Answer to Question #157889 in Geometry for Michael

Question

ABCD is a square and P, Q are the midpoints of BC, CD respectively. If AP = a

and AQ = b, find in terms of a and b, the directed line segments (i) AB

Answer 1

ABCD is a square: $\overrightarrow{AB}=\overrightarrow{DC}, \overrightarrow{BC}=\overrightarrow{AD}$

$P, Q$ are the midpoints of $BC, CD$ respectively: $\overrightarrow{BP}=\dfrac{1}{2}\overrightarrow{BC}, \overrightarrow{DQ}=\dfrac{1}{2}\overrightarrow{DC}$

$$\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}$$ $$\overrightarrow{AQ}=\overrightarrow{AD}+\overrightarrow{DQ}=\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{DC}$$

If $\overrightarrow{AP}=\vec{a}$ and $\overrightarrow{AQ}=\vec{b}$

$$\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}=\vec{a}$$ $$\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{AB}=\vec{b}$$

(i)

$$\overrightarrow{AD}=\vec{b}-\dfrac{1}{2}\overrightarrow{AB}$$ $$\overrightarrow{AB}+\dfrac{1}{2}(\vec{b}-\dfrac{1}{2}\overrightarrow{AB})=\vec{a}$$ $$\overrightarrow{AB}=\dfrac{4}{3}\vec{a}-\dfrac{2}{3}\vec{b}$$