Question #157889

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


1
Expert's answer
2021-01-26T04:03:00-0500


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

P,QP, Q are the midpoints of BC,CDBC, CD respectively: BP=12BC,DQ=12DC\overrightarrow{BP}=\dfrac{1}{2}\overrightarrow{BC}, \overrightarrow{DQ}=\dfrac{1}{2}\overrightarrow{DC}



AP=AB+BP=AB+12BC\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{BC}AQ=AD+DQ=AD+12DC\overrightarrow{AQ}=\overrightarrow{AD}+\overrightarrow{DQ}=\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{DC}

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



AB+12AD=a\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}=\vec{a}AD+12AB=b\overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{AB}=\vec{b}


(i)



AD=b12AB\overrightarrow{AD}=\vec{b}-\dfrac{1}{2}\overrightarrow{AB}AB+12(b12AB)=a\overrightarrow{AB}+\dfrac{1}{2}(\vec{b}-\dfrac{1}{2}\overrightarrow{AB})=\vec{a}AB=43a23b\overrightarrow{AB}=\dfrac{4}{3}\vec{a}-\dfrac{2}{3}\vec{b}




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