Point B divides the arc oversetfrownAC = 90^° of a circle…

Point $B$ divides the arc $\overset{\frown}{AC} = 90^\circ$ of a circle with center $O$ in the ratio $1:2$. Express analytically and geometrically the vector $\overrightarrow{OC} = \vec{c}$ in terms of vectors $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$.

Step 1: Setup

Since $\overset{\frown}{AB} : \overset{\frown}{BC} = 1:2$ and the whole arc is $90^\circ$: \[\overset{\frown}{AB} = \tfrac{1}{3}\cdot 90^\circ = 30^\circ, \qquad \overset{\frown}{BC} = \tfrac{2}{3}\cdot 90^\circ = 60^\circ,\] so the central angles are $\angle AOB = 30^\circ$, $\angle BOC = 60^\circ$, $\angle AOC = 90^\circ$. With radius $r$, $|\vec{a}| = |\vec{b}| = |\vec{c}| = r$. Place $O$ at the origin and $A$ on the positive $x$-axis:

\[ A = r(1,\,0), \qquad B = r\left(\tfrac{\sqrt{3}}{2},\,\tfrac{1}{2}\right), \qquad C = r(0,\,1). \]

Step 2: Analytical decomposition

We seek $\alpha, \beta$ with $\vec{c} = \alpha\,\vec{a} + \beta\,\vec{b}$. The radius $r$ cancels, so take $r=1$: \[\vec{a} = (1,\,0), \qquad \vec{b} = \left(\tfrac{\sqrt{3}}{2},\,\tfrac{1}{2}\right), \qquad \vec{c} = (0,\,1)\]

\[ \begin{cases} \alpha + \dfrac{\sqrt{3}}{2}\beta = 0 \\[6pt] \dfrac{1}{2}\beta = 1 \end{cases} \quad\Longrightarrow\quad \beta = 2, \qquad \alpha = -\sqrt{3}. \]

\[\vec{c} = -\sqrt{3}\,\vec{a} + 2\,\vec{b}\]

Step 3: Verification

\[ -\sqrt{3}(1,0)+2\left(\tfrac{\sqrt{3}}{2},\tfrac{1}{2}\right) = (-\sqrt{3},0)+(\sqrt{3},1) = (0,1) = \vec{c} \;\checkmark \]

Step 4: Alternative check by rotation

$\vec b$ is $\vec a$ rotated by $30^\circ$, and $\vec c$ is $\vec b$ rotated by $60^\circ$. The rotation by $60^\circ$ is

\[ R(60^\circ)\colon (x,y)\mapsto(x\cos60^\circ-y\sin60^\circ,\ x\sin60^\circ+y\cos60^\circ), \] and applied to $\vec b$ it gives

\[ \left(\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2}-\tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2},\ \tfrac{\sqrt{3}}{2}\cdot\tfrac{\sqrt{3}}{2}+\tfrac{1}{2}\cdot\tfrac{1}{2}\right) = (0,\,1) = \vec c \;\checkmark \]

Step 5: Geometric construction

$\vec c$ is the diagonal of the parallelogram built on $-\sqrt{3}\,\vec a$ and $2\vec b$: from $O$ go $\sqrt3$ radii in the direction opposite to $\vec a$, then twice $\vec b$; you arrive at $C$.