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3D Geometry question

2010 · Shift 0 · Q48
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3D Geometry question

2010 · Shift 0 · Q48

JEE MainMathematics3D GeometryMCQ+4 / −1
A line ABABAB in three-dimensional space makes angles 45∘{45^ \circ }45∘ and 120∘{120^ \circ }120∘ with the positive xxx-axis and the positive yyy-axis respectively. If ABABAB makes an acute angle θ\thetaθ with the positive zzz-axis, then θ\thetaθ equals :
  1. A
    45∘{45^ \circ }45∘
  2. B
    60∘{60^ \circ }60∘
  3. C
    75∘{75^ \circ }75∘
  4. D
    30∘{30^ \circ }30∘
View written solutionFree

Correct answer: B

  1. Use direction cosines

If a line makes angles α,β,γ\alpha, \beta, \gammaα,β,γ with the positive xxx-, yyy-, and zzz-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,

  • α=45∘\alpha = 45^\circα=45∘
  • β=120∘\beta = 120^\circβ=120∘
  • γ=θ\gamma = \thetaγ=θ

So,

cos⁡245∘+cos⁡2120∘+cos⁡2θ=1.\cos^2 45^\circ + \cos^2 120^\circ + \cos^2 \theta = 1.cos245∘+cos2120∘+cos2θ=1.
  1. Compute the known cosine values
cos⁡45∘=12  ⟹  cos⁡245∘=12\cos 45^\circ = \frac{1}{\sqrt{2}} \implies \cos^2 45^\circ = \frac{1}{2}cos45∘=2​1​⟹cos245∘=21​ cos⁡120∘=−12  ⟹  cos⁡2120∘=14\cos 120^\circ = -\frac{1}{2} \implies \cos^2 120^\circ = \frac{1}{4}cos120∘=−21​⟹cos2120∘=41​

Substitute:

12+14+cos⁡2θ=1\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 121​+41​+cos2θ=1 34+cos⁡2θ=1\frac{3}{4} + \cos^2 \theta = 143​+cos2θ=1 cos⁡2θ=14\cos^2 \theta = \frac{1}{4}cos2θ=41​
  1. Use the fact that θ\thetaθ is acute

Since θ\thetaθ is acute, cos⁡θ>0\cos \theta > 0cosθ>0. Therefore,

cos⁡θ=12\cos \theta = \frac{1}{2}cosθ=21​

Hence,

θ=60∘.\theta = 60^\circ.θ=60∘.
  1. Check options
  • A: 45∘45^\circ45∘ ❌
  • B: 60∘60^\circ60∘ ✅
  • C: 75∘75^\circ75∘ ❌
  • D: 30∘30^\circ30∘ ❌

Therefore, the correct answer is B.

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