THREE DIMENSIONS · Web viewMathematics is the way you think THREE DIMENSIONS STANDARD COMPETENCE:6. Determine position, distance and angles related to point, line and plane in three

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Mathematics is the way you think

THREE DIMENSIONS

THREE DIMENSIONS

SMA Negeri 2 Sekayu Mathematics is The Way You Think

STANDARD COMPETENCE:6. Determine position, distance and angles related to point, line and plane

in three dimensionsBASIC COMPETENCE:

Determine the point position, line and the plane in three dimensionsDetermine the distance from point to line and from point to plane in three dimensions.Determine angle between line and plane in solid figure

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SMA NEGERI 2 SEKAYU3-D (or 3D) means three dimensional, or having three dimensions. For example the picture above, it has up and down, left and night and front and back. 3D The three dimensions are called length (or depth), width (or breadth) and height. 3D is also used to make video games or animated movies.

THE GREAT PYRAMIDS

The Great Pyramid of Khufu, at Giza, Egypt, is 751 feet long on each side at the base, is 450 feet high, and is composed of approximately 2 million blocks of stone, each weighing more than 2 tons. The maximum error between side lengths is less than 0.1%.

The sloping angle of its sides is 51°51'. Each side is oriented with the compass points of north, south, east, and west. Each cross section of the pyramid (parallel to the base) is a square.

Until the 19th century, the Great Pyramid at Giza was the tallest building in the world. At over 4500 years in age, it is the only one of the famous Seven Wonders of the Ancient World that remains standing. According to the Greek historian Herodotus, the Great Pyramid was built as a tomb for the Pharaoh Khufu. (www.regentsprep.com)


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7.1 INTRODUCING OF SOLID FIGURE

Polyhedral(they must have flat faces )

Platonic SolPids

Prisms

Pyramids

Non-Polyhedra:(if any surface is not flat)

Sphere Torus

Cylinder Cone

7.2 POINT, LINE, PLANEIn geometry, definition are formed using known words or t describe a terms to describe a new word. There are three words in geometry that are not formally defined. These three undefined terms are point, line and plane.


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http://www.mathsisfun.com/geometry/cone.html

http://www.mathsisfun.com/geometry/cylinder.html

http://www.mathsisfun.com/geometry/torus.html

http://www.mathsisfun.com/geometry/sphere.html

http://www.mathsisfun.com/geometry/tetrahedron.html

http://www.mathsisfun.com/geometry/hexahedron.html

http://www.mathsisfun.com/geometry/octahedron.html

http://www.mathsisfun.com/geometry/dodecahedron.html

http://www.mathsisfun.com/geometry/icosahedron.html

A. POINT

In geometry, a point has no dimension (actual size). Even though we represent a point with a dot, the point has no length, width, or thickness. Our dot can be very tiny or very large and it still represents a point. A point is usually named with a capital letter. In the coordinate plane, a point is named by an ordered pair, (x,y).

B. LINE

In geometry, a line has no thickness but its length extends in one dimension and goes on forever in both directions. Unless otherwise stated a line is drawn as a straight line with two arrowheads indicating that the line extends without end in both directions. A line is named by a single lowercase letter, l, or by any two points on the line, .

C. PLANE Look at the picture

Undecagon Triangle Square Pentagon

Octagon Dodecagon Circle Ellipse


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From the picture above which one is plane? Ok, I believe that your answer will be all of them. Yes, there are two kinds of plane. Flat surface and not flat. On the next , what we mean a plane is a flat plane

In geometry, a plane has no thickness but extends indefinitely in all directions. Planes are usually represented by a shape that looks like a tabletop or a parallelogram. Even though the diagram of a plane has edges, you must remember that the plane has no boundaries. A plane is named by a single letter (plane m) or by three non-collinear points (plane ABC).

Our world has three dimensional, can you imagine, you live in a two dimensional world ?

There are a few basic concepts in geometry that need to be understood, but are seldom used as reasons in a formal proof.

Collinear Points: points that lie on the same line. Coplanar points: points that lie in the same plane.erOpposite rays : 2 rays that lie on the same line, with a common

endpoint and no other points in common. Opposite rays form a straight line and/or a straight angle (180º).

Parallel lines : two coplanar lines that do not intersect

Skew lines : two non-coplanar lines that do not intersect.

7.3 THE POSITION OF POINT

A. THE POSITON OF POINT TO LINEThe position of point to line are :1. The point is lie on the line


6

.P q

. A

. B

l

. A

. B

l

A point P is called lie on the line q, if point P can be though by line q. The point is lie on the line if there is no distance at all.

Line q

2. The point is lie out of line

Point P is lie out of line q, if point P cannot be trough by line q. The p

AXIOM

Passing through 2 arbitrary points and do not dense each other, can be made only 1 line.

If a line has two line which are located on a plane, that line is full on that plane.

Passing through 2 point in a line, can be made a lot of planes


.P

7

.AA

Q. Q.

.P.

. p

Passing through 3 arbitrary point that is not located on a line, can be made only 1 plane.

B. THE POSITION OF POINT TO PLANEThere are two possibilities of the position point to plane:

(1). The Point is On Plane

The point A is on plane α

(2). The Point is Out of Plane

The point A is out of plane α

THE POSITION OF LINE

A. THE POSITION OF LINE TO LINEThe position of two line has four possibilities. They are:

1). Two lines are tight each other.Two lines are called tight each other if that lines have two common points.

2).Intersection Of Two LinesTwo lines are called intersect each other, if both of them have one common point. We call it intersection point

3). Parallel Lines


α

.A

8

l

.

If two of them have no intersection point, we call parallel.

4). Skew Lines

Lines that are not coplanar and do not intersect.

B. THE POSITION OF LINE TO PLANEThere are three possibilities of the position of line to plane. They are :section

The three possible plane-line intersections.: No intersection

Point intersectionLine intersection

1). The line is located on plane If there are two point that are located on line and in plane, we call that line is located on plane.

2).The line parallel to planeIf line and plane do not have intersection point even trough the line and plane are extended

3).The line intersect planeIf and only if there is one common point.


9

),(

,

A B

CD

E F

GH

D

7.4 THE POSITION OF PLANE TO THE OTHER PLANES

v

There are three possibility of the position a plane to the other planes.(!). Two Planes are dense each other(2). Two planes are parallel(3). Two planes intersect each other

(1) (2) (3)

Take a look the examples

Consider the following cube ABCD.EFGH.

1). The points that are located on plane ABCD are A, B, C, D

2). All of points which are on line (a). AG are A and G

(b). DH are D and H

3). All of the points which are on plane (a). ADHE are A, D, H, E(b). BDHF are B,D, H, F.

(4). The position of line AF to plane: (a). ABFE

AF is on plane ABFE. It is because there are two common points between line AF and plane ABFE

(b). CDHG AF is parallel with plane CDHG because if we extend the line and plane there will no common line. It means AF is parallel CDHG

(c). BCHE AF intersects BCHE at the point P.

5. In cube ABCD. EFGH, find:(a). Plane that is parallel to ADHE

ans: BCGF(b). Planes that are perpendicular to ADHE

ans: ABCD and EFGH(c). Diagonal planes that intersect ADHE

Ans: BCHE and ADGF


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EXERCISE 1“Your mind is the generator of failure, and also the generator of success”.

1. In cube EFGH.IJKL find:a. Sides that are located on EFGHb. Sides that are parallel EFGHc. Sides that intersect plane EFGH

2. In cube ABCD.EFGH find the common line betweena. ADHE and CDHGb. Diagonal ABGH and EFGHc. Diagonal BCGH and ABGHd. EFGH and BCGF

3. Given ABCD.EFGH, point Q is in the middle of diagonal AF. Find the intersectional point line QH toa. plane ACGEb. plane ACFc. plane DEG

7.5 THE DISTANCE IN SOLID FIGURE

Introduction: The Line Perpendicular Plane

Proposition 1:If a line is perpendicular to plane, this line will be perpendicular to each line which is on that plane.

Proposition 2:Line g is perpendicular to plane α , at least line g is perpendicular to two lines which is intersect in plane α .

Proposition 3:


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. B

g

A BCD

E F

GH

SP

If one of from two parallel lines is perpendicular in a plane, the other line will perpendicular to that plane.

1. The Distance from Point to Point

The distance between two points is the length of line that connects both of them.

The distance between 2 points is the length of line that connect both of them.The distance between point A to point B is AB = √AC 2+CB2=√x2+ y2

2. The Distance from Point to Line

The distance from point to line is taken from the length of line that intersect perpendicularly. It is taken because that distance is the nearest distance between point and line.

The distance from point A to line g is AB because line AB

start from point A and perpendicular to line g.

Take a look at this example:DISTANCEIn Cube ABCD.EFGH of edge 6 cm, find:(a). The distance from point A to point H(b). The distance from point A and Point P(c). Point A to line CE(d). Point A to line CG(e). Point A to line BC

Answer:Given the edge 6 cm(a). The distance from point A to pint H is AH


.A

12

.A

.T

ADH is a right angled triangle , D = 900 SO we use Pythagoras’ formula

AH = √ (AD )2 + (DH )2 = √62 + 62 = 6√2 cm(b). The distance from point A to point P is AP

AP = 12 AG

AG = √ (AC )2 + (CG )2 = √(6√2 )2 + (6 )2 = 6√3 cm(c). The distance from point A to line CE is AS, since AS is

perpendicular to line CE.Notice triangle CAE, ∠ A = 900

The area of Δ CAE= AC x AE

2= CE x AS

2

AS =

AC x AECE

= 6√2 x 66√3

= 6√2√3

. √3√3

= 2√6cm

(d) The distance from point A to line CG is AC = 6√2

(e) The distance from point A to line BCFrom point A, pull the line that is perpendicular to line BCIt is AB = 6 cm

3. The Distance from Point to Plane

The distance from point to plane that is pulled from a point until intersect perpendicularly a plane.Step:

From point A, pull line that intersects

perpendicularly. Remember line g if line g intersect perpendicularly to two lines that intersectional in plane α .Find the intersectional point of line g to plane α . Suppose it is T so AT is the distance from point A to plane α .

Take a look at this example:DISTANCE1. In block ABCD.EFGH of edge AB = 8 cm, AD = 6 cm and AE = 4

cm. Point O is as the intersection point of the edge diagonal AC and BD. Find the distance from point:(a). A to plane BCGF d) O to plane EFGH


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A B

CD

E F

GHT

O

GH

A BCD

E . FS P

(b) A to plane CDHG e) O to plane BCGF(c). A to plane EFGH f) O to plane EFGH

2. In Cube ABCD.EFGH of edge 6 cm. Find the distance from point C to plane AFH.

Answer:

1.

(a). The distance from A to plane BCGH is AB = 8 cm (b). The distance from point A to plane CDHG is AD = 6 cm (c). The distance from point A to plane EFGH is AE = 4 cm (d). The distance from point O to plane EFGH is TO. TO = AE = 4

cm

(e). The distance from point O to plane BCGF is 12AE= 1

2(4 ) = 2

2.

The distance from point C to plane AFH is the line that is pulled perpendicularly to plane AFH.Point C is located in plane ACGE. The intersection point between plane ACGE and plane AFH is line AP. We know that plane AFH is formed by edge diagonal AF and AH. Line CE is space diagonal which is crossed to edge diagonal AF and AH. We know that edge diagonal and space diagonal are crossed perpendicularly. ThenCE ⊥ AFCE ⊥ AH } CE ⊥ plane AFH

S is the alliance point between CE and plane AFH. Then , the distance between point C to plane AFH is CS. P is in the middle of HF so

HP = 12HF

= 12

6√2= 3√2

ΔAPH is right angled triangle in P then


P

14

GH

A BCD

E . FQ

P

S

AP2 = AH 2 − HP2= (6√2 )2 − (3√2 )2 = 72 − 18 = 54

AP = 3√6

Then, Using the formula of the area of triangle AEP

L Δ AEP =

AP x EQ2

⇔ EQ= AE x AP2

= 6 x 3√23√6

= 2√3

So The length of CS isCS = CE – EQ = 6√3 − 2√3 = 4√3

4. The Distance between Two Parallel PlaneIf a plane is parallel to the other plane, then there is the distance between two of them. The distance between two parallel plane can be found by these steps:1). Find an arbitrary point in plane v .2). Make line g through point Q and

perpendicular plane w .Line g intersects plane w in point P.

3). QP is the distance between two parallel plane α and β .

Take a look at this exampleDISTANCEIn cube ABCD. EFGH of edge a = 9 cm. Find the distance between plane AFH to plane DBG.

Answer:


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A

GPE

C

Q

S

CE ⊥ AFCE ⊥ AH} CE ⊥ plane AFH

CE ⊥ BGCE ⊥ DG} CE ⊥ plane BDG

HP = 12HF

= 12

9√2 = 3√2

ΔAPH , right angle in P then AP2 = AH2 – HP2

= (a√2 )2 − ( a

2 √2)2 = 6a2

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AP = a2 √6

We calculate EQ by using the area of Δ AEP = AP x EQ

2= AE x EP

AP= 1

3a√3

With the same ways, we get SC = 12a √3

.

QS = CE – EQ – SC = a√3 − 1

3a√3 − 1

3a√3 = 1

3a√3 = 1

3(9)√3 = 3√3

cm

EXERCISE 2“I make use of what I am given”.

1. In the cube ABCD.EFGH of edge 7 cm, find:a. Point E to point Bb. distance from point C to line HBc. distance from point F line ADd. distance from point E to K, where BK : KH = 2 : 3e. distance from midpoint of GF to line DF.

2. In cube ABCD.EFGH of edge 8 cm, find: a. distance from point F to plane BGE.b. distance from point F to plane ACH

3. Given cube ABCD.EFGH. The length of each edge is 6 cm. Find the distance point E to :a. Plane AFH


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M

A BCD

E . F

GHN

l

g

GH

A BCD

E . F

ABED,G

b. Plane BDG

4. The distance between points M and N as shown in the figure beside, where MH

= 34a

, KN = 13a

. Diagonals HF and EG meet at the point K, the point A, K, N are collinear.

5. Given pyramid T.ABCD. AB = TA = 6 cm. Find:a. The length of Ab. The volume of T. ABCDc. The distance from point T to plane ABCD.

6. Given that pyramid T. ABC. ABC is right-angled triangle and the base of the pyramid. TA¿ ABC , ∠A = 900. TA = AB = AC = 5 cm, TE is the altitude on TBC.a. Draw the projection TE on ABCb. Find the length of projectionc. Project point A to plane TBC and calculate the length of it.

7.6 ANGLE IN SOLID FIGURE1. The Angle between Two Lines

Angle between two lines is the smallest angle between the two lines (line l and line g ). Since α < β , then angle between the line l and line g is α

Take a look at this exampleANGLE1. Notice to line AB and ED are crossed each other in cube ABCD.EFGH

To find the angle between AB and ED, we have to move one of that line parallel so it intersects the other point. Suppose we move AB to the line that is parallel to itself and intersect


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A BCD

E . F

EBAC,

GH

g

.K

1g

g

. P

Q

Q’

ED. We get DC. Then the ang;e between line AB and ED is ∠ ( AB , ED ) =∠ ( AB, DC ) = 900

2. Find the angle between line AC and line EB. AC and EB are two crossed lines in cube ABCD. EFGH.We have to move one of that line parallel so it intersects the other line. Suppose we move line AC to line EG (parallel each other and intersect EB). So, the angle between line AC and EB is ∠ ( AC , EB ) =∠ (EG , EB ) = 600

since ΔABG is equilateral triangle.

2. The Angle between Line and PlaneIn the previous lesson, we have learned about the angle beween two lines. Now, we are going to learn how to find the angle between line and plane. Are you ready?Lets start!It will better for you to know about projection.

THE PROJECTION OF LINE TO PLANEThere are three possibilities of the projection of line to plane.

1. If line g perpendicular to plane α , the projection line g to planeα is a point K.

2. If line g parallel plane α then the projection line g to planeα is line g1 .

3. If line g trough planeα , the steps to draw the its projection :


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A BCD

E . F

GH

A BCD

E . F

GH

a. Find point P which is located on line g and trough the planeα .

b. Make the projection of point Q which is located on line g and out of plane α .

c. ∠ QPQ'is the angle between line g and plane α .

Take a look at this exampleANGLEGiven that cube ABCD.EFGH of edge 6 cm.(a) Find ∠ (BG, plane ABCD ) .

(b) If ∠ (BH , plane ABCD ) is α , find

(i) sin α (ii) cosα(iii) tan α

ANSWER

(a) ∠ (BG, plane ABCD ) .= ∠ CBG , is the angle that formed of line BG and line BC. It is because line BC is the projection of line BG to plane ABCD. Since Δ BCG is isosceles right angled triangle, then ∠ CBG = 450.

(b) ∠ (BH , plane ABCD ) =∠ DBH = α , is the angle that formed of line BH and line BD, because BD is the projection of BH to plane ABCD. ΔBDH is right angled triangle on D, BD = 6√2 , BH = 6√3cm and DH= 6 cm.We get :

(i) sin α =

DHBH

= 66√3

= 13 √3

(ii)osα =

BDBH

= 6√26√3

= 13 √6

(iii) tan α =

DHBD

= 66√2

= 12 √2

3. The Angle between Two Intersectional Plane.


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P

Q

R

S

A

UB

V

A BCD

E . F

GH

T

A B

CD

13

6

8

If line AB is the common line of plane U and V, the point P is on AB, the point Q is on plane U, the point S is on plane V, QP ¿ AB, SP ¿ AB then ∠ QPS is angle between plane U and plane V. Angle QPS is also called dihedral angle.

Take a look at this exampleANGLE1. In the cube ABCD. EFGH, find

angle between the planes ABCD and BCHE.

ANSWER

{CB ⊥ BECB ⊥ AB

The line BC is common lie of planes ABCD and BCHE Angle between ABCD and BCHE = ∠ EBA = 450

Since tan ∠ EBA = EA

BA= 1

2. Given a pyramid T.ABCD, the base ABCD is a rectangle. The length of AB = 8 cm, BC = 6 cm and TA = TB = TC = 13.(a). If the angle between plane TBC and ABCD is α , fine tanα(b). If the angle between plane TAB and plane TCD is β , find cos β .

ANSWER (a). The angle between plane TBC and

ABCD is ∠ TPO= α , P is in the middle of BC, and O is the intersection of diagonal AC and BD

Tan ∠ TPO = tan α = TO

POWe can find TO = 12 cm and PO = 4 cm.

Tan α = 12

4= 3


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T

A B

CD

133

6

(b). The angle between plane TAB and plane TCD is ∠ QTR= β , Q is the midpoint of and R is the midpoint of CD.We can find that QR = BC = 6 cm, TQ = TR = 3√17 cmQR2 = TQ2 + TR2 – 2. TQ. TR cos β

Cos β = TQ2 + TR2 −QR 2

2 . TQ . TR

Cos β =

(3√17 )2 + ( 3√17 )2 − (6 )2

2 . (3√17 ) . (3√17 )

Cos β =

153 + 153 − 362 (153)

= 270306

Cos β = 1517

So, Cos β = 1517

3. If plane p ¿ plane q , the line l 1 is common line of plane p and plane q, the line l 2 forms an angle θ to l 1 , the l 2 forms an angle α to plane p and l2 forms an angle β to plane q, then :sin2 θ= sin 2α + sin 2 β

ANSWER:PROVE

In the cuboid ABCD. EFGH beside:(a). p ¿ BCGF, q ¿ ABCD , p¿ q(b). The line BC is common line of p and q, BC = l 1 .(c). l2 ¿ the line BH(d). angle between l 1 and l2 = ∠ HBC = θ(e). angle between l2 and p(f). angle between l2 and q

If length of BH = r, then:(i). GC = DH = r sin β(ii). BG = r cos α(iii).BC2 = BG2 – BC2 = r2(cos2 α - sin2 β

(iv). Sin θ = HCBH

There fore

Sin2 θ = HC2

BH 2 = r2 ( sin2 α + sin2 β )

r2 or sin2 θ= sin 2α + sin 2 β

Since sin2x = 1 – cos2x, then this formula also means


8

21

D

B

C

A

cos2 θ = cos2 α + cos2 β −1

EXERCISE 3“Having a true friend is more valuable than thousand friends that are egoist”.

1. Given that balk ABCD.EFGH. The length of AB = 10 cm , BC = 8 cm and BF = 4 cm. Finda. sin ∠ (FG, BG ) b. cos ∠ (CF , DE )

2. Given that cube ABCD.EFGH of edge 6 cm. Point S is the intersectional point of diagonal EG and FH. Show and calculate ∠( SA , ABCD)

3. The figure beside shows the tetrahedron A.ABC, where DA ¿ ABC, ΔABC is equilateral. The length of DA = 12 and AB = 5 cm. Find tan of angle between plane DBC and plane ABC.

4. Given a pyramid T.ABCD. ABCD is a square. The length of AB = 8 cm . The altitude of pyramid is 6 cm. Show and calculate the angle between ∠ (TBC , ABCD )

5. In cube ABCD. EFGH of edge a cm, show and calculate angle between plane AHF and plane CHF.

EXERCISE 4“Welcome for new day where we find new hope and new optimism”.

7.6 DRAWING INTERSECTION OF PLANES


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As usual drawing planes , drawing the intersection of planes in solid figure can be done if :(1). Given three points which are through that plane(2). Given one point and one line which are through that plane(3). The afinity axis has the important role to draw intersection of

planes.

THE AFINITY AXES(1). The Afinity axis is helping line in finding the point of intersection

plane and solid figure. The afinity line is common line between intersection plane that intersect solid figure and its base.

(2). Even though afinity axis is on base, in the special condition this axis can be on the other plane. We suppose that plane as base, then the axis is called an usual afinity axis.

(3). The afinity axis is got if it have found 2 common point between intersection and base.

(4). If it have been found , next the afinity can be used to determine the point of intersection plane

Note: Before continuing to draw intersection of planes, we have to understand about :(1) The lines which is coplanar and the common line between two

planes.(2) Two points can be connected to be useful line if and only if two

points is coplanar.

Intersection plane and base.

A solid figure can be cut by a plane so it become 2 part with the intersection plane.

PQRS is the intersection plane that is got since W intersects cube ABCD> EFGH

Take a look at this exampleANGLE1. Given cube ABCD. EFGH. The point P is on BF so that BP = FP,

point Q is on GC , GQ : QC = 1 : 4 and point R is on AE so that ER : RA=5:1. Show the intersection between plane that is passing through points P, Q, R and the cube ABCD.EFGH.

ANSWER:1. Make a line trough P and Q so that intersects the extension

of CB in the point K2. Make the line trough P and R so that intersects the

extensions of AB.3. Connecting K and L, line KL is afinity axis4. Extend DA so that intersects KL in M5. Make a line through M and R so that intersects DH in S.


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A BCD

E . F

GH M

W

N

A BCD

E . F

GH

.X

R

W..Y

6. connecting the points P, Q, R, S so that the plane PQRS is the intersection of plane and cube ABCD.EFGH

2. In cube ABCD.EFGH, the point W is the midpoint of CG. Passing trough the

points E, B, W is drawn a plane that id the plane α . Show the intersection between plane α and the cube.

ANSWER:Since B and W on α , then the line BW is on α .Now extend BW so that intersects the extension of FG at point N.The point N is on BW, BW is onα , then N is on α

Sin ce E is on α , then EN interects HG at M, M is on EN, EN is on α . Thus the points E, B, W, M is the intersection of plane and the cube.

EXERCISE 4“When it is obvious that the goals cannot be reached, don’t adjust the goals, adjust the

action steps”.

1. In cube ABCD.EFGH, the point P is the midpoint of BF, point Q is on line CG so that CQ : QC = 1 : 3 and point R is the midpoint of HG. If plane V passing trough points P, Q, R, draw the intersection of plane V and cube ABCD.EFGH.

2. In the cube beside, the plane passes trough W, X, Y, where: EW : WA = 1 : 3GY : YC = 1 : 1FX : XB = 2 : 3Show the intersection between planeα and the cube.


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3. Given the pyramid T. ABCD. ABCD is square. Point P is on TA so that AP : PT = 1 : 2, Q is on TB, so that BQ : QT = 1 : 3 and R is on TC, so that CR : RT = 1 : 2 Show the intersection between plane that is passing trough P, Q, R and pyramid T. ABCD.

4. Draw the intersection of plane with the following solid figure, if the intaesection of plane trough point P, Q and R.

5. Given that prism ABCDE.FGHIJ. Point P is on FA, point Q is on GB and point R is on HC. Draw ugh intersection of plane trough points P, Q and R and prism ADCDE.FGHIJ.


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THREE DIMENSIONS · Web viewMathematics is the way you think THREE DIMENSIONS STANDARD COMPETENCE:6. Determine position, distance and angles related to point, line and plane in three