Answer to Question #156916 in Geometry for PRINCE KPOLU

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, (ii) AD,

(iii) BD and (iv) AC

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)

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

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

$$\overrightarrow{AB}=\dfrac{4}{3}\vec{a}-\dfrac{2}{3}\vec{b}$$

(ii)

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

$$\overrightarrow{AD}=\dfrac{4}{3}\vec{b}-\dfrac{2}{3}\vec{a}$$

(iii)

$$\overrightarrow{AD}-\overrightarrow{AB}=\overrightarrow{BD}$$

$$\overrightarrow{BD}=\dfrac{4}{3}\vec{b}-\dfrac{2}{3}\vec{a}-(\dfrac{4}{3}\vec{a}-\dfrac{2}{3}\vec{b})$$

$$\overrightarrow{BD}=2\vec{b}-2\vec{a}$$

(iv)

$$\overrightarrow{AD}+\overrightarrow{AB}=\overrightarrow{AC}$$

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

$$\overrightarrow{AC}=\dfrac{2}{3}\vec{a}+\dfrac{2}{3}\vec{b}$$