos⁡2β+cos⁡2γ=1.\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1.cos2α+cos2β+cos2γ=1.

Here the line makes the same angle θ\thetaθ with the xxx- and zzz-axes, so

α=θ,γ=θ.\alpha=\theta,\quad \gamma=\theta.α=θ,γ=θ.

Hence,

cos⁡2θ+cos⁡2β+cos⁡2θ=1\cos^2\theta+\cos^2\beta+\cos^2\theta=1cos2θ+cos2β+cos2θ=1

which gives

2cos⁡2θ+cos⁡2β=1.2\cos^2\theta+\cos^2\beta=1.2cos2θ+cos2β=1.
  1. Use the given relation involving sin⁡2β\sin^2\betasin2β

Given,

sin⁡2β=3sin⁡2θ.\sin^2\beta=3\sin^2\theta.sin2β=3sin2θ.

Now,

sin⁡2β=1−cos⁡2β,\sin^2\beta=1-\cos^2\beta,sin2β=1−cos2β,

so

1−cos⁡2β=3(1−cos⁡2θ).1-\cos^2\beta=3(1-\cos^2\theta).1−cos2β=3(1−cos2θ).

Thus,

1−cos⁡2β=3−3cos⁡2θ.1-\cos^2\beta=3-3\cos^2\theta.1−cos2β=3−3cos2θ.

Rearranging,

cos⁡2β=3cos⁡2θ−2.\cos^2\beta=3\cos^2\theta-2.cos2β=3cos2θ−2.
  1. Substitute into the direction cosine equation

From step 1,

2cos⁡2θ+cos⁡2β=1.2\cos^2\theta+\cos^2\beta=1.2cos2θ+cos2β=1.

Substitute cos⁡2β=3cos⁡2θ−2\cos^2\beta=3\cos^2\theta-2cos2β=3cos2θ−2:

2cos⁡2θ+(3cos⁡2θ−2)=1.2\cos^2\theta+(3\cos^2\theta-2)=1.2cos2θ+(3cos2θ−2)=1.

So,

5cos⁡2θ−2=15\cos^2\theta-2=15cos2θ−2=1 5cos⁡2θ=35\cos^2\theta=35cos2θ=3 cos⁡2θ=35.\cos^2\theta=\frac{3}{5}.cos2θ=53​.
  1. Check the options
cos⁡2θ=35\cos^2\theta=\frac{3}{5}cos2θ=53​

which corresponds to Option C.

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