AP DSC School Assistant Math Model Test 3

Concept:This problem uses vector geometry and the mid‑point formula. The position vector of a midpoint is the average of the two vertex vectors.Explanation:Let the position vectors of vertices A, B, C be $\vec{a}, \vec{b}, \vec{c}$. Since D, E, F are midpoints of BC, CA, AB respectively,$\vec{d} = \frac{\vec{b}+\vec{c}}{2},\;\vec{e} = \frac{\vec{c}+\vec{a}}{2},\;\vec{f} = \frac{\vec{a}+\vec{b}}{2}$.The given expression is:$\overline{\mathrm{AD}} + \frac{2}{3}\overline{\mathrm{BE}} + \frac{1}{3}\overline{\mathrm{CF}} = (\vec{d}-\vec{a}) + \frac{2}{3}(\vec{e}-\vec{b}) + \frac{1}{3}(\vec{f}-\vec{c})$.Substitute the midpoint vectors:$\left(\frac{\vec{b}+\vec{c}}{2} - \vec{a}\right) + \frac{2}{3}\left(\frac{\vec{c}+\vec{a}}{2} - \vec{b}\right) + \frac{1}{3}\left(\frac{\vec{a}+\vec{b}}{2} - \vec{c}\right)$.Simplify each bracket:$\frac{\vec{b}+\vec{c}-2\vec{a}}{2} + \frac{\vec{c}+\vec{a}-2\vec{b}}{3} + \frac{\vec{a}+\vec{b}-2\vec{c}}{6}$.Express with common denominator 6:$\frac{3(\vec{b}+\vec{c}-2\vec{a}) + 2(\vec{c}+\vec{a}-2\vec{b}) + (\vec{a}+\vec{b}-2\vec{c})}{6}$.Combine the numerators:$3\vec{b}+3\vec{c}-6\vec{a} + 2\vec{c}+2\vec{a}-4\vec{b} + \vec{a}+\vec{b}-2\vec{c}$$= (-6\vec{a}+2\vec{a}+\vec{a}) + (3\vec{b}-4\vec{b}+\vec{b}) + (3\vec{c}+2\vec{c}-2\vec{c})$$= -3\vec{a} + 0\vec{b} + 3\vec{c} = 3(\vec{c} - \vec{a})$.Hence the expression equals $\frac{3(\vec{c}-\vec{a})}{6} = \frac{1}{2}(\vec{c}-\vec{a}) = \frac{1}{2}\overline{\mathrm{AC}}$.Answer:Option B: $\frac{1}{2} \overline{\mathrm{AC}}$.