Question

Cho hình vuông ABCD tâm O, cạnh bằng 1. Gọi M là điểm trên cạnh AB sao cho BM = 2MA; N trên AC sao cho AN = 3NC. Tích vô hướng của \(\overrightarrow{MN}.\overrightarrow{DN}\) bằng:

A. \(-\dfrac{1}{8}\)

B. \(\dfrac{1}{9}\)

C. \(\dfrac{1}{8}\)

D. \(\dfrac{3}{4}\)

Answer 1

\(BM=2MA\Rightarrow\overrightarrow{AM}=\dfrac{1}{3}\overrightarrow{AB}\); \(AN=3NC\Rightarrow\overrightarrow{AN}=\dfrac{3}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\)

Do đó:

\(\overrightarrow{MN}.\overrightarrow{DN}=\left(\overrightarrow{MA}+\overrightarrow{AN}\right)\left(\overrightarrow{DA}+\overrightarrow{AN}\right)\)

\(=\left(-\dfrac{1}{3}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\right)\left(-\overrightarrow{AD}+\dfrac{3}{4}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\right)\)

\(=\left(\dfrac{5}{12}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\right)\left(\dfrac{3}{4}\overrightarrow{AB}-\dfrac{1}{4}\overrightarrow{AD}\right)\)

\(=\dfrac{5}{16}AB^2-\dfrac{3}{16}AD^2=\dfrac{1}{8}AB^2=\dfrac{1}{8}\) (chú ý rằng \(\overrightarrow{AB}.\overrightarrow{AD}=0\) và \(AB=AD=1\))