\(\overrightarrow{AC}-\overrightarrow{AD}=\overrightarrow{AC}-\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CB}=\overrightarrow{AB}\)

Đáp án A đúng

Ta có: \(AB = BC = CD = DA = 1;\)

            \(AC = BD = \sqrt {A{B^2} + B{C^2}}  = \sqrt {{1^2} + {1^2}}  = \sqrt 2 \)

a) \(\overrightarrow a  = \overrightarrow {OB}  - \overrightarrow {OD}  = \overrightarrow {OB}  + \overrightarrow {DO}  = \left( {\overrightarrow {DO}  + \overrightarrow {OB} } \right) = \overrightarrow {DB} \)

\( \Rightarrow \left| {\overrightarrow a } \right| = \left| {\overrightarrow {DB} } \right| = DB = \sqrt 2 \)

b)  \(\overrightarrow b = \left( {\overrightarrow {OC}  - \overrightarrow {OA} } \right) + \left( {\overrightarrow {DB}  - \overrightarrow {DC} } \right)\)

   \( = \left( {\overrightarrow {OC}  + \overrightarrow {AO} } \right) + \left( {\overrightarrow {DB}  + \overrightarrow {CD} } \right) = \left( {\overrightarrow {AO}  + \overrightarrow {OC} } \right) + \left( {\overrightarrow {CD}  + \overrightarrow {DB} } \right)\)

   \( = \overrightarrow {AC}  + \overrightarrow {CB}  = \overrightarrow {AB} \)

\( \Rightarrow \left| {\overrightarrow b } \right| = \left| {\overrightarrow {AB} } \right| = AB = 1\)

Chú ý khi giải:

Khi có dấu trừ phía trước ta thường thay bằng vectơ đối của nó và ngược lại

\(\left|\overrightarrow{OA}-\overrightarrow{CB}\right|=\left|\overrightarrow{OA}+\overrightarrow{BC}\right|=\left|\overrightarrow{OA}+\overrightarrow{AD}\right|=\left|\overrightarrow{OD}\right|=OD=\dfrac{1}{2}BD=\dfrac{a\sqrt{2}}{2}\)

\(\left|\overrightarrow{AB}+\overrightarrow{DC}\right|=\left|\overrightarrow{AB}+\overrightarrow{AB}\right|=2\left|\overrightarrow{AB}\right|=2AB=2a\)

\(\left|\overrightarrow{CD}-\overrightarrow{DA}\right|=\left|\overrightarrow{CD}+\overrightarrow{AD}\right|=\left|\overrightarrow{BA}+\overrightarrow{AD}\right|=\left|\overrightarrow{BD}\right|=BD=a\sqrt{2}\)

1.

Gọi G là trọng tâm tam giác

\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}\)

\(\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{0}\)

\(\Leftrightarrow O\equiv G\)

\(\Rightarrow O\) là trọng tâm tam giác ABC

\(\Rightarrow\Delta ABC\) đều

Gọi độ dài các cạnh tam giác là a

\(\overrightarrow{BN}.\overrightarrow{AM}=\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=-\dfrac{1}{4}a^2-\dfrac{1}{8}a^2-\dfrac{1}{8}a^2+\dfrac{1}{2}a^2=0\)

Mặt khác \(\overrightarrow{BN}.\overrightarrow{AM}=BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)\)

\(\Rightarrow BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow\left(\overrightarrow{AM};\overrightarrow{BN}\right)=90^o\)

\(BD=\dfrac{AB}{cos45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)

\(\overrightarrow{BQ}.\overrightarrow{BP}=\dfrac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)

\(=\dfrac{1}{4}BA.BC.cos90^o+\dfrac{1}{4}BA.BD.cos45^o+\dfrac{1}{4}BD.BC.cos45^o+\dfrac{1}{4}BD^2\)

\(=\dfrac{1}{4}a^2+\dfrac{1}{4}a^2+\dfrac{1}{2}a^2=a^2\)