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Vectors and configurations

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Season 01 • Episode 14 • Vectors and configurations

0

Vectors and configurations

Season 01

Episode 14 Time frame 1 period

Prerequisites :

Basinotionsaboutvetors.

Objectives :

•

Disover some simplebut importantongurations involvingvetors.

•

Pratise the voabulary about geometry.

Materials :

•

Hand-outs withall the ongurations and properties.

•

Properties involvingvetorsand gures.

1 – Matching game 10 mins

Eah student is handed out a ard with either a property involving vetors or a gure.

Students mingle tond who has the rightproperty for their gureand vie-versa.

2 – Writing a property 15 mins

Eahpairhastowriteapreiseruleabout thepropertyand gurethey have.The teaher

heks eah one of them.

3 – Presentations 30 mins

Eah pair goes to the board to explain their property. A hand-out with all properties is

given at the end of the session.

(2)

Vectors and configurations

Document Summary

A quadrilateral

ABCD

is a parallelogram if and onlyif

− − − − − →

AB =

− − − − − →

DC.

b

A

b B

b

C

b D

A quadrilateral

ABDC

is a parallelogram if and onlyif

− − − − − →

AB =

− − − − − →

CD.

b

A

b B

b D

b C

For any three points

A

,

B

and

C

,

− − − − − →

AB +

− − − − − →

BC =

− − − − − →

AC.

b

A

b

B

b

C

A quadrilateral

ABCD

is a parallelogram if and onlyif

− − − − − →

AB +

− − − − − →

AD =

− − − − − →

AC.

b

A

b

B

b

D

b

C

Apoint

A

isthemidpointofasegment

BC

if and onlyif

− − − − − →

AB +

− − − − − →

AC = → 0. b

B

b

C

b

A

(3)

Season 01 • Episode 14 • Vectors and configurations

2

Apoint

C

isthemidpointofasegment

AB

if and onlyif

− − − − − →

AB = 2

− − − − − →

AC. b

A

b

B

b

C

The fat that the lines

CD

and

AB

are parallel,that thevetorshavethe same di-

retionandthat

CD = 2AB

anbewritten as the equality

− − − − − →

CD =

− − − − − − − − →

2AB.

b

A b

B

b

C

b

D

Ifpoint

B

isthemidpointofasegment

AC

, then for any point

M

,

− − − − − − →

M A +

− − − − − − − →

M C = 2

− − − − − − − →

M B.

b

M

b

A

b C

b

B

Ifthe points

B

and

C

are the midpointsof the legs of atriangle

ADE

, then

− − − − − →

DE = 2

− − − − − →

BC.

b

A

b

D b

E

b

B

b

C

If

A

is the intersetion of two segments

BC

and

DE

suh that

AB = 1

3 AC

and

AD = 1

3 AE

then

− − − − − →

BD = − 1 3

− − − − − →

CE.

b A

b E

b C

b B

b D

(4)

Document 1

Properties and gures

− − − − − →

AB =

− − − − − →

DC − AB − − − − → =

− − − − − →

C D

− − − − − →

AB +

− − − − − →

BC =

− − − − →

AC − AB − − − − → +

− − − − − →

AD =

− − − − →

AC

− − − − − →

AB +

− − − − →

AC = → 0

− − − − − →

AB = 2

− − − − →

AC

− − − − − →

C D =

− − − − − − − →

2 AB − M A − − − − − → +

− − − − − − →

M C = 2

− − − − − − − →

M B

− − − − − →

DE = 2

− − − − − →

BC − BD − − − − → = − 1

3

− − − − − →

C E

(5)

Season 01 • Episode 14 • Vectors and configurations

4

b

A

b B

b C

b D

b

A

b B

b D

b C

b

A

b

B

b

C

b

A

b

B

b

D

b C

b

B

b

C

b

A

b

A

b

B

b

C

b

M

b

A

b

C

b

B b

A b

B

b

C

b

D

b

A

b

D b

E

b

B

b

C

b A

b E

b C

b B

b D

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