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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

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2 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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3 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 Jarak dua titik

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4 e. g. : Given that the edge length of a cube ABCD.EFGH is a cm. Determine the distance of : a) Point A to point C b) Point A to point G c) The distance of point A to the middle of plane EFGH A B C D H E F G a cm P

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5 Solution: Consider Δ ABC which has right angle at B AC = = = = Thus, the diagonalof AC = cm A B C DH E F G a cm

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6 Distance of AG Consider Δ ACG which has right angle at C AG = = = == Thus, the diagonal of AG = cm A B C D H E F G a cm

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7 A B C D H E F G P Distance of AP Consider Δ AEP which has right angle at E AP = = = = = Thus distance of A to P = cm

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8 Distance Point to Line A g distance point 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.

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9 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 B C D H E F G 5 cm

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10 Solutio The distance from point A to the edge of HG is length of the line segment AH, ( AH HG ) A B C D H E F G 5 cm AH = (AH is a side diagonal) AH = Thus, the distance from point A to the edge of HG= 5√2 cm

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11 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 B C D H E F G 6 cm

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12 Solution The distance from point B to AG = the distance from point B to P ( BP AG ) The side diagonal of BG = 6√2 cm The space diagonal of AG = 6√3 cm Consider a triangle ABG ! A B C D H E F G 6√2 cm 6 cm P 6√3 cm A B G P 6√3 6 6√2 ?

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13 Consider a triangle ABG Sin A = = = BP = BP = 2√6 A B G P 6√3 6 6√2 ? Thus, the distance from point B to AG= 2√6 cm 2

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14 e.g. 3 Given that T.ABCD is 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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15 Solution The distance from A to TC= AP AC is a cube’s diagonal AC = 12√2 AP = = = = 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

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16 e.g. 4 : Given that the edge length of a cube ABCD.EFGH is 6 cm and A B C D H E F G 6 cm Point P is in the middle of FG. The distance from point A to line DP is… P

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17 A B C D H E F G 6 cm P Solution Q 6√2 cm R P AD GF 6 cm 3 cm DP = = =

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18 Solution Q 6√2 cm R P AD GF 6 cm 3 cm DP = 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 4

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19 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 b g a, g b, Thus g V

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20 The Distance of a Point to a Plane This 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

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21 e.g. 1 : Given that the edge length of a cube ABCD.EFGH is 10 cm. Thus the distance from point A to plane is…. A B C D H E F G 10 cm P

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22 Solution The distance from point A to plane BDHF is representated by the length of AP (AP BD) AP = ½ AC (AC BD) = ½.10√2 = 5√2 A B C D H E F G 10 cm P Thus the distance from A to plane BDHF = 5√2 cm

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23 e.g. 2 : Given that T.ABCD is a pyramid. The length of AB = 8 cm and TA = 12 cm. The distance from point T to plane ABCD is…. 8 cm T C A B D 12 cm

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24 Solution The distance from T to ABCD = The distance from T to the intersection of AC and BD= TP AC is a cube’ss diagonal AC = 8√2 AP = ½ AC = 4√2 8 cm T C A B D 12 cm P

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25 AP = ½ AC = 4√2 TP = = = = = 4√7 8 cm T C A B D 12 cm P Thus the distance from T to ABCD = 4√7 cm

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26 e.g. 3 : Given that the edge length of a cube ABCD.EFGH is 9 cm. The distance from point C to plane BDG is…. A B C D H E F G 9 cm

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27 Solution The distance from point C to plane BDG = CP That is the line segment which is drawn through point C and perpendicular to GT A B C D H E F G 9 cm P T CP = ⅓CE = ⅓.9√3 = 3√3 Thus the distance from C to BDG = 3√3 cm

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28 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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29 e.g. Given that the edge length of a cube ABCD.EFGH is 4 cm. Determine the distance of: A B C D H E F G 4 cm a.Line AB to line HG b.Line AD to line HF c.Line BD to line EG

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30 Solution The 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 B C D H E F G 4 cm

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31 Solution The distance of: b.BD to line EG = PQ ( PQ BD, PQ EG = AE = 4 cm A B C D H E F G 4 cm P Q

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32 The Distance of Line to Plane This display shows the distance of line g to plane V is length of line segment which connect that line and plane perpendicularly. V g g

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33 e.g. 1 Given that the edge length os a cobe ABCD.EFGH is 8 cm The distance of line AE to plane BDHF is…. A B C D H E F G 8 cm P

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34 Solution The distance of line AE to plane BDHF Is represented by the length of AP.(AP AE AP BDHF) AP = ½ AC (AC BDHF ) = ½.8√2 = 4√2 A B C D H E F G 8 cm P Thus the distance from A to BDHF = 4√2 cm

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35 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 Bidang

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36 e.g. 1 : Given that the edge length of a cube ABCD.EFGH is 6 cm. The distance of plane AFH to plane BDG is…. A B C D H E F G 6 cm

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37 Solution The distance of plane AFHto plane BDG Is represented by PQ PQ = ⅓ CE (CE is a space diagonal) PQ = ⅓. 6√3 = 2√3 A B C D H E F G 6 cm P Q Thus the distance of AFH to BDG = 2√3 cm

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38 e.g. 2 : Given that the edge length of a cube ABCD.EFGH is 12 cm. A B C D H E F G 12 cm Points K, L and M are the middle point of BC, CD dan CG. The distance of plane AFH and KLM is…. K L M

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39 Solution Diagonal EC = 12√3 The distance from E to AFH = distance from AFH to BDG = distance from BDG to C A B C D H E F G 12 cm Thus the distance from point E to AFH = ⅓EC =⅓.12√3 = 4√3 So that the distance from BDG to C is 4√3 too. L

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40 A B C D H E 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 K L 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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41 Have a nice try !

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Essential Cell Biology

Essential Cell Biology

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