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प्रश्न
If \[\vec{a} \text{ and } \vec{b}\] are two unit vectors inclined at an angle θ, such that \[\left| \vec{a} + \vec{b} \right| < 1,\] then
विकल्प
(a) \[\theta < \frac{\pi}{3}\]
(b) \[\theta > \frac{2\pi}{3}\]
(c) \[\frac{\pi}{3} < \theta < \frac{2\pi}{3}\]
(d) \[\frac{2\pi}{3} < \theta < \pi\]
उत्तर
(d) \[\frac{2\pi}{3} < \theta < \pi\]
\[\text{ We have }\]
\[\left| \vec{a} + \vec{b} \right| < 1\]
\[ \Rightarrow \sqrt{\left| \vec{a} \right|^2 + \left| \vec{b} \right|^2 + 2\left| \vec{a} \right| \times \left| \vec{b} \right| \cos\theta} < 1\]
\[ \Rightarrow \sqrt{1^2 + 1^2 + 2 \times 1 \times 1 \times \cos\theta} < 1\]
\[ \Rightarrow \sqrt{2 + 2 \cos\theta} < 1\]
\[ \Rightarrow \sqrt{2\left( 1 + \cos\theta \right)} < 1\]
\[ \Rightarrow \sqrt{2 \times 2 \cos^2 \frac{\theta}{2}} < 1\]
\[ \Rightarrow 2\left| \cos\frac{\theta}{2} \right| < 1\]
\[ \Rightarrow \left| \cos\frac{\theta}{2} \right| < \frac{1}{2}\]
\[ \Rightarrow \frac{\pi}{3} < \frac{\theta}{2} < \frac{2\pi}{3}\]
\[ \Rightarrow \frac{2\pi}{3} < \theta < \frac{4\pi}{3}\]
\[\text{ But here } \theta \text{ cannot be more than } \pi . \]
\[ \therefore \frac{2\pi}{3} < \theta < \pi\]
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