What is the angle between these unit vectors?

Question

Let $a$ and $b$ be two unit vectors and $p$ is the angle between them. If $a+b$ is also a unit vector, then:

1. $p = \pi/3$

2. $p = \pi/4$

3. $p = \pi/2$

4. $p = 2\pi/3$

Answer 1

Let $\boldsymbol{a} = (a_1, \ldots, a_n)$ and $\boldsymbol{b} = (b_1, \ldots, b_n)$ be two unit vectors in $\mathbb{R}^n$. If $\boldsymbol{a} + \boldsymbol{b}$ is also a unit vector, then we have $\langle \boldsymbol{a + b}, \boldsymbol{a+ b} \rangle = 1$, where $\langle \cdot, \cdot \rangle$ is the standard scalar product on $\mathbb{R}^n$. Therefore, \begin{gather*} \sum_{i = 1}^n (a_i + b_i)^2 = 1 \\ \sum_{i = 1}^n a_i^2 + \sum_{i = 1}^n b_i^2 + 2 \sum_{i = 1}^n a_i b_i = 1 \\ 1 + 1 + 2 \langle \boldsymbol{a}, \boldsymbol{b} \rangle = 1 \\ \langle \boldsymbol{a}, \boldsymbol{b} \rangle = -\frac{1}{2}. \end{gather*} We know that \begin{equation*} \cos p = \frac{\langle \boldsymbol{a}, \boldsymbol{b} \rangle}{\Vert \boldsymbol{a} \Vert \Vert \boldsymbol{b} \Vert}. \end{equation*} Since $\Vert \boldsymbol{a} \Vert = 1$ and $\Vert \boldsymbol{b} \Vert = 1$, we have $\cos p = -\dfrac{1}{2}$, and $p = \dfrac{2}{3} \pi$.

Answer 2

i think it is $p= 120^\circ.$

here is a the reason: the vectors $a=i, b = -1/2 i + \sqrt3/2j, a+b = 1/2i+\sqrt 3/2 j $ satisfy the requirements and the angle between the vectors $a$ and $b$ is $120^\circ.$

Answer 3

The vectors with non-zero $y$ coordinate are the candidates. The unit vector $(1,0)$ is the addition partner.

Now consider that the sum of two vectors is the diagonal of the parallelogram formed by the two vectors and that it must lie on the unit circle.