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.

A sai

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

Khẳng định A 

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

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

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

Chọn B

B

a) Chứng minh \(\overrightarrow{AC}-\overrightarrow{BA}=\overrightarrow{AD}\)

Ta có: \(\overrightarrow{AC}-\overrightarrow{CD}=\overrightarrow{AD}\left(đpcm\right)\) ( vì \(\overrightarrow{BA}=\overrightarrow{CD}\) )

b) Chứng minh \(\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=AC\)

Ta có: \(\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}\) ( theo quy tắc hình bình hành )

\(\Rightarrow\left|\overrightarrow{AB}+\overrightarrow{AD}\right|=\left|\overrightarrow{AC}\right|=AC\left(đpcm\right)\)

bài này chả khó áp dụng 1 bước là ra ngay điều cần chứng minh rồi

\(\overrightarrow{AD}=\overrightarrow{BC}\Rightarrow T=\left|\overrightarrow{AB}+3\overrightarrow{AD}\right|\)

\(T^2=AB^2+9AD^2+6\overrightarrow{AB}.\overrightarrow{AD}\) (để ý rằng AB, AD vuông góc nên \(\overrightarrow{AB}.\overrightarrow{AD}=0\))

\(T^2=AB^2+9AD^2=2^2+9.3^2=85\)

\(\Rightarrow T=\sqrt{85}\)

1.

Đặt \(P=\left|\overrightarrow{AD}+3\overrightarrow{AB}\right|\Rightarrow P^2=AD^2+9AB^2+6\overrightarrow{AD}.\overrightarrow{AB}\)

\(=AD^2+9AB^2=10AB^2=10a^2\)

\(\Rightarrow P=a\sqrt{10}\)

2.

Tam giác ABC đều nên AM là trung tuyến đồng thời là đường cao \(\Rightarrow AM\perp BM\)

\(AM=\dfrac{a\sqrt{3}}{2}\) ; \(BM=\dfrac{a}{2}\)

\(T=\left|\overrightarrow{MA}+2\overrightarrow{MB}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|\overrightarrow{MA}+2\overrightarrow{MB}\right|\)

\(\Rightarrow T^2=MA^2+4MB^2+4\overrightarrow{MA}.\overrightarrow{MB}=MA^2+4MB^2\)

\(=\left(\dfrac{a\sqrt{3}}{2}\right)^2+4\left(\dfrac{a}{2}\right)^2=\dfrac{7a^2}{4}\Rightarrow T=\dfrac{a\sqrt{7}}{2}\)

3.

\(T=\left|\overrightarrow{AB}+\overrightarrow{CG}\right|=\left|\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CB}\right|=\left|\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{CA}+\dfrac{1}{3}\overrightarrow{AB}\right|\)

\(=\left|\dfrac{4}{3}\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AC}\right|\Rightarrow T^2=\dfrac{16}{9}AB^2+\dfrac{4}{9}AC^2-\dfrac{16}{9}\overrightarrow{AB}.\overrightarrow{AC}\)

\(=\dfrac{20}{9}AB^2-\dfrac{16}{9}AB^2.cos60^0=\dfrac{20}{9}a^2-\dfrac{16}{9}a^2.\dfrac{1}{2}=\dfrac{4}{3}a^2\)

\(\Rightarrow T=\dfrac{2a}{\sqrt{3}}\)

Bài 2:

\(\left|\overrightarrow{BC}+\overrightarrow{BA}\right|=\left|\overrightarrow{AC}\right|=AC=a\sqrt{2}\)

\(\left|\overrightarrow{AB}-\overrightarrow{AC}\right|=\left|\overrightarrow{CA}+\overrightarrow{AB}\right|=CB=a\)