Dimensional Geometry notes

INTRODUCTION TO THREE-DIMENSIONAL GEOMETRY
MAIN CONCEPTS AND RESULTS
**Coordinate Axes and Coordinate Planes in Three Dimensional Space : In three dimensions, the
coordinate axes of a rectangular Cartesian coordinate system are three mutually perpendicular
lines. The
axes are called the x, y and z-axes.
The three planes determined by the pair of axes are the coordinate planes, called XY, YZ and
ZX-planes.




The three coordinate planes divide the space into eight parts known as octants.
** Coordinates of a Point in Space : The coordinates of a point P in three
dimensional geometry is always written in the form of triplet like (x, y, z).
Here x, y and z are the distances from the YZ, ZX and XY-planes.
Any point (i) on x-axis is of the form (x, 0, 0)
(ii) on y-axis is of the form (0, y, 0)
(iii) on z-axis is of the form (0, 0, z).
Any point (i) in XY-plane is of the form (x, y, 0)
(ii) in YZ-plane is of the form (0, y, z)
(iii) on ZX-plnane is of the form (x, 0, z).
** The three coordinate planes divide the space into eight parts known as octants.

Octants  I II III IV V VI VII VIII
Coordinates XOYZ X′OYZ X′OY′Z XOY′Z XOYZ′ X′OYZ′ X′OY′Z′ XOY′Z′

x + – – + + – – +
y + + – – + + – –
z + + + + – – – –


** Distance between two points (x1 , y1, z1) and (x2 , y2 , z2) = x 2  x1 2  y 2  y1 2  z 2  z1 2 .
** Section Formula : The coordinates of the point R which divides the line segment joining two
points
P (x1 , y1, z1) and Q (x1 , y1, z1) internally and externally in the ratio m : n are given by
 mx 2  nx1 my2  ny1 mz 2  nz1  ,  mx 2  nx1 my2  ny1 mz 2  nz1  respectively.
 , ,   , , 
 mn mn mn   mn mn mn 
** The coordinates of the mid point of the line joining P (x1 , y1, z1) and Q (x2 , y2 , z2) is
 x1  y1  z1 x 2  y 2  z 2 
 , 
 2 2 
** The coordinates of the centroid of the triangle, whose vertices are (x1 , y1, z1), (x2 , y2, z2) & (x3 , y3,
z3)
 x  y  z x  y 2  z 2 x 3  y3  z 3 
are  1 1 1 , 2 , .
 3 3 3 
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, II .Illustrations/Examples:
Example 1: Find the distance between P(2, -3, 4) and Q( – 1, 2, 1).
Solution: By distance formula PQ = 𝑥2 − 𝑥1 2 + 𝑦2 − 𝑦1 2 + 𝑧2 − 𝑧1 2

We get PQ = (1)  2)2  2  (3)2  1  42
=  32  52   32
= 43 units
Example 2: Determine the point in yz-plane which is equidistant from three points A (2, 0, 3), B (0, 3,
2) and C (0, 0, 1).
Solution: As x-coordinate of every point in yz-plane is zero. So, let P(0, b, c)be any point in yz plane.
The given points are A(2, 0, 3), B (0, 3, 2) and C (0, 0, 1)
From the given condition, we have
PA = PB = PC
Now,
PA2 = PB2
(0 – 2)2 + (b – 0)2 + (c – 3)2 = (0 – 0)2 + (b – 3)2 + (c – 2)2
Therefore,
4 + b2 + c2 – 6c + 9 = b2 – 6b + 9 + c2 – 4c + 4
– 6c + 6b + 4c = 0
3b – c = 0 -----(i)
Again PB = PC
PB2 = PC2
(0 – 0)2 + (b – 3)2 + (c – 2)2 = (0 – 0)2 + (b – 0)2 + (c – 1)2
Therefore, b2 – 6b + 9 + c2 – 4c + 4 = b2 + c2 – 2c + 1
3b + c = 6 ------ (ii)
On solving (i) and (ii) we get,
b = 1 and c =3
Required point is (0, 1, 3).
III:-Practice Questions:
MCQ
Q1 A point lie on the x-axis, then its y coordinate and z –coordinate is
a) b,0 b) 0,c c) b,c d)0, 0
Q2 The point (3, -4, -5) lies in the octant
a) Second Octant b) Fourth Octant c) Sixth Octant d)Eighth Octant
Q3 The locus of a point for which x = 0 is
a) xy-plane b) yz-plane c)zx-plane d)None of these
Q4 y-axis the intersection of two planes
a) xy and yz b)yz and zx plane c) xy and zx d)None of these
Assertion-and-Reason Type
Each question consists of two statements, namely,Assertion (A) and Reason (R).For selecting the correct
answer, use the following code:
(a) Both Assertion (A) and Reason (R) are the true and Reason (R) is a correct explanation of Assertion
(A).
(b) Both Assertion (A) and Reason (R) are the true but Reason (R) is not a correct explanation of
Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.
Q5 Statement I: The point A(–2, 3, 5), B(1, 2, 3) and C(7, 0, – 1) are collinear.
Statement II: Three points A, B, C are collinear when AB – BC = AC


Q6 Statement I: The distance between the points (3, 2, – 4) and (1, 2, 3) is 5 unit.
Statement I: The coordinates of points in XY-plane are of the form (a, b, 0).
Short Answer type Questions:
Q7 Show that the points ( – 2, 3, 5), (1, 2, 3) and (7, 0, – 1) are collinear.
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