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2-D & 3-D GEOMETRY

2-D & 3-D GEOMETRY

We all have some amount of geometry. We know that any line can be represented on the Cartesian plane. Any figure can be drawn on it. But can we represent a 3-d object on it. Yes we can. A Cartesian plane has 2 axis. While representing in 3-D we need to add a third axis. This axis does not come in between the axis or in the same plane. It appears to be coming out of the paper as we cannot represent a 3-d object on a 2-d surface.


This new z-axis represents a line coming out of the screen. Before understanding 3-d geometry you need to imagine this axis coming out of the screen. 

REMEMBER: all the three axis are perpendicular .i.e there an angle 0f 90 between them and they meet at the origin
If you are unable to imagine you can take a thick book as an example. Any corner becomes it origin and the three edges as the three axis

REPRESENTING 3-D GEOMETRY

Like in 2-d geometry we represent the value of the different axis as (x,y) we use the same method to write in 3-d geometry. (x,y,z). You can imagine that more the value of z the more it will come out of the paper.

SIGN CONVENTIONS IN GEOMETRY

In 2-d geometry we have 4 quadrants or the x-axis and the y-axis divide the plane into 4 parts. In each part different quadrants have different conventions. They are in the following way



In 3-d geometry we have octants that is we have 8 parts. Dividing a cube from the  center to form 8 parts.


We take xy-plane as the surface and z-axis as the height. Remember you need to imagine the above diagram as 3-d
Sign conventions are as following 
1 octant – (X,Y,Z) 
2 octant – (-X,Y,Z)
3 octant – (-X,-Y,Z)
4 octant – (X,-Y,Z)
5 octant – (X,Y,-Z)
6 octant – (-X,Y,-Z)
7 octant – (-X,-Y,-Z)
8 octant – (X,-Y,-Z)

LOCATING A POINT

While locating a point in xy-plane we measure the distances from the axis. The distance away from the y-axis is our x-coordinate and distance from x-axis becomes our y-coordinate. 
   


In 3-d geometry the same method follows but with height. In the below diagram it seems that the distance from x axis is not perpendicular but it is. As I told earlier we cannot exactly show 3-d we can only give 3-d effects 


DISTANCE BETWEEN TWO POINTS

  Let there be two points A and B. they are plotted in a xy-plane then how do we calculate their distance. We use Pythagoras theorem to find the formula. 

Now ,
OL= x1, OM = x2, AL = y1, BM = y2
Therefore 
AC=LM=OM-OL= x2 - x1
BC=BM-BC=BM-AL=y2-y1
Since ABC is a right-triangle then 
AB2 = AC2 + BC2
AB = equation.pdf
Similarly the formula of distance in 3-d is
 equation_1.pdf

SECTION FORMULA 

Let there be line AB. A point P divides the line in the ratio m: n. If coordinates of the two points (A and B) then what will be the coordinates of the P. To find this we use the below section formula . 


equation_2.pdf=equation_3.pdf=equation_4.pdf 
AK=LM=OM – OL= x – x1
PT=MN=ON – OM= x2 – x 
PK=MP – AL= y – y1
BT= BN – PM= y2 – y
equation_5.pdf=equation_6.pdf=equation_7.pdf=equation_8.pdf
Therefore
equation_9.pdf=equation_10.pdf
mx2 – mx = nx – nx1
mx + nx = mx2 + nx1
x(m + n) = mx2 + nx1
x = equation_11.pdf
Similarly
y = equation_12.pdf 
in a 3-d graph the formula remains the same but the ‘x’ becomes z ..that is
z = equation_13.pdf 

CENTROID OF A TRIANGLE

Let there be a triangle with vertices A, B, and C. let AD be the median of the triangle and let the centroid be G. We know that centroid divides the median in the ratio 2: 1. Also D divides the line BC in the ratio 1: 1.

D=equation_14.pdf,equation_15.pdf
D=equation_16.pdf,equation_17.pdf
G =equation_18.pdf,equation_19.pdf
G =equation_20.pdf,equation_21.pdf
Similarly, in 3-d the third axis z comes in the formula
G =equation_22.pdf,equation_23.pdf,equation_24.pdf

AREA OF TRIANGLE

Let there be a triangle ABC. Coordinates of all the vertices are given. So how we can find its area. We can do so by using the formula of trapezium .i.e
Area of trapezium=1/2(sum of parallel sides) X (distance between them)



Now area of ABC will be:-
Area (ABC) = Area (ABLM) + Area (ACNM) – Area (BCNL)
Area (ABC) = ½ [(BL + AM) (LM) + (AM + CN)(MN) – (BL + CN)(LN)
Area (ABC) = ½ [(y2 + y1)(x1 - x2) + (y1 + y3)(x3 – x1) - (y2 + y3)(x3 - x2) ]
Area (ABC) =½ [y2 x1 - y2 x2 + y1 x1 – y1 x2 + y1 x3 y1 x1 + y3 x3 – y3 x1 - y2 x3 + y2 x2 - y3 x3 + y3 x2]

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