The angle between two unit vectors $q$ and $b$ is $ heta$ and $|a + b| < 1$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2022 - Paper 1

To determine the possible range of values for $heta$, we start by understanding the implications of the given condition $|a + b| < 1$. This condition suggests that the sum of the magnitudes of the vectors $a$ and $b$ is limited, which occurs when the angle between the two vectors is less than rac{2 ext{π}}{3}.

Thus, since $|a| = |b| = 1$, the angle between them, denoted as $heta$, must satisfy the following condition:

1. Cosine Rule Applied: The cosine of the angle formed between the two unit vectors is given by: $|a + b|^2 = |a|^2 + |b|^2 + 2|a||b| ext{cos}( heta)$ Hence, $|a + b|^2 = 1 + 1 + 2( ext{cos}( heta))$ Simplifying, $|a + b|^2 = 2 + 2 ext{cos}( heta) < 1$ The inequality indicates: $2 + 2 ext{cos}( heta) < 1$ From this, rearranging gives us: ext{cos}( heta) < -rac{1}{2}

2. Finding Angles: The angles where this condition holds are in the range: rac{2 ext{π}}{3} < heta < rac{4 ext{π}}{3}

Thus, the best description of the range of $heta$ based on the provided options is:

• Option B: 0 eq heta < rac{2 ext{π}}{3} is true since $heta$ can take values from (0, rac{2 ext{π}}{3}).
• The other options do not satisfy the conditions derived from the inequality.