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Mathematics Question on Vector Algebra

If a⃗\vec{a} and b⃗\vec{b} are unit vectors enclosing an angle θ\theta and ∣a⃗+b⃗∣<1 \left| \vec{a} + \vec{b}\right| < 1, then

A

θ=π2\theta = \frac{\pi}{2}

B

θ<π3\theta < \frac{\pi}{3}

C

π≥θ>2π3\pi \ge \theta > \frac{2\pi}{3}

D

π3<θ<2π3\frac{\pi}{3} < \theta < \frac{2\pi}{3}

Answer

π≥θ>2π3\pi \ge \theta > \frac{2\pi}{3}

Explanation

Solution

∣a⃗+b⃗∣<1\left|\vec{a} + \vec{b}\right| < 1 ⇒∣a⃗+b⃗∣2<1\Rightarrow \left|\vec{a} + \vec{b}\right|^{2} < 1 ⇒∣a⃗∣2+∣b⃗∣2+2a⃗⋅b⃗<1\Rightarrow \left|\vec{a}\right|^{2} + \left|\vec{b}\right|^{2} + 2\vec{a}\cdot\vec{b} < 1 ⇒a⃗⋅b⃗<−12\Rightarrow \vec{a}\cdot \vec{b} < -\frac{1}{2} ⇒∣a⃗∣∣b⃗∣cos θ<−12\Rightarrow \left|\vec{a}\right|\left|\vec{b}\right|cos\,\theta < -\frac{1}{2} ⇒1×1×cos θ<−12\Rightarrow 1 \times 1\times cos\,\theta < -\frac{1}{2} ⇒cos θ<−12\Rightarrow cos\,\theta < -\frac{1}{2} ⇒−1≤cos θ<−12\Rightarrow -1 \le cos\,\theta < -\frac{1}{2} ⇒π≥θ>2π3\Rightarrow \pi \ge \theta > \frac{2\pi}{3}