ABCD is a square: A B → = D C → , B C → = A D → \overrightarrow{AB}=\overrightarrow{DC}, \overrightarrow{BC}=\overrightarrow{AD} A B = D C , BC = A D
P , Q P, Q P , Q are the midpoints of B C , C D BC, CD BC , C D respectively: B P → = 1 2 B C → , D Q → = 1 2 D C → \overrightarrow{BP}=\dfrac{1}{2}\overrightarrow{BC}, \overrightarrow{DQ}=\dfrac{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}+\dfrac{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}+\dfrac{1}{2}\overrightarrow{DC} A Q = A D + D Q = A D + 2 1 D C If A P → = a ⃗ \overrightarrow{AP}=\vec{a} A P = a and A Q → = b ⃗ \overrightarrow{AQ}=\vec{b} A Q = b
A B → + 1 2 A D → = a ⃗ \overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AD}=\vec{a} A B + 2 1 A D = a A D → + 1 2 A B → = b ⃗ \overrightarrow{AD}+\dfrac{1}{2}\overrightarrow{AB}=\vec{b} A D + 2 1 A B = b
(i)
A D → = b ⃗ − 1 2 A B → \overrightarrow{AD}=\vec{b}-\dfrac{1}{2}\overrightarrow{AB} A D = b − 2 1 A B A B → + 1 2 ( b ⃗ − 1 2 A B → ) = a ⃗ \overrightarrow{AB}+\dfrac{1}{2}(\vec{b}-\dfrac{1}{2}\overrightarrow{AB})=\vec{a} A B + 2 1 ( b − 2 1 A B ) = a A B → = 4 3 a ⃗ − 2 3 b ⃗ \overrightarrow{AB}=\dfrac{4}{3}\vec{a}-\dfrac{2}{3}\vec{b} A B = 3 4 a − 3 2 b
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