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JEE Mains · Maths · STD 12 - 11. three dimension geometry

त्रिविमीय आकाश (space) में एक रेखा \(x\) तथा \(y\), दोनों अक्षों के साथ कोण \(\theta\left(0<\theta \leq \frac{\pi}{2}\right)\) बनाती है, तो \(\theta\) के सभी मानों का समुच्चय निम्न अंतराल है

  1. A \(\left( {0,\frac{\pi }{4}} \right]\)
  2. B \(\left[ {\frac{\pi }{6},\frac{\pi }{3}} \right]\)
  3. C \(\left[ {\frac{\pi }{4},\frac{\pi }{2}} \right]\)
  4. D \(\left( {\frac{\pi }{3},\frac{\pi }{2}} \right]\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[ {\frac{\pi }{4},\frac{\pi }{2}} \right]\)

Step-by-step Solution

Detailed explanation

It makes \(\theta\) with \(x\) and \(y\) -axes. \(l=\cos \theta, m=\cos \theta, n=\cos (\pi-2 \theta)\) we have \(l^{2}+m^{2}+n^{2}=1\) \(\Rightarrow \cos ^{2} \theta+\cos ^{2} \theta+\cos ^{2}(\pi-2 \theta)=1\) \(\Rightarrow 2 \cos ^{2} \theta+(-\cos 2 \theta)^{2}=1\)…
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