Post on 01-Apr-2015
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
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
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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
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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
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
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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
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
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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 !