A B C D i s a s q u a r e : A B → = D C → , B C → = A D → ABCD is a square: \overrightarrow{AB}=\overrightarrow{DC}, \overrightarrow{BC}=\overrightarrow{AD} A BC D i s a s q u a re : A B = D C , BC = A D
P , Q a r e t h e m i d p o i n t s o f B C , C D r e s p e c t i v e l y : P, Qare the midpoints of BC, CD respectively: P , Q a re t h e mi d p o in t so f BC , C Dres p ec t i v e l y :
B P → = 1 2 B C → , D Q → = 1 2 D C → \overrightarrow{BP} =\frac{1} {2} \overrightarrow{BC}, \overrightarrow{DQ} =\frac{1} {2} \overrightarrow{DC} BP = 2 1 BC , D Q = 2 1 D C
A P → = A B → + B P → = A B → + 1 2 B C → \overrightarrow{AP} =\overrightarrow{AB} +\overrightarrow{BP} =\overrightarrow{AB} +\frac{1} {2} \overrightarrow{BC} A P = A B + BP = A B + 2 1 BC
A Q → = A D → + D Q → = A D → + 1 2 D C → \overrightarrow{AQ} =\overrightarrow{AD} +\overrightarrow{DQ} =\overrightarrow{AD} +\frac{1} {2} \overrightarrow{DC} A Q = A D + D Q = A D + 2 1 D C
If A P → = a → \overrightarrow{AP} =\overrightarrow{a} A P = a A Q → = b → \overrightarrow{AQ} =\overrightarrow{b} A Q = b
A B → + 1 2 A D → = a → \overrightarrow{AB} +\frac{1} {2} \overrightarrow{AD} =\overrightarrow{a} A B + 2 1 A D = a
A D → + 1 2 A B → = b → \overrightarrow{AD} +\frac{1} {2} \overrightarrow{AB} =\overrightarrow{b} A D + 2 1 A B = b
i) A D → = b → − 1 2 A B → \overrightarrow{AD} =\overrightarrow{b} - \frac {1} {2} \overrightarrow{AB} A D = b − 2 1 A B
A B → + 1 2 b → − 1 2 A B → = a → \overrightarrow{AB} +\frac{1} {2} \overrightarrow{b} - \frac {1} {2} \overrightarrow{AB}=\overrightarrow{a} A B + 2 1 b − 2 1 A B = a
= A B → = 4 3 a → − 2 3 b → =\overrightarrow{AB} =\frac{4}{3} \overrightarrow{a} - \frac {2}{3} \overrightarrow{b} = A B = 3 4 a − 3 2 b
ii) A D → = b → − 1 2 ( 4 3 a → − 2 3 b → ) \overrightarrow{AD} =\overrightarrow{b} - \frac {1} {2} (\frac{4}{3} \overrightarrow{a} - \frac {2}{3} \overrightarrow{b}) A D = b − 2 1 ( 3 4 a − 3 2 b )
= A D → = 4 3 b → − 2 3 a → =\overrightarrow{AD} =\frac {4} {3} \overrightarrow{b} - \frac{2}{3} \overrightarrow{a} = A D = 3 4 b − 3 2 a
iii) A D → − A B → = B D → \overrightarrow{AD}-\overrightarrow{AB }=\overrightarrow{BD} A D − A B = B D
B D → = \overrightarrow{BD} = B D = 4 3 b → − 2 3 a → \frac {4} {3} \overrightarrow{b} - \frac{2}{3} \overrightarrow{a} 3 4 b − 3 2 a − - − ( 4 3 a → − 2 3 b → ) (\frac{4}{3} \overrightarrow{a} - \frac {2}{3} \overrightarrow{b}) ( 3 4 a − 3 2 b )
B D → = 2 b → − 2 a → \overrightarrow{BD} =2\overrightarrow{b} -2\overrightarrow{a} B D = 2 b − 2 a
iv) A D → + A B → = A C → \overrightarrow{AD}+\overrightarrow{AB }=\overrightarrow{AC} A D + A B = A C
A C → = \overrightarrow{AC} = A C = 4 3 b → − 2 3 a → \frac {4} {3} \overrightarrow{b} - \frac{2}{3} \overrightarrow{a} 3 4 b − 3 2 a + ( 4 3 a → − 2 3 b → ) +(\frac{4}{3} \overrightarrow{a} - \frac {2}{3} \overrightarrow{b}) + ( 3 4 a − 3 2 b )
A C → = 2 3 a → + 2 3 b → \overrightarrow{AC} =\frac {2}{3} \overrightarrow{a} + \frac {2}{3} \overrightarrow{b} A C = 3 2 a + 3 2 b
