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

ત્રિ-પરિમાણીય અવકાશમા રેખાએ \(x\) અને \(y\) અક્ષ સાથે બનાવેલ ખૂણો  \(\theta \left( {0 < \theta  \le \frac{\pi }{2}} \right)\)  હોય તો  \(\theta \) ની બધીજ કિમંતો નો ગણ એ  . . . . અંતરાલમાં આવેલ છે.

  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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