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Three Dimension (Distance). After learning this slide, you’ll be able to determine the distance between the elements in the space of three dimension. We will study the distance : point to point point to line point to plane line to line line to plane, and plane to plane. - PowerPoint PPT Presentation
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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
33
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
44
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
66
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
88
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
1010
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
1212
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
?
1313
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
1414
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
1515
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
1616
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
2323
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
2424
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
2525
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
2626
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
2727
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
2828
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
2929
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
3131
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
3232
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
3434
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
3535
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
3737
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
3838
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
3939
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
4040
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
4141
Have a nice try !Have a nice try !