1

Three Dimension

(Distance)After learning this slide, you’ll be able to determine the distance between the elements in the space of three dimension

22

We will study the distance :

point to point

point to line

point to plane

line to line

line to plane, and

plane to plane

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The distance of point to point

This display, shows that the

distance of point A to B is the length

of line segment which connect

point A to point B A

B

Jara

k du

a tit

ik

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e. g. :Given that the edge

length of a cube ABCD.EFGH is a cm.

Determine the distance of :

a) Point A to point Cb) Point A to point G

c) The distance of point A to

the middle of plane EFGH

A BCD

HE F

G

a cm

a cm

a cm

P

55

Solution:

Consider Δ ABC which has right angle at B

AC = = =

= Thus, the diagonal of AC = cm

A BCD

HE F

G

a cm

a cm

a cm

22 BCAB 22 aa

22a

2a

2a

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Distance of AG

Consider Δ ACG which has right angle at C

AG = = = = =

Thus, the diagonal of AG = cm

A BCD

HE F

G

a cm

a cm

a cm

22 CGAC 22 a)2a(

2a3 3a

3a

22 aa2

77

A BCD

HE F

G

a cm

P

Distance of AP

Consider Δ AEP which has right angle at E

AP =

=

=

= =

Thus distance of A to P = cm

22 EPAE

2

212 2aa

2212 aa

223 a 6a2

1

6a21

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Distance Point to Line

A

g

dist

ance

poi

nt to

line

This display shows the distance from point A to line g is length of the line segment which is connected from point A and is perpendicular to line g.

99

e.g. 1:

Given that the edge length of a cube ABCD.EFGH is 5 cm.The distance from point A to the edge of HG is…

A BCD

HE F

G

5 cm

5 cm

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SolutioThe distance from point A to the edge of HG is length of the line segment AH, (AH HG)

A BCD

HE F

G

5 cm

5 cm

AH = (AH is a side diagonal)

AH = Thus, the distance from point A to the edge of HG= 5√2 cm

2a

25

1111

e.g. 2:

Given that the edge length of a cube ABCD.EFGH is 6 cm.The distance from point B to the diagonal of AG is…

A BCD

HE F

G

6 cm

6 cm

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Solution

The distance from point B to AG = the distance from point B to P (BP AG)The side diagonal of BG = 6√2 cmThe space diagonal of AG = 6√3 cmConsider a triangle ABG !

A BCD

HE F

G

6√2

cm6 cm

P6√

3 cm

A B

G

P

6√3

6

6√2

?

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Consider a triangle ABGSin A = = =

BP =

BP = 2√6

A B

G

P

6√3

6

6√2AG

BG

AB

BP

36

26

6

BP

36

)6)(26(

?

Thus, the distance from point B to AG= 2√6 cm

3

66

3

3x 2

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e.g. 3

Given that T.ABCDis a pyramid. The edge length of its base is 12 cm, and the edge length of its upright is 12√2 cm. The distance from A to TC is...12 cm

12√2

cm

T

C

A B

D

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SolutionThe distance from A to TC= APAC is a cube’s diagonalAC = 12√2AP = = = = Thus, the distance from A to TC= 6√6 cm

12 cm

12√2

cm

T

C

A B

D

P

12√2

6√2

6√2

22 PCAC 22 )26()212( 108.2)36 144(2

6636.3.2

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e.g. 4 :

Given that the edge length of a cube ABCD.EFGH is 6 cm and

A BCD

HE F

G

6 cm6 cm

Point P is in the middle of FG.

The distance from point A to line DP is…

P

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A BCD

HE F

G

6 cm6 cm

P Solution

Q

6√2

cm

R

P

AD

G F

6 cm

3 cm

DP =

=

=

22 GPDG 22 3)26(

9972

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Solution

Q

6√2

cm

R

P

AD

G F

6 cm

3 cmDP =

Area of ADP

½DP.AQ = ½DA.PR

9.AQ = 6.6√2

AQ = 4√2

Thus the distance from point A to line DP= 4√2 cm

9972

4

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Perpendicular Line toward a plane

Perpendicular line toward a plane means that line is perpendicular to two intersecting lines which are located on a plane..

V

g

a

bg a, g b,

Thus g V

2020

The Distance of a Point to a PlaneThis display shows

the distance between point A and plane V is length of line segment which connect point A to plane V perpendicularly.

A

V

2121

e.g. 1 :

Given that the edge length of a cube ABCD.EFGHis 10 cm.Thus the distance from point A to plane is….

A BCD

HE F

G

10 cm

P

2222

SolutionThe distance from point A to plane BDHF is representated by the length of AP (APBD)AP = ½ AC (ACBD) = ½.10√2 = 5√2

A BCD

HE F

G

10 cm

P

Thus the distance from A to plane BDHF = 5√2 cm

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e.g. 2 :Given that T.ABCD is a pyramid.The length of AB = 8 cmand TA = 12 cm.The distance from point T to plane ABCD is….8 cm

T

C

A B

D

12 c

m

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SolutionThe distance from T to ABCD = The distance from T to the intersection of AC and BD= TP AC is a cube’ss diagonalAC = 8√2AP = ½ AC = 4√2

8 cm

T

C

A B

D

12 c

m

P

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AP = ½ AC = 4√2 TP = = = = = 4√7 8 cm

T

C

A B

D

12 c

m

P

2 2 AP AT 2 2 )24( 12

32 144 112

Thus the distance from T to ABCD = 4√7 cm

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e.g. 3 :

Given that the edge length of a cube ABCD.EFGHis 9 cm.The distance from point C to plane BDG is….

A BCD

HE F

G

9 cm

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SolutionThe distance from point C to plane BDG = CPThat is the line segment which is drawn through point C and perpendicular to GT

A BCD

HE F

G

9 cm

PT

CP = ⅓CE = ⅓.9√3 = 3√3

Thus the distance from C to BDG = 3√3 cm

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The Distance of line to line

This display explains the distance of line g and line h h is the length of line segment which connect those lines perpendicularly.

P

Q

g

h

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e.g.Given that the edge length of a cube ABCD.EFGHis 4 cm.Determine the distance of:A B

CD

HE F

G

4 cm a.Line AB to line HG

b.Line AD to line HF

c.Line BD to line EG

3030

SolutionThe distance of line:a.AB to line HG = AH (AH AB, AH HG) = 4√2 (a side

diagonal)b.AD to line HF = DH (DH AD, DH HF = 4 cm

A BCD

HE F

G

4 cm

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Solution

The distance of:b.BD to line EG = PQ (PQ BD, PQ EG = AE = 4 cm

A BCD

HE F

G

4 cm

P

Q

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The Distance of Line to Plane

This display shows the distance of line g to plane V islength of line segment which connect that line and plane perpendicularly.

V

g

g

3333

e.g. 1

Given that the edge length os a cobe ABCD.EFGH is 8 cmThe distance of line AE to planeBDHF is….

A BCD

HE F

G

8 cm

P

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SolutionThe distance of line AE to plane BDHF Is represented by the length of AP.(AP AEAP BDHF)AP = ½ AC(ACBDHF) = ½.8√2 = 4√2

A BCD

HE F

G

8 cm

P

Thus the distance from A to BDHF = 4√2 cm

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V

W

The Distance of Plane to Plane

This display explains the distance of plane W and plane V is length of line segment which is perpendicuar to plane W and is perpendicular to plane V.

W

Jarak Dua B

idang

3636

e.g. 1 :

Given that the edge length of a cubeABCD.EFGH is6 cm.The distance of plane AFH to plane BDG is….

A BCD

HE F

G

6 cm

6 cm

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SolutionThe distance of plane AFHto plane BDGIs represented by PQPQ = ⅓ CE(CE is a space diagonal)PQ = ⅓. 6√3 = 2√3

A BCD

HE F

G

6 cm

6 cm

P

Q

Thus the distance of AFH to BDG = 2√3 cm

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e.g. 2 :Given that the edge length of a cubeABCD.EFGH is 12 cm.

A BCD

HE F

G

12 cm

Points K, L and M are the middle point of BC, CDdan CG. The distance of plane AFH and KLM is….

KL

M

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Solution• Diagonal EC = 12√3• The distance from E to AFH = distance from AFH to BDG = distance from BDG to CA B

CD

HE F

G

12 cm

Thus the distance from point E to AFH = ⅓EC =⅓.12√3 = 4√3So that the distance from BDG to C is 4√3 too.

L

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A BCD

HE F

G

12 cm

The distance of BDG to point C is 4√3.The distance of BDG to KLM = distance of KLM to point C = ½.4√3 = 2√3

KL

M

Thus the distance of AFH to KLM = Distance of AFH to BDG + distance of BDG to KLM = 4√3 + 2√3 = 6√3 cm

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Have a nice try !Have a nice try !