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HOME WORK :
IF DOUBTS IN HOMEWORK QUESTIONS IT WILL BE DISCUSSED IN THE CLASS NEXT DAY

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CENTROID OF A TRIANGLE :
WE KNOW THAT THE POINT WHERE THE 3 MEDIANS MEET IS CALLED  CENTROID ,THESE MEDIANS INTERSECT EACH OTHER IN 2:1 RATIO .
USING SECTION FORMULA WE CAN DERIVE THE FORMULA FOR CENTROID LETS UNDERSTAND THAT WHEN VERTICES OF TRIANGLE ABC ARE A(X, ₁ Y 1 Z ₁) ,  B(X ₂ Y 2 Z ₂) ,   AND    C(X, ₃ Y 3 Z ₃₎ ,AND RATIO IN WHICH MEDIANS DIVIDE EACH OTHER IS 2:1 THEN AS WE SUBSTITUTE THESE VALUES IN SECTION FORMULA HENCE WE GET  CENTROID

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1) FIND THE DISTANCE BETWEEN THE POINT (2,3,-5) AND (1,4,7)
2) SHOW THAT THW POINT (1,2,3), (2,3,1) , AND ( 3,1,2) FORM A EQUILATERAL TRIANGLE
3)FIND THE POINT ON XY PLANE WHICH IS EQUIDISTANT FROM THREE POINTS (2,0,3) (0,3,2) AND ( 0,0,1)
4) FIND THE LOCUS OF A POINT P SUCH THAT
PA 2  + PB ²  = 2 K ²
WHERE A( 1,2,3) AND   ( -1,4,-3)

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CONCLUSION :
1)POINT ON X AXIS WILL BE (X,0,0) AND Y AXIS WILL BE (0,Y,0) AND Z AXIS WILL BE (0,0,Z)
2) MIRROR IMAGE OF (X,Y,Z) ON XY PLANE ( X,Y,-Z)AND ON YZ PLANE IT WILL BE (-X,Y,Z) SIMILARLY IMAGE FOR XZ PLANE  WILL BE (X,-Y,Z)
3)FOOT OF PERPENDICULAR FROM (X,Y,Z) ON XY PLANE          ( X,Y,0),SIMILARELY WE CAN FIND IT ON OTHER TWO PLANE
4) OCTANT  AND COORDINATE ARE SUMMARIZE AS BELOW :

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HOW TO DERIVE MID POINT FORMULA :                                    NOW IF LINE AB HAVING COORDINATES  GIVEN BELOW AND LINE IS DIVIDED AT MIDPOINT IN RATIO 1:1 , USING THESE VALUES IN SECTION FORMULA WE GET MID POINT FORMULA

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(I)FIND OUT THE COORDINATE OF A POINT  'C ' WHICH DIVIDES THE LINE AB IN 2: 3 RATIO WHEN A IS (2,3,4)  AND B IS (5,6,7)
2) FIND THE CENTROID OF TRIANGLE WHOSE VERTCES ARE (1,5,-2), (3,7,1) AND (2,-3,-5)
3) FIND THE RATION IN WHICH THE LINE JOINING THE POINTS (2,5,4) AND (3,4,5)IS DIVIDED IN YZ PLANE . ALSO FIND THE COORDINATE S OF THE POINT OF DIVISION .
4)SHOW THAT THE POINTS(1,3,0) , (-5,5,2) , ( -9,-1,2) AND (-3,-3,0) ,TAKEN IN ORDER ARE THE VERTICES OF THE PARALLELOGRAM .
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LETS UNDERSTAND THE SECTION FORMULA IN 3D

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STEP 13:CALCULATE THE DISTANCE BETWEEN 2 POINTS IN 3D 

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STEP 13:CALCULATE THE DISTANCE BETWEEN 2 POINTS IN 3D 

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DISTANCE FORMULA IS GIVEN BY

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1)NAME THE OCTANTS IN WHICH  THE FOLLOWING POINTS LIE
(i)(2,3,4)
(II) (-1,9,4)
(III) (-2,-3,-6)
2) FIND THE COORDINATE OF A POINT LIE ON X AXIS , Y AXIS AND Z AXIS
3) WRITE THE COORDINATES OF THE PERPENDICULAR FROM ( 3,-4,5) ON X AXIS , Y AXIS AND Z AXIS
3)WRITE THE COORDINATES OF THE PERPENDICULAR FROM ( 3,-4,5) ON XY PLANE , YZ PLANE  AND ZY PLANE
4)  WRITE THE COORDINATES OF THE MIRROR  IMAGE OF ( 3,-4,5) IN  XY PLANE , YZ PLANE  AND ZY PLANE

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STEP 6: LETS UNDERSTAND THAT LIKE WISE QUADRANT WE HAVE OCTANT AND DIFFERENT PLANES AND 3 AXISES   IN 3 DIMENSION 

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STEP 6: LETS UNDERSTAND THAT LIKE WISE QUADRANT WE HAVE OCTANT AND DIFFERENT PLANES AND 3 AXISES   IN 3 DIMENSION 

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THERE IS XY PLANE , YZ PLANE AND XZ PLANE ..REPRESENTED BY EQUATION Z=0 , X=0 AND Y=0.
ANY PLANE PARALLEL TO XY PLANE IS GIVEN BY Z=a AND PARALLEL TO YZ PLANE WILL BE X=b AND PARALLEL TO ZX PLANE WILL BE Y=c

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LETS UNDERSTAND MORE ABOUT OCTANT AND  POINTS REPRESENTED ON 3D

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Calculus III - Three Dimensional Coordinate Systems (Level 1 of 10) _ Basics-vZNtoloJbEI_Segment_0

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FORMULA OF DISTANCE BETWEEN 2 POINTS IN 3 DIMENSION

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introduction to three dimensional geometry

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introduction to three dimensional geometry

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Distance formula To find distance of line joining two given points in three dimensional space.

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WE ARE AWARE THAT THERE ARE 4 QUADRANTS IN 2 DIMENSION , AS SHOWN BELOW

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THESE OBJECTS HAVE LENGTH , BREADTH AND HEIGHT ,WHICH CAN NOT BE SHOWN ON 2-DIMENSION OR ON COORDINATE PLANE , HENCE WE NEED 3 AXIS

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STEP 4: WHAT ARE  3 AXIS ,                  ... X AXIS , Y AXIS AND Z AXIS ..FOR EXAMPLE :  CORNER OF THE ROOM .   LET US CONSIDER A POINT (a,b,c) IS IN 3 DIMENSION WHERE X = a UNIT , Y =b UNIT AND Z= c UNIT AS SHOWN IN THE FIGURE 

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STEP 4: WHAT ARE  3 AXIS ,                  ... X AXIS , Y AXIS AND Z AXIS ..FOR EXAMPLE :  CORNER OF THE ROOM .   LET US CONSIDER A POINT (a,b,c) IS IN 3 DIMENSION WHERE X = a UNIT , Y =b UNIT AND Z= c UNIT AS SHOWN IN THE FIGURE 

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STEP 2: AS WE SAW IN THE PREVIOUS   DIAGRAMS ,IT IT IS CLEAR THAT THESE FIGURES HAVE ONLY TWO DIMENSIONS  SO WE CAN PLOT THEM   (FIGURES) ON COORDINATE PLANE LIKE WE CAN PLOT THESE POINTS  (2,3) (3,3),(0,0) AND (2,0)  ON COORDINATE PLANE AND GET A RECTANGLE .THIS PLOTTING OF POINTS WE DID IT IN PREVIOUS CLASSES . WE KNOW TWO AXIS i.e  X AXIS AND Y AXIS 

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STEP 2: AS WE SAW IN THE PREVIOUS   DIAGRAMS ,IT IT IS CLEAR THAT THESE FIGURES HAVE ONLY TWO DIMENSIONS  SO WE CAN PLOT THEM   (FIGURES) ON COORDINATE PLANE LIKE WE CAN PLOT THESE POINTS  (2,3) (3,3),(0,0) AND (2,0)  ON COORDINATE PLANE AND GET A RECTANGLE .THIS PLOTTING OF POINTS WE DID IT IN PREVIOUS CLASSES . WE KNOW TWO AXIS i.e  X AXIS AND Y AXIS 

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STEP 1 : WE ARE FAMILIAR WITH 2 DIMENSIONS FIGURES WHICH CAN BE REPRESENTED  ON WHITE BOARD                                     FOR EXAMPLE :THIS WHITE  BOARD IN CLASS ROOM  IS ALSO TWO DIMENSIONAL FIGURE WHICH HAS LENGTH AND BREADTH.

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STEP 1 : WE ARE FAMILIAR WITH 2 DIMENSIONS FIGURES WHICH CAN BE REPRESENTED  ON WHITE BOARD                                     FOR EXAMPLE :THIS WHITE  BOARD IN CLASS ROOM  IS ALSO TWO DIMENSIONAL FIGURE WHICH HAS LENGTH AND BREADTH.

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AND SOME FIGURES GIVEN HERE EXPLAIN 2 D