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3 Dimension (Distance)

3 Dimension (Distance). SMA 5 Mtr. BASE COMPETENCE : D etermine distance between two points, a point to line and a point to plane in the 3 dimension. SMA 5 Mtr. THE MATERIAL Distance in 3 Dimension : distance between two points distance point to line distance point to plane.

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3 Dimension (Distance)

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  1. 3 Dimension (Distance) SMA 5 Mtr

  2. BASE COMPETENCE : Determine distance between two points, a point to line and a point to plane in the 3 dimension. SMA 5 Mtr

  3. THE MATERIAL Distance in 3 Dimension : distance between two points distance point to line distance point to plane SMA 5 Mtr

  4. Distance between two points Distance from points A to point B is the shortest way from A to B B Distance between two points A SMA 5 Mtr

  5. H G E F D C A B Example 1 Given : The cube ABCD – EFGH With 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 P a cm a cm a cm SMA 5 Mtr

  6. H G E F a cm D C a cm A B a cm Solving: Look at the right triangle ABC that right at B, so AC = = = = So distance A to C= cm (AC is side diagonal) SMA 5 Mtr

  7. H G E F a cm D C a cm A B a cm Distance A to G = ? Look at the triangle ACG that right at C, so AG = = = = = So distance A to G = cm (AG is space diagonal) SMA 5 Mtr

  8. H G E F D C A B DistanceA to P = ? Look at the triangle AEP that Right at E, so AP = = = = = So distance A to P = cm P a cm SMA 5 Mtr

  9. Distance point to line A Distance point A to line g is length of perpendicular segment from point A to line g Distance from point A to line g g SMA 5 Mtr

  10. H G E F D C A B Example 1 Given cube ABCD-EFGH, with length of edge 5 cm. Distance point A to segment HG is…. 5 cm 5 cm SMA 5 Mtr

  11. H G E F D C A B Solving Distance point A to segment HG is segment AH, (AH  HG) 5 cm 5 cm AH = (AH side diagonal) AH = So distance A to HG = 5√2 cm SMA 5 Mtr

  12. H G E F D C A B Example 2 Given cube ABCD-EFGH With length of edge 6 cm. Distance point B to diagonal AG is…. 6 cm 6 cm SMA 5 Mtr

  13. H G E F G 6√3 6√2 D P C A B B A 6 Solving Distance B to AG = Distance B to P (BPAG) Side diagonal BG = 6√2 cm Space diagonal AG = 6√3 cm Look at triangle ABG 6√3 cm P 6√2 cm 6 cm ? SMA 5 Mtr

  14. G 6√3 6√2 P B A 6 Look at the triangle ABG Sin A = = = BP = BP = 2√6 ? 2 So distance B to AG = 2√6 cm SMA 5 Mtr

  15. T D C A B Example 3 Given T.ABCD Uniform 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√2 cm 12 cm SMA 5 Mtr

  16. T 12√2 cm D C A 12 cm B • Solving Distance A to TC = AP AC = side diagonal = 12√2 AP = = = = So distance A to TC = 6√6 cm 6√2 P 6√2 12√2 SMA 5 Mtr

  17. H G E F D C A B Example 4 Given cube ABCD-EFGH With length of edge 6 cm P 6 cm 6 cm point P at the middle of FG. Distance point A to line DP is…. SMA 5 Mtr

  18. P 3 cm G F H G E F D A 6 cm D C A B Solving  P Q 6√2 cm 6 cm R 6 cm DP = = = SMA 5 Mtr

  19. P 3 cm G F Q 6√2 cm D A R 6 cm Solving DP = Area of triangle ADP ½DP.AQ = ½DA.PR 9.AQ = 6.6√2 AQ = 4√2 So distance A to DP = 4√2 cm 4 SMA 5 Mtr

  20. V Distance point to plane Distance between point A to plane V is length of segment that connect perpendicular point A to plane V A  SMA 5 Mtr

  21. V Line perpendicular Plane Line g said perpendicular plane V if line g perpendicular at two lines that intersection in plane V g  a b g  a, g  b, So g  V SMA 5 Mtr

  22. H G E F D C A B Example 1 Given cube ABCD-EFGH With edge length 10 cm Distance point A to plane BDHF is…. P 10 cm SMA 5 Mtr

  23. H G E F D C A B Solving Distance point A to Plane BDHF Represented by length of line AP. (APBD) AP = ½ AC (ACBD) = ½.10√2 = 5√2 P 10 cm So distance A to BDHF = 5√2 cm SMA 5 Mtr

  24. T D C A B Example 2 Given uniform pyramid T-ABCD. Length of AB = 8 cm and TA = 12 cm. Distance point T to plane ABCD is… 12 cm 8 cm SMA 5 Mtr

  25. T D C A B Solving Distance T to ABCD = Distance T to intersection AC and BD = TP AC is side diagonal AC = 8√2 AP = ½ AC = 4√2 12 cm P 8 cm SMA 5 Mtr

  26. T 12 cm P D C 8 cm A B AP = ½ AC = 4√2 TP = = = = = 4√7 So distance T to ABCD = 4√7 cm SMA 5 Mtr

  27. H G E F D C A B Example 3 Given cube ABCD-EFGH With length of edge 9 cm. Distance point C to plane BDG is… 9 cm SMA 5 Mtr

  28. H G E F D C A B Solving Distance point C to plane BDG = CP CP perpendicular with GT P T 9 cm CP = ⅓CE = ⅓.9√3 = 3√3 So distance C to BDG = 3√3 cm SMA 5 Mtr

  29. GOOD LUCK ! SMA 5 Mtr

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