In an isosceles trapezoid OACB : angle BOA = 60^°…

In an isosceles trapezoid $OACB$: $\angle BOA = 60^\circ$, $\overline{OB} = \overline{BC} = \overline{CA} = 2$, and $M$, $N$, $P$ are midpoints of sides $BC$, $AC$ and $OA$ respectively. Express vectors $\overrightarrow{AC}$, $\overrightarrow{OM}$, $\overrightarrow{ON}$, $\overrightarrow{MN}$, $\overrightarrow{PM}$ and $\overrightarrow{NP}$ in terms of unit vectors $\vec{m}$ and $\vec{n}$ of sides $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

Step 1: Setup and unit vectors

Let $\vec{m}$ and $\vec{n}$ be the unit vectors along $\overrightarrow{OA}$ and $\overrightarrow{OB}$: \[\vec{m} = \frac{\overrightarrow{OA}}{|\overrightarrow{OA}|}, \qquad \vec{n} = \frac{\overrightarrow{OB}}{|\overrightarrow{OB}|}, \qquad |\vec{m}| = |\vec{n}| = 1.\]

Finding $\overline{OA}$

With $BC \parallel OA$, projecting the broken line $O\to B\to C\to A$ onto the base: \[OA = OB\cos60^\circ + BC + CA\cos60^\circ = 2\cdot\tfrac{1}{2} + 2 + 2\cdot\tfrac{1}{2} = 4.\]

Therefore $\overrightarrow{OA} = 4\,\vec{m}, \qquad \overrightarrow{OB} = 2\,\vec{n}$.

Step 2: Coordinates of all points

Using $\vec{m} = (1,0)$ and $\vec{n} = (\cos60^\circ, \sin60^\circ) = \left(\tfrac{1}{2}, \tfrac{\sqrt{3}}{2}\right)$: \[O = (0,\,0), \quad A = (4,\,0), \quad B = (1,\,\sqrt{3}), \quad C = (3,\,\sqrt{3}).\]

Check: $|CA| = \sqrt{(4-3)^2+(0-\sqrt{3})^2} = 2$ $\checkmark$. The midpoints are

\[ M = \text{mid}(BC) = (2,\,\sqrt{3}), \qquad N = \text{mid}(AC) = \left(\tfrac{7}{2},\,\tfrac{\sqrt{3}}{2}\right), \qquad P = \text{mid}(OA) = (2,\,0). \]

Step 3: General method

For any vector $\vec{v} = (v_x, v_y)$ we seek $\vec{v} = \alpha\,\vec{m} + \beta\,\vec{n}$, i.e.

\[ \alpha(1,0) + \beta\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right) = (v_x,\, v_y). \]

From the $y$-component $\beta = \dfrac{2v_y}{\sqrt{3}}$; from the $x$-component $\alpha = v_x - \dfrac{\beta}{2}$.

Step 4: Express each vector

$\overrightarrow{AC}$

\[ \overrightarrow{AC} = C - A = (-1,\,\sqrt{3}), \qquad \beta = \frac{2\sqrt{3}}{\sqrt{3}} = 2, \qquad \alpha = -1 - \frac{2}{2} = -2 \]

\[\overrightarrow{AC} = -2\,\vec{m} + 2\,\vec{n}\]

Check: $-2(1,0)+2\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right) = (-1,\sqrt{3})$ $\checkmark$

$\overrightarrow{OM}$

\[ \overrightarrow{OM} = (2,\,\sqrt{3}), \qquad \beta = 2, \qquad \alpha = 2 - 1 = 1 \]

\[\overrightarrow{OM} = \vec{m} + 2\,\vec{n}\]

Check: $(1,0)+(1,\sqrt{3}) = (2,\sqrt{3})$ $\checkmark$

$\overrightarrow{ON}$

\[ \overrightarrow{ON} = \left(\tfrac{7}{2},\,\tfrac{\sqrt{3}}{2}\right), \qquad \beta = \frac{2\cdot\frac{\sqrt{3}}{2}}{\sqrt{3}} = 1, \qquad \alpha = \frac{7}{2} - \frac{1}{2} = 3 \]

\[\overrightarrow{ON} = 3\,\vec{m} + \vec{n}\]

Check: $(3,0)+\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right) = \left(\tfrac{7}{2},\tfrac{\sqrt{3}}{2}\right)$ $\checkmark$

$\overrightarrow{MN}$

\[ \overrightarrow{MN} = N - M = \left(\tfrac{3}{2},\;-\tfrac{\sqrt{3}}{2}\right), \qquad \beta = -1, \qquad \alpha = \frac{3}{2} + \frac{1}{2} = 2 \]

\[\overrightarrow{MN} = 2\,\vec{m} - \vec{n}\]

Check: $(2,0)-\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right) = \left(\tfrac{3}{2},-\tfrac{\sqrt{3}}{2}\right)$ $\checkmark$

$\overrightarrow{PM}$

\[ \overrightarrow{PM} = M - P = (0,\,\sqrt{3}), \qquad \beta = 2, \qquad \alpha = 0 - 1 = -1 \]

\[\overrightarrow{PM} = -\vec{m} + 2\,\vec{n}\]

Check: $(-1,0)+(1,\sqrt{3}) = (0,\sqrt{3})$ $\checkmark$

$\overrightarrow{NP}$

\[ \overrightarrow{NP} = P - N = \left(-\tfrac{3}{2},\;-\tfrac{\sqrt{3}}{2}\right), \qquad \beta = -1, \qquad \alpha = -\frac{3}{2} + \frac{1}{2} = -1 \]

\[\overrightarrow{NP} = -\vec{m} - \vec{n}\]

Check: $-(1,0)-\left(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}\right) = \left(-\tfrac{3}{2},-\tfrac{\sqrt{3}}{2}\right)$ $\checkmark$