Three Dimentional Geometry

8/8/2019 Three Dimentional Geometry

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You may recall that to locate the position of apoint in a plane, we need two intersectingmutually perpendicular lines in the plane.These lines are called the coordinates of thepoint with respect to the axes.

Similarly , if we were to locate the position

of lower tip of an electric bulb hanging fromthe ceiling of the room or the position of thecentral tip of the ceiling fan in a room, we willnot only require the perpendicular distance of



the point to be located from two perpendicular

wall of room but also the height of the pointfrom the floor of the room. Therefore weneed not only two but three numbers

representing the perpendicular distance of

the point from three mutually perpendicularplanes, namely the floor of the room and twoadjacent walls of the room. The three

numbers representing the three distances are

called the coordinates of the point with

reference to the three coordinate planes. So,a point in space has three coordinates.



Consider three planesintersecting at a point O suchthat these three planes are

mutually perpendicular toeach other (12.1). These threeplanes intersect along thelines XOX, YOY and ZOZ,

called the x, y and z axis,respectively. We may notethat these lines are mutuallyperpendicular to each other.



These lines constitute the rectangular coordinate system. The planes XOY,YOZ &ZOX called respectively the XY plane ,YZplane & the ZX plane are known as the threecoordinate planes. We take the XOY plane asthe plane of the paper & the line ZOZ asperpendicular to the plane XOY. If the planeof the paper is considered as horizontal, thenthe line ZOZ will be vertical. The distancemeasured from XY-plane upward in the

direction of OZ are taken as positive & thosemeasured downward in the direction of OZare taken as negative . Similarly , the distance



measured to the right of ZX-plane along OY aretaken as positive to the left of ZX- plane &

along OY as negative in front of the YZ- plane

along OX as positive & to the back of it along

OX as negative. The point O is called the

origin of the coordinate system. The threecoordinate planes divide the space into eight

parts known as octants. These octants could

be named as XOYZ, XOYZ, XOYZ, XOYZ,

XOYZ, XOYZ, XOYZ & XOYZ & denotedby I, II, III, .,VIII, respectively.



Having chosen a fixedcoordinate system in thespace, consisting of coordinate

axes, coordinate planes andthe origin, we now explain, asto how, given a point in thespace, we associate with itthree coordinates (x, y, z) &conversely given a triplet of three numbers (x, y, z) how, welocate a point in the space.



Given a point P in space, we drop a

perpendicular PM on the XY-plane with M asthe foot of this perpendicular (Fig:12.2). Then

from the point M, we draw a perpendicular

ML to the x-axis, meeting it at L. let OL be x,

LM be y & MP be z. Then x, y & z are calledthe x, y & z coordinates, respectively, of the

point P in the space. In (Fig:12.2) we may

note that the point P (x, y, z) lies in the octant

XOYZ & so all x, y, z are positive. If P was inany other octant, the sign of x, y & z would

change accordingly.



Conversely, given any triplet (x, y, z) we would

first fix the point L on the x-axiscorresponding to x, then locate the point M inthe XY-plane such that (x, y) are the

coordinate of the point M in the XY-plane.

Note that LM is perpendicular to the x-axis orparallel to y-axis. Having reached the point Mwe draw a perpendicular MP to the XY- plane

& locate on it the point P corresponding to z.

The point P so obtain has then the coordinate

(x, y, z) . Thus there is a one to onecorrespondence between the points in space



And ordered triplet (x, y, z)

of real numbers.Alternatively through

the point P in the space wedraw the three plane parallel

to the coordinate planes,meeting the x-axis, y-axis &z-axis in the point A,B & Crespectively (Fig: 12.3) . LetOA =x, OB= y, OC= z. thenthe point P will have thecoordinate axes. Throughthe point A, B, C we drawplanes parallel to the



YZ- plane, ZX- plane & XY- plane respectively.

The point of intersection of these threeplanes, namely, ADPF, BDPE, CEPF is

obviously the point P corresponding to the

ordered triplet (x, y, z). We observe that if P(x,

y, z) is any point in the space, then x, y, z areperpendicular distance from YZ, ZX & XY

planes, respectively.





with one diagonal PQ (Fig 12.4)

Now, since <PAQ is a right angle,

it follow that in triangle PAQ,

PQ2 = PA2 + AQ2

(1)

Also, triangle ANQ is right angle

triangle with <ANQ a right

angle.Therefore, AQ2 = AN2 + NQ2

.(2)



From (1) & (2) we have

PQ2

= PA

2

+ AN2

+ NQ2

Now, PA = y2 y1 , AN = x2-x1 & NQ =z2 z1

Hence, PQ2 = (x2 x1)2 + (y2 y1)2 + (z2 z1)2

:. PQ= [

(x2 x1)

2

+ (y2 y1)

2

+ (z2 z1)

2

]

1/2

This gives us the distance between two points

(x1 , y1 , z1) & ( x2 , y2 , z2).

In particular, if x1 = y1 = z1 = o, i.e, point P is

origin O, then OQ =( x22 + y2

2 + z22)1/2

This gives us the distance between two points

(x1 , y1 , z1) & ( x2 , y2 , z2).



In particular, if x1 = y1 = z1 = o

i.e, point P is origin O,

Then, OQ =( x22 + y2

2 + z22)1/2

Which give the distance between the origin O &any point Q(x2, y2, z2).



In two dimensional geometry, we have learnthow to find the coordinate of a point dividing

a line segment in a given ratio internally .

Now , we extend this to three dimensionalgeometry as follow:

Let the two given point be P(x1, y1, z1) &Q(x

2

, y2

, z2

). Let the point R (x, y, z) divided

PQ in the given ratio m:n internally . Draw PL,

QM & RN perpendicular to the XY- plane.

Obviously PL ll RN ll QM & feet of these



perpendiculars lie in a XY- plane.

The points L, M &N will lie on aline which is the intersection of

the plane containing PL, RN &

QM with the XY- plane. Through

the point R draw a line STparallel to the line LM. Line ST

will intersect the line LP

externally at the point S& the line MQ at T, as shown in

fig 12.5



Also note that quadrilateral LNRS & NMTR are

parallelogram.The triangles PSR & QTR are similar.

Therefore ,

This implies

Similarly, by drawing the perpendiculars to theXZ &YZ-plane, we get



Hence the coordinate of the point R whichdivides the line segment joining two points

P(x1, y1,z1) & Q(x2, y2, z2) internally in the ratio

m:n are

If the point R divide PQ externally in the ratio

m:n then its coordinates are obtain by

replacing n by n so that coordinate of pointR will be



Case1- Coordinate of the midpoint. In case R isthe midpoint of PQ, then m:n =1 :1 so that

These are the coordinate of the midpoint of thesegment joining P (x

1

, y1

, z1

) & Q (x2

, y2

, z2

)

Case 2- The coordinate of the point R whichdivides PQ in the ratio k:1 are obtained bytaking k= m/n which are given below:

Generally, this result is used in solving problem involving ageneral point on the line passing through two given

point.



In three dimension the coordinate axes of arectangular Cartesian coordinate system are

three mutually perpendicular line. The axes

are called the x, y & z axes. The three plane determined by the pair of

axes are the coordinate planes, called XY, YZ,

& ZX planes. The three coordinate plane divide the space

into eight parts known are octants.



The coordinate of a point P in three

dimensional geometry is always written inthe front of triplet like (x, y, z). Here x, y &zare the distance from theYZ, ZX & XY planes.

Any point on x- axis is of the form (x,o,o)

Any point on y- axis is of the form (o,y,o)

Any point on z-axis is of the form of (o,o,z)

Distance between two point P(x1,y1,z1) &

Q(x2,y2,z2) is given by



The coordinate of the point R which dividesthe line segments joining two point P(x

1

,y1

,z1

)

& Q(x2,y2,z2) internally & externally in the

ratio m:n are given by :

The coordinates of the midpoint of the linesegment joining two points P(X1, y1, z1) and

Q(x2, y2, z2) are



Three Dimentional Geometry

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