Question

Bài 3. (1 điểm) Cho hình vuông $ABCD$ cạnh $a$ có tâm $O$ là giao điểm của hai đường chéo.

a) Tính độ dài của $\overrightarrow{OA}-\overrightarrow{CB}$ .

b) Chứng minh rằng $\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}=\overrightarrow{0}$.

Answer 1

1.Ta có: $ABCD$ là hình vuông

$\to \overrightarrow{CB}=\overrightarrow{DA}$

$\to \overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OA}-\overrightarrow{DA}=\overrightarrow{OA}+\overrightarrow{AD}=\overrightarrow{OD}$

→|→OA−→CB|=|→OD|=OD=a√2→|OA→−CB→|=|OD→|=OD=a2

2.Ta có:

$\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}$

$=\overrightarrow{BA}+\overrightarrow{DC}$

$=\overrightarrow{BA}-\overrightarrow{CD}$

$=\overrightarrow{BA}-\overrightarrow{BA}$ vì $ABCD$ là hình vuông 
$\overrightarrow{CD}=\overrightarrow{BA}$

=0⃗

Answer 2

e không biết lun

 

Answer 3

a) BC = AC = AB√ 2 = a√ 2

O - tđ AC và BD

vecter OA = vecter CO

 

Answer 4

a) BC = AC = AB√ 2 = a√ 2

O - tđ AC và BD

vecter OA = vecter CO

Answer 5

C

Answer 6

Answer 7

hi

Answer 8

a) Ta có ��→−��→=��→−��→=��→+��→=��→OA−CB=OA−DA=OA+AD=OD

⇒∣��→−��→∣=∣��→∣=��=12��=12��2+��2=12�2+�2=�22⇒​OA−CB​=​OD​=OD=21​BD=21​AB2+AD2​=21​a2+a2​=2a2​​

b) ��=��→−��→+��→=(��→−��→)+��→=��→+��→=��→+��→=0→=��VT=DA−DB+DC=(DA−DB)+DC=BA+DC=CD+DC=0=VP.

Answer 9

em làm vở

Answer 10

ko bt

Answer 11

 

Answer 12

a.Ta có: $ABCD$ là hình vuông

$\to \overrightarrow{CB}=\overrightarrow{DA}$

$\to \overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OA}-\overrightarrow{DA}=\overrightarrow{OA}+\overrightarrow{AD}=\overrightarrow{OD}$

→|→OA−→CB|=|→OD|=OD=a√2→|OA→−CB→|=|OD→|=OD=a2

b.Ta có:

$\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}$

$=\overrightarrow{BA}+\overrightarrow{DC}$

$=\overrightarrow{BA}-\overrightarrow{CD}$

$=\overrightarrow{BA}-\overrightarrow{BA}$ vì $ABCD$ là hình vuông 
$\overrightarrow{CD}=\overrightarrow{BA}$

Answer 13

a) Ta có ��→−��→=��→−��→=��→+��→=��→OA−CB=OA−DA=OA+AD=OD

⇒∣��→−��→∣=∣��→∣=��=12��=12��2+��2=12�2+�2=�22⇒​OA−CB​=​OD​=OD=21​BD=21​AB2+AD2​=21​a2+a2​=2a2​​

b) ��=��→−��→+��→=(��→−��→)+��→=��→+��→=��→+��→=0→=��VT=DA−DB+DC=(DA−DB)+DC=BA+DC=CD+DC=0=VP.

Answer 14

vâng ạ

Answer 15

a) Ta có ��→−��→=��→−��→=��→+��→=��→OA−CB=OA−DA=OA+AD=OD

⇒∣��→−��→∣=∣��→∣=��=12��=12��2+��2=12�2+�2=�22⇒​OA−CB​=​OD​=OD=21​BD=21​AB2+AD2​=21​a2+a2​=2a2​​

b) ��=��→−��→+��→=(��→−��→)+��→=��→+��→=��→+��→=0→=��VT=DA−DB+DC=(DA−DB)+DC=BA+DC=CD+DC=0=VP.

Answer 16

ad

Answer 17

  

Answer 18

 

Answer 19

 

Answer 20

 

Answer 21

 

Answer 22

0

Answer 23

 

Answer 24

 

Answer 25

 

Answer 26

Oa-cb =
da-db+dc=0

Answer 27

a) Ta có \(\overrightarrow{OA}-\overrightarrow{CB}=\overrightarrow{OA}-\overrightarrow{DA}=\overrightarrow{OA}+\overrightarrow{AD}=\overrightarrow{OD}\)    

\(\rightarrow|\overrightarrow{OA}-\overrightarrow{CB}|=|\overrightarrow{OD|}=OD=\dfrac{1}{2}BD=\dfrac{1}{2}\sqrt{AB^2+AD^2}=\dfrac{1}{2}\sqrt{a^2+a^2}=\dfrac{a\sqrt{2}}{2}\)

b) \(VT=\overrightarrow{DA}-\overrightarrow{DB}=(\overrightarrow{DA}-\overrightarrow{DB})+\overrightarrow{DC}=\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{CD}+\overrightarrow{DC}=\overrightarrow{0}=VP\)

Answer 28

Ta có ��→−��→=��→−��→=��→+��→=��→OA−CB=OA−DA=OA+AD=OD

⇒∣��→−��→∣=∣��→∣=��=12��=12��2+��2=12�2+�2=�22⇒​OA−CB​=​OD​=OD=21​BD=21​AB2+AD2​=21​a2+a2​=2a2​​

b) ��=��→−��→+��→=(��→−��→)+��→=��→+��→=��→+��→=0→=��VT=DA−DB+DC=(DA−DB)+DC=BA+DC=CD+DC=0=VP

Answer 29

File: undefined 

Answer 30

a) OA-CB= OA+BC

Mà BC= AD

=>OA+BC= OA+AD= OD

b) DA-DB+DC= DA+BD+DC= BA+DC