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Three-Dimensional Space can also be understood as 3-space or tri-dimensional space. It is a geometric medium that contains three values that are needed to decide the position of an element. In Mathematics and Physics, a sequence of n figures can be reasoned as a position in n-dimensional space. When n = 3, it's known as three-dimensional Euclidean space. This acts as a three-parameter model of the physical universe in which all given matter exists. This space is only one case of a large space in three confines known as 3-manifolds. In this case, these three values are taken from the terms width, height, depth, and length.
Key Words: Three Dimensional Coordinate System, Coordinates, Three-Dimensional Geometry, Distance Formula, Plane, x-Coordinate, y-Coordinate, Quadrants, Line Segment
Three Dimensional Coordinate System
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In order to discover the location of a point in space, we need a rectangular coordinate system. After taking a particular coordinate system in 3D, the coordinates of any point P in that system can be given by an arranged 3-tuple (x, y, z). Similarly, if the coordinates (x, y, z) are formally understood then we can effortlessly discover the point P in space.
On a two dimensional plane a point in the XY- plane by an arranged pair that consists of two real figures, an x-coordinate and y- coordinate, which signify signed distances along the X-axis and y-axis, independently, from the origin, which is the point (0, 0). These axes, which are known as the coordinate axes, divided the plane into four quadrants.
Three Dimensional Coordinate System
The properties of three-dimensional space are as follows:
- A point is represented by an ordered triadic (x, y, and z) that consists of three figures, an x-coordinate, a y- coordinate,
- A z- coordinate in the two-dimensional XY- plane, these indicate the signed distance along the coordinate axes,
- The x-axis, y-axis and z-axis, independently, from the origin, expressed by o, which has coordinates (0, 0, and 0).
The video below explains this:
Coordinate Geometry Detailed Video Explanation:
Read More: Coordinate Geometry
Example of Three Dimensional Coordinate System
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There's a one-to-one correspondence between a point in XYZ- space and a triple in R3, which is the set of all arranged triples of real figures. This is understood as a three-dimensional rectangular coordinate system.
The figure displays the point (2, 3, and 1) in XYZ- space, expressed by the letter P, along with its projections onto the coordinate planes. The origin is expressed by the letter O.
The point (2, 3, 1) in XYZ- space, expressed by the letter P. The origin is expressed by the letter O. The projections of P onto the coordinate planes are denoted by the diamonds. The dashed lines are line parts perpendicular to the coordinate planes that join P to its projections. Just as the X-axis and y-axis divide the XY- plane into four quadrants, these three planes divide XYZ- space into eight octants. Within each octant, all x-coordinates hold the identical sign, as do all y- equals, and all z-coordinates.
Read More: Abscissa, Plotting Ordinates, Relation, Solved Examples
Steps to Find Coordinates of a Point in a Three Dimensional Space
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Locating a point in the (x,y,z) space can be tricky because, unlike graphing in the xy plane, depth perception is needed. The projection of a point (x, y, z) onto the x,y plane is attained by joining the point to the x,y plane by a line segment that's perpendicular to the plane and calculating the intersection of the line segment with the plane.
Also, the projection of this point onto the xy plane is the point (0, y, z), and the projection of this point onto the xz- plane is the point (x, 0, z). The figure shows these projections, and how they can be used to put up a point in x,y,z space. One can first put up the point’s projections, which is resembling the task of putting up points in the x,y plane, and additionally use line segments forming from these projections and perpendicular to the coordinate planes to “ detect” the point in the x,y,z space.
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Distance Formula
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The distance between two points P2 = (x2, y2) and P1 = (x1, y1) in the xy-plane is given by the following distance formula,
d (P1, P2) = √(x2 - x1)2 + (y2 - y1)2
In the same way, the distance between two points P2 = (x2, y2, z2) and P1 = (x1, y1, z1) in XYZ-space is given by the following generalisation of the distance formula,
d (P1, P2) = √(x2 - x1)2 + (y2 + y1)2 + (z2 - z1)2
Read More: Distance Between Two Points
Solved Examples
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| Example 1: Find the distance between P2 = (8, −5, 0) and P1 = (2, 3, 1). Solution: From the distance formula, we have. d(P1, P2) = √(8-2)2 + (-5-3)2 + (0-1)2 = √36 + 64 + 1 = 101 ≈ 10.05 Example 2: Find the distance between the points (3, 4, 5) and (2,-5, and 7). Solution: From the distance formula, we get d = √(3-2)2 + (4-(-5))2 + (5-7)2 =√1 + 81 + 4 = √86 |
Read More: Introduction to Three-Dimensional Geometry
Things to Remember
- In order to discover the location of a point in space, we need a rectangular coordinate system.
- After taking a particular coordinate system in 3D, the coordinates of any point P in that system can be given by an arranged 3-tuple (x, y, z).
- Similarly, if the coordinates (x, y, z) are formally understood then we can effortlessly discover the point P in space.
- On a two dimensional plane a point in the XY- plane by an arranged pair that consists of two real figures, an x-coordinate and y- coordinate, which signify signed distances along the X-axis and y-axis, independently, from the origin, which is the point (0, 0).
- These axes, which are known as the coordinate axes, divided the plane into four quadrants. The properties of three-dimensional space.
- Three-dimensional space can also be understood as 3- space or tri-dimensional space. It's a geometric medium that contains three values that are needed to decide the position of an element.
- In mathematics and physics, a sequence of n figures can be reasoned as a position in n-dimensional space. When n = 3 it's known as three-dimensional Euclidean space. It's generally shown by the symbol ?3.
Sample Questions
Ques. On what axis does the point (4, 0, 0) lie? (1 Mark)
Ans. In a 3d coordinate system, a point with z and y coordinate zero and x-coordinate having non-zero value must lie on the x-axis. So, (4, 0, 0) lie on the x-axis.
Ques. Find the distance of points (2, 3, 5) from the YZ plane. (1 Mark)
Ans. We know, the distance of a point from the Y-Z plane is equal to the value of its x-coordinate. So, the distance of points (2, 3, 5) from the Y-Z plane is 2 units.
Ques. In which octant does the point (1, 5, 7) lie? (1 Mark)
Ans. Since the point with all positive coordinates i.e. of the form (+, +, +) so lie in 1st octant. So, (1, 5, 7) lies in 1st octant.
Ques. In which octant does the point (-1, – 5, -7) lie? (1 Mark)
Ans. Since the point with all negative coordinates i.e. of the form (-, -, -) lies in the 7th octant. So, (-1, – 5, -7) lies in the 7th octant.
Ques. In which octant does the point (-1, 5, -7) lie? (1 Mark)
Ans. Since the point with only y coordinate positive i.e. of the form (-, +, -) lies in the 6th octant. So, (-1, 5, -7) lies in the 6th octant.
Ques. Point D has coordinates as (3,4,5). Referring to the given figure, find the coordinates of point B. (3 Marks)
Ans.
From the given figure, D has x-coordinate =3, y-coordinate =4 and z-coordinate =5.
Point B lies in the xy-plane. So, the z-coordinate will be 0.
B is in the same line of point D.
∴ x and y coordinates will be the same as of point D.
Hence, the coordinates of point B is (3,4,0).
Ques. Point D has coordinates as (3,4,5). Find the coordinates of the point F. (3 Marks)
Ans.
From the given figure, D has x-coordinate =3, y-coordinate =4 and z-coordinate =5.
Lines GF and DE are parallel and have the same height.
∴ All four points D, E, F and G have z-coordinate same i.e. 5.
Point F lie on the z-axis â?¹ x-coordinate and y-coordinate are 0.
Hence, the coordinates of F is (0,0,5).
Ques. Name the octants in which the following points lie (1, 2, 3), (4, -2, 3), (4, -2, -5), (4, 2, -5), (-4, 2, -5), (-4, 2, 5), (-3, -1, 6), (2, -4, -7) (5 Marks)
Ans. The octants for the given points are as follows:
- The x-coordinate, y-coordinate, and z-coordinate of point (1, 2, 3) are all positive. Therefore, this point lies in octant I.
- The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, 3) are positive, negative, and positive respectively. Therefore, this point lies in octant IV.
- The x-coordinate, y-coordinate, and z-coordinate of point (4, -2, -5) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
- The x-coordinate, y-coordinate, and z-coordinate of point (4, 2, -5) are positive, positive, and negative respectively. Therefore, this point lies in octant V.
- The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, -5) are negative, positive, and negative respectively. Therefore, this point lies in octant VI.
- The x-coordinate, y-coordinate, and z-coordinate of point (-4, 2, 5) are negative, positive, and positive respectively. Therefore, this point lies in octant II.
- The x-coordinate, y-coordinate, and z-coordinate of point (-3, -1, 6) are negative, negative, and positive respectively. Therefore, this point lies in octant III.
- The x-coordinate, y-coordinate, and z-coordinate of point (2, -4, -7) are positive, negative, and negative respectively. Therefore, this point lies in octant VIII.
Ques. Find the distance between the following pairs of points:
(i) (2, 3, 5) and (4, 3, 1)
(ii) (-3, 7, 2) and (2, 4, -1) (3 Marks)
Ans. The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by
PQ = √(x2 -x1)2 + (y2-y1)2 + (z2 - z1)2
(i) Distance between points (2, 3, 5) and (4, 3, 1)
√(4-2)2 + (3-3)2 + (1-5)2
√(2)2 + (0)2 + (-4)2
√4 + 16
√20
2√5
(ii) Distance between points (-3, 7, 2) and (2, 4, -1)
√(2+3)2 + (4-7)2 + (-1-2)2
√(5)2 + (-3)2 + (-3)2
√25 + 9 + 9
√43
Ques. Show that the points (-2, 3, 5), (1, 2, 3) and (7, 0, -1) are collinear. (5 Marks)
Ans. Let points (-, 3, 5), (1, 2, 3), and (7, 0,-1) be denoted by P, Q, and R respectively.
Points P, Q, and R are collinear if they lie on a line.
PQ = √(1+2)2 + (2-3)2 + (3-5)2
= √(3)2 + (-1)2 + (-2)2
√9 + 1 + 4
√14
QR = √(7-1)2 + (0-2)2 + (-1-3)2
= √(6)2 + (-2)2 + (-4)2
= √36+4+16
= √56
PR = √(7+2)2 + (0-3)2 + (-1-5)2
= √(9)2 + (-3)2 + (-6)2
= √81 + 9 + 36
= √126
= 3√14
Here, PQ + QR = √14 + 2√14 = 3√14 = PR
Hence, points P(-2, 3, 5), Q(1, 2, 3), and R(7, 0, -1) are collinear.
Ques.Find the distance between the following pairs of points: (3 Marks)
(i) (-1, 3, -4) and (1, -3, 4)
(ii) (2, -1, 3) and (-2, 1, 3)
Ans. The distance between points P(x1, y1, z1) and P(x2, y2, z2) is given by
PQ = √(x2 -x1)2 + (y2-y1)2 + (z2 - z1)2
(i) Distance between points (-1, 3, -4) and (1, -3, 4)
= √(1+1)2 + (-3-3)2 + (4+4)2
= √(2)2 + (-6)2 + (8)2
= √4+36+64 = √104 = 2√26
(ii) Distance between points (2, -1, 3) and (-2, 1, 3)
= √(-2-2)2 + (1+1)2 + (3-3)2
= √(-4)2 + (2)2 + (0)2
= √16 + 4
= √20
Ques. Find the equation of the set of points which are equidistant from the points (1, 2, 3) and (3, 2, -1). (3 Marks)
Ans. Let P (x, y, z) be the point that is equidistant from points A(1, 2, 3) and B(3, 2,1).
Accordingly, PA = PB
= PA2 = PB2
= (x-1)2 + (y-2)2 + (z-3)2 = (x-3)2 + (y-2)2 + (z+1)2
⇒ x2 – 2x + 1 + y2 – 4y + 4 + z2 – 6z + 9 = x2 – 6x + 9 + y2 – 4y + 4 + z2 + 2z + 1
⇒ -2x -4y – 6z + 14 = -6x – 4y + 2z + 14
⇒ – 2x – 6z + 6x – 2z = 0
⇒ 4x – 8z = 0
⇒ x – 2z = 0
Thus, the required equation is x – 2z = 0.
Ques. Find the equation of the set of points P, the sum of whose distances from A (4, 0, 0) and B (-4, 0, 0) is equal to 10. (5 Marks)
Ans. Let the coordinates of P be (x, y, z).
The coordinates of points A and B are (4, 0, 0) and (-4, 0, 0) respectively.
It is given that PA + PB = 10.
On squaring both sides, we obtain
On squaring both sides again, we obtain
25 (x2 + 8x + 16 + y2 + z2) = 625 + 16x2 + 200x
⇒ 25x2 + 200x + 400 + 25y2 + 25z2 = 625 + 16x2 + 200x
⇒ 9x2 + 25y2 + 25z2 – 225 = 0
Thus, the required equation is 9x2 + 25y2 + 25z2 – 225 = 0.
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