11

3 Dimension(Distance)

22

BASE COMPETENCE :

Determine distance between two points, a point to line and a point to

plane in the 3 dimension.

33

THE MATERIAL

Distance in 3 Dimension : distance between two points

distance point to line

distance point to plane

44

Distance between two points

Distance from points A to

point B is the shortest way

from A to B

A

B

Distan

ce b

etwee

n tw

o po

ints

55

Example 1Given :The cube ABCD –EFGHWith the long of the edge is a cm.

Find distance : 1. point A to C, 2. Point A to G,3. Point A to centre of line FH

A BCD

HE F

G

a cm

a cm

a cm

P

66

Solving: Look at the right

triangle ABC

that right at B, so

AC = = = = So distance A to C= cm(AC is side diagonal)

A BCD

HE F

G

a cm

a cm

a cm

22 BCAB 22 aa

2a2

2a

2a

77

Distance A to G = ?Look at the triangle

ACG thatright at C, so

AG =

= = = =So distance A to G = cm(AG is space diagonal)

A BCD

HE F

G

a cm

a cm

a cm

22 CGAC 22 a)2a(

2a3 3a

3a

22 aa2

88

A BCD

HE F

G

a cm

P

Distance A to P = ?Look at the triangle

AEP thatRight at E, so

AP =

=

=

= =So distance A to P = cm

22 EPAE

2

212 2aa

2212 aa

223 a 6a2

1

6a21

99

Distance point to line

A

g

Dis

tanc

e fr

om p

oint

A

to li

ne g

Distance point A to line g is length of perpendicular segment from point A to line g

1010

Example 1

Given cubeABCD-EFGH, with length of edge5 cm.Distance point A to segment HG is….

A BCD

HE F

G

5 cm

5 cm

1111

Solving

Distance point A tosegment HG is segment AH, (AH HG)

A BCD

HE F

G

5 cm

5 cm

AH = (AH side diagonal)

AH =

So distance A to HG = 5√2 cm

2a

25

1212

Example 2

Given cubeABCD-EFGHWith length of edge 6 cm.Distance point B to diagonal AG is….

A BCD

HE F

G

6 cm

6 cm

1313

Solving

Distance B to AG =Distance B to P (BPAG)Side diagonal BG =6√2 cmSpace diagonal AG= 6√3 cmLook at triangle ABG

A BCD

HE F

G

6√2

cm6 cm

P6√

3 cm

A B

G

P

6√3

6

6√2

?

1414

Look at the triangle ABGSin A = = =

BP =

BP = 2√6

A B

G

P6√

3

6

6√2AG

BGAB

BP

36

26

6

BP

36

)6)(26(

?

So distance B to AG = 2√6 cm

3

66

3

3x

2

1515

Example 3

Given T.ABCDUniform pyramid.Length of the based side is 12 cm, and Length of the vertical side is 12√2 cm. Distance A to TC is…12 cm

12√2

cm

T

C

A B

D

1616

Solving

Distance A to TC = APAC = side diagonal = 12√2AP = = = =

So distance 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

1717

Example 4

Given cubeABCD-EFGHWith length of edge6 cmA B

CD

HE F

G

6 cm6 cm

point P at the middle of FG.

Distance point A to line DP is….

P

1818

A BCD

HE F

G

6 cm6 cm

P

Solving

Q

6√2

cm

R

P

AD

G F

6 cm

3 cm

DP =

=

=

22 GPDG 22 3)26(

9972

1919

Solving

Q

6√2

cm

R

P

AD

G F

6 cm

3 cmDP =

Area of triangle ADP

½DP.AQ = ½DA.PR

9.AQ = 6.6√2

AQ = 4√2

So distance A to DP = 4√2 cm

9972

4

2020

Distance point to plane

Distance between point A to plane V is length of segment that connect perpendicular point A to plane V

A

V

2121

Line perpendicular Plane

Line g said perpendicular plane V if line g perpendicular at two lines that intersection in plane V

V

g

a

bg a, g b,

So g V

2222

Example 1

Given cubeABCD-EFGHWith edge length 10 cmDistance point A toplane BDHF is….

A BCD

HE F

G

10 cm

P

2323

Solving

Distance point A toPlane BDHF Represented by length of line AP. (APBD)AP = ½ AC (ACBD) = ½.10√2 = 5√2

A BCD

HE F

G

10 cm

P

So distance A to BDHF = 5√2 cm

2424

Example 2

Given uniform pyramid T-ABCD.Length of AB = 8 cmand TA = 12 cm.Distance point T to plane ABCD is…

8 cm

T

C

A B

D

12 c

m

2525

Solving

Distance T to ABCD = Distance T to intersection AC and BD = TP AC is side diagonal AC = 8√2AP = ½ AC = 4√2

8 cm

T

C

A B

D

12 c

m

P

2626

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

So distance T to ABCD = 4√7 cm

2727

Example 3

Given cubeABCD-EFGHWith length of edge 9 cm.Distance point C to plane BDG is…

A BCD

HE F

G

9 cm

2828

Solving

Distance point C to plane BDG = CPCP perpendicular with GT

A BCD

HE F

G

9 cm

PT

CP = ⅓CE = ⅓.9√3 = 3√3

So distance C to BDG = 3√3 cm

2929

GOOD LUCK !GOOD LUCK !