1. Magnitude ka formula lagayye:
Hume pata hai ki:
∣a−b∣2=∣a∣2+∣b∣2−2∣a∣∣b∣cos(2θ)
2. Values substitute karein:
Sawal ke mutabiq |\vec{a}-\vec{b}| < 1, toh:
|\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}| \cos(2\theta) < 1^2
(1)^2 + (1)^2 - 2(1)(1) \cos(2\theta) < 1
1 + 1 - 2 \cos(2\theta) < 1
2 - 2 \cos(2\theta) < 1
3. Equation ko simplify karein:
2(1 - \cos(2\theta)) < 1
Trigonometry ka formula use karein: 1−cos(2θ)=2sin2θ
\sin^2\theta < \frac{1}{4}
4. Sine ki range nikalein:
Dono side square root lene par:
|\sin\theta| < \frac{1}{2}
-\frac{1}{2} < \sin\theta < \frac{1}{2}
5. Interval check karein:
Hume diya gaya hai ki 0≤θ≤π. Is range mein sinθ hamesha positive ya zero hota hai, isliye:
0 \leq \sin\theta < \frac{1}{2}
Ab, sinθ=1/2 tab hota hai jab θ=π/6.
Since θ ki range 0 se π tak hai, toh \sin\theta < 1/2 do jagah ho sakta hai:
-
0 \leq \theta < \frac{\pi}{6}
-
\frac{5\pi}{6} < \theta \leq \pi
Final Answer
θ in do intervals mein lie karega:
