The angle between two unit vectors $ extbf{a}$ and $ extbf{b}$ is $ heta$ and $| extbf{a} + extbf{b}| < 1$ - HSC - SSCE Mathematics Extension 1 - Question 8 - 2022 - Paper 1

To determine the angle $heta$ between two unit vectors $extbf{a}$ and $extbf{b}$, we can use the property of the dot product, which relates to the cosine of the angle:

$| extbf{a} + extbf{b}|^2 = | extbf{a}|^2 + | extbf{b}|^2 + 2| extbf{a}|| extbf{b}|\cos(\theta)$

Since $extbf{a}$ and $extbf{b}$ are unit vectors, $| extbf{a}| = | extbf{b}| = 1$, simplifying the expression:

$| extbf{a} + extbf{b}|^2 = 1 + 1 + 2\cos(\theta) = 2 + 2\cos(\theta)$

Given that the condition states $| extbf{a} + extbf{b}| < 1$, we have:

$2 + 2\cos(\theta) < 1$

This can be rearranged to:

$2\cos(\theta) < -1\implies \cos(\theta) < -\frac{1}{2}$

The angles that satisfy this condition occur in the ranges:

• Between rac{2\pi}{3} and rac{4\pi}{3} (in radians).

This specifically leads to:

$\frac{2\pi}{3} < \theta < \pi$

Thus, the correct answer that describes the possible range of values of $heta$ is option D: $\frac{2\pi}{3} < \theta < \pi$.