a) Do ABCD cũng là một hình bình hành nên \(\overrightarrow {DA}  + \overrightarrow {DC}  = \overrightarrow {DB} \)

\( \Rightarrow \;|\overrightarrow {DA}  + \overrightarrow {DC} |\; = \;|\overrightarrow {DB} |\; = DB = a\sqrt 2 \)

b) Ta có: \(\overrightarrow {AD}  + \overrightarrow {DB}  = \overrightarrow {AB} \) \( \Rightarrow \overrightarrow {AB}  - \overrightarrow {AD}  = \overrightarrow {DB} \)

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

c) Ta có: \(\overrightarrow {DO}  = \overrightarrow {OB} \)

\( \Rightarrow \overrightarrow {OA}  + \overrightarrow {OB}  = \overrightarrow {OA}  + \overrightarrow {DO}  = \overrightarrow {DO}  + \overrightarrow {OA}  = \overrightarrow {DA} \)

\( \Rightarrow \left| {\overrightarrow {OA}  + \overrightarrow {OB} } \right| = \left| {\overrightarrow {DA} } \right| = DA = a.\)

Ta có: \(AC = BD = \sqrt {A{B^2} + B{C^2}}  = \sqrt {{a^2} + {a^2}}  = a\sqrt 2 \)

+) \(AB \bot AD \Rightarrow \overrightarrow {AB}  \bot \overrightarrow {AD}  \Rightarrow \overrightarrow {AB} .\overrightarrow {AD}  = 0\)

+) \(\overrightarrow {AB} .\overrightarrow {AC}  = \left| {\overrightarrow {AB} } \right|.\left| {\overrightarrow {AC} } \right|.\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = a.a\sqrt 2.\cos 45^\circ  = a^2\)

+) \(\overrightarrow {AC} .\overrightarrow {CB}  = \left| {\overrightarrow {AC} } \right|.\left| {\overrightarrow {CB} } \right|.\cos \left( {\overrightarrow {AC} ,\overrightarrow {CB} } \right) = a\sqrt 2 .a.\cos 135^\circ  =  - {a^2}\)

+) \(AC \bot BD \Rightarrow \overrightarrow {AC}  \bot \overrightarrow {BD}  \Rightarrow \overrightarrow {AC} .\overrightarrow {BD}  = 0\)

Chú ý

\(\overrightarrow {a}  \bot \overrightarrow {b}  \Leftrightarrow \overrightarrow {a} .\overrightarrow {b}  = 0\)

A sai

\(\overrightarrow{AB}-\overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{DA}=\overrightarrow{DA}+\overrightarrow{AB}=\overrightarrow{DB}=-\overrightarrow{BD}\) mới đúng

Khẳng định A 

Gọi N là trung điểm BC

\(\left|\overrightarrow{MA}+\overrightarrow{MC}+2\overrightarrow{MB}+2\overrightarrow{OC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow\left|2\overrightarrow{MO}+2\overrightarrow{MB}+2\overrightarrow{OC}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow\left|2\overrightarrow{MC}+2\overrightarrow{MB}\right|=\left|\overrightarrow{AB}-\overrightarrow{AD}\right|\)

\(\Leftrightarrow4\left|\overrightarrow{MN}\right|=\left|\overrightarrow{BD}\right|\)

\(\Rightarrow\left|\overrightarrow{BD}\right|=4\left|\overrightarrow{MN}\right|=4\left|\overrightarrow{DN}+\overrightarrow{MD}\right|\ge4MD-4DN\)

\(\Rightarrow4MD\le BD+4DN\)

\(\Leftrightarrow MD\le\dfrac{BD+4DN}{4}=\dfrac{a\sqrt{2}+2a\sqrt{5}}{4}=\dfrac{2\sqrt{5}+\sqrt{2}}{4}a\)

a, \(AC=\dfrac{AB}{sin45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)

\(\overrightarrow{AB}.\overrightarrow{AC}=AB.AC.cos\widehat{BAC}=a.a\sqrt{2}.cos45^o=a^2\)

b, \(\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{BD}+\overrightarrow{BC}\right)=\overrightarrow{AC}\left(\overrightarrow{BD}+\overrightarrow{BC}\right)\)

\(=\overrightarrow{AC}.\overrightarrow{BD}+\overrightarrow{AC}.\overrightarrow{BC}\)

\(=AC.BD.cos90^o+AC.AD.cos45^o\)

\(=a\sqrt{2}.a\sqrt{2}.0+a\sqrt{2}.a.\dfrac{\sqrt{2}}{2}=a^2\)

c, \(\overrightarrow{AB}.\overrightarrow{BD}=AB.BD.cos135^o=-a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=-a^2\)

d, \(\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\left(2\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{BC}.\left(\overrightarrow{AD}+\overrightarrow{BD}\right)\)

\(=\overrightarrow{BC}.\overrightarrow{AD}+\overrightarrow{BC}.\overrightarrow{BD}\)

\(=AD^2+BC.BD.cos45^o\)

\(=a^2+a.a\sqrt{2}.\dfrac{\sqrt{2}}{2}=2a^2\)

e, \(\left(\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}\right)\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{DC}\right)\)

\(=\left(\overrightarrow{AC}+\overrightarrow{AC}\right)\left(\overrightarrow{DB}+\overrightarrow{DB}\right)\)

\(=4.\overrightarrow{AC}.\overrightarrow{DB}=4.AC.DB.cos90^o=0\)

+) ABCD là hình thoi nên cũng là hình bình hành

 Áp dụng quy tắc hình bình hành ta có:

 \(\overrightarrow p  = \overrightarrow {AB}  + \overrightarrow {AD}  = \overrightarrow {AC} \)

 \(\Rightarrow  |\overrightarrow p|  = | \overrightarrow {AC}| =AC \)

+) \(\overrightarrow u  = \overrightarrow {AB}  - \overrightarrow {AD}  = \overrightarrow {DB} \)

 \(\Rightarrow  |\overrightarrow u|  = | \overrightarrow {DB}| =DB\)

+) \(\overrightarrow v  = 2\overrightarrow {AB}  - \overrightarrow {AC}  = \overrightarrow {AB}  + \left( {\overrightarrow {AB}  - \overrightarrow {AC} } \right) = \overrightarrow {AB}  + \overrightarrow {CB} \)\( = \overrightarrow {AB}  + \overrightarrow {DA}  = \overrightarrow {DB} \)

 \(\Rightarrow  |\overrightarrow v|  = | \overrightarrow {DB}| =DB\)

+ Tính \(AC, DB\)

Tam giác ABD có \(AB=AD=a, \widehat A = 60^o\) nên nó là tam giác đều. Do đó DB = a.

Gọi O là giao điểm hai đường chéo.

Ta có: \(AO = AB. \sin B = a. \sin 60^o = \frac {a \sqrt 3}{2} \Rightarrow  AC = a \sqrt 3\)

Vậy \(|\overrightarrow p|  =  a \sqrt 3 ,|\overrightarrow u|  =  a, |\overrightarrow v|  =  a.\)

a) \(\overrightarrow {BD}  = \overrightarrow {AD}  - \overrightarrow {AB} ;\;\overrightarrow {AC}  = \overrightarrow {AB}  + \overrightarrow {AD} .\)

b) \(\overrightarrow {AB} .\overrightarrow {AD}  = 4.6.\cos \widehat {BAD} = 24.\cos {60^o} = 12.\)

\(\begin{array}{l}\overrightarrow {AB} .\overrightarrow {AC}  = \overrightarrow {AB} (\overrightarrow {AB}  + \overrightarrow {AD} ) = {\overrightarrow {AB} ^2} + \overrightarrow {AB} .\overrightarrow {AD}  = {4^2} + 12 = 28.\\\overrightarrow {BD} .\overrightarrow {AC}  = (\overrightarrow {AD}  - \overrightarrow {AB} )(\overrightarrow {AB}  + \overrightarrow {AD} ) = {\overrightarrow {AD} ^2} - {\overrightarrow {AB} ^2} = {6^2} - {4^2} = 20.\end{array}\)

c) Áp dụng định lí cosin cho tam giác ABD ta có:

\(\begin{array}{l}\quad \;B{D^2} = A{B^2} + A{D^2} - 2.AB.AD.\cos A\\ \Leftrightarrow B{D^2} = {4^2} + {6^2} - 2.4.6.\cos {60^o} = 28\\ \Leftrightarrow BD = 2\sqrt 7 .\end{array}\)

Áp dụng định lí cosin cho tam giác ABC ta có:

\(\begin{array}{l}\quad \;A{C^2} = A{B^2} + B{C^2} - 2.AB.BC.\cos B\\ \Leftrightarrow A{C^2} = {4^2} + {6^2} - 2.4.6.\cos {120^o} = 76\\ \Leftrightarrow AC = 2\sqrt {19} .\end{array}\)

Tham khảo:

A. Ta có: \(\left( {\overrightarrow {AB} ,\overrightarrow {BD} } \right) = \left( {\overrightarrow {BE} ,\overrightarrow {BD} } \right) = {135^o} \ne {45^o}.\) Vậy A sai.

B. Ta có: \(\left( {\overrightarrow {AC} ,\overrightarrow {BC} } \right) = \left( {\overrightarrow {CF} ,\overrightarrow {CG} } \right) = {45^o}\) và  \(\overrightarrow {AC} .\overrightarrow {BC}  = AC.BC.\cos {45^o} = a\sqrt 2 .a.\frac{{\sqrt 2 }}{2} = {a^2}.\)

Vậy B đúng.

Chọn B

C. Dễ thấy \(AC \bot BD\) nên \(\overrightarrow {AC} .\overrightarrow {BD}  = 0 \ne {a^2}\sqrt 2.\) Vậy C sai.

D. Ta có: \(\left( {\overrightarrow {BA} .\overrightarrow {BD} } \right) = {45^o}\) \( \Rightarrow \overrightarrow {BA} .\overrightarrow {BD}  = BA.BD.\cos {45^o} = a.a\sqrt 2 .\frac{{\sqrt 2 }}{2} = {a^2} \ne  - {a^2}.\) Vậy D sai.