Question

ABCD is a square and P, Q are the mid points of BC, CD respectively. If AP = a and AQ = b, find in terms of a and b, the directed line segment AB, AD, BD, and AC.

Answer 1

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

$P, Qare the midpoints of BC, CD respectively:$

$\overrightarrow{BP} =\frac{1} {2} \overrightarrow{BC}, \overrightarrow{DQ} =\frac{1} {2} \overrightarrow{DC}$

$\overrightarrow{AP} =\overrightarrow{AB} +\overrightarrow{BP} =\overrightarrow{AB} +\frac{1} {2} \overrightarrow{BC}$

$\overrightarrow{AQ} =\overrightarrow{AD} +\overrightarrow{DQ} =\overrightarrow{AD} +\frac{1} {2} \overrightarrow{DC}$

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

$\overrightarrow{AB} +\frac{1} {2} \overrightarrow{AD} =\overrightarrow{a}$

$\overrightarrow{AD} +\frac{1} {2} \overrightarrow{AB} =\overrightarrow{b}$

i) $\overrightarrow{AD} =\overrightarrow{b} - \frac {1} {2} \overrightarrow{AB}$

$\overrightarrow{AB} +\frac{1} {2} \overrightarrow{b} - \frac {1} {2} \overrightarrow{AB}=\overrightarrow{a}$

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

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

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

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

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

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

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

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

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