Coordinates of a Point in Space and Distance Between Two Points

Introduction to Three-Dimensional Geometry

Three-dimensional (3D) geometry extends the concepts of two-dimensional geometry into the third dimension. It deals with objects that have length, width, and height. This branch of mathematics is crucial for understanding various physical phenomena and is widely applied in fields like engineering, physics, computer graphics, and architecture.

In 2D geometry, we use two axes (x and y) to define the position of a point. In 3D geometry, we introduce a third axis, the z-axis, which is perpendicular to both the x and y axes at their intersection point, the origin. These three mutually perpendicular axes form a coordinate system in space.

Coordinate System in Three Dimensions

The three mutually perpendicular lines are called the coordinate axes: the x-axis, the y-axis, and the z-axis. Their intersection point is called the origin (O), denoted by (0, 0, 0).

These three axes divide the space into eight regions called octants. The position of any point P in space can be uniquely determined by an ordered triplet of real numbers (x, y, z), where:

• 'x' is the perpendicular distance of the point from the yz-plane (the plane containing the y and z axes).
• 'y' is the perpendicular distance of the point from the xz-plane (the plane containing the x and z axes).
• 'z' is the perpendicular distance of the point from the xy-plane (the plane containing the x and y axes).

The values x, y, and z are called the coordinates of the point P. The x-coordinate is also called the abscissa, the y-coordinate is called the ordinate, and the z-coordinate is called the applicate.

Locating a Point in Space

To locate a point P(x, y, z) in space:

1. Start from the origin (0, 0, 0).
2. Move along the x-axis by a distance 'x' units. If x is positive, move in the positive x-direction; if x is negative, move in the negative x-direction.
3. From this position, move parallel to the y-axis by a distance 'y' units. If y is positive, move in the positive y-direction; if y is negative, move in the negative y-direction.
4. Finally, from this new position, move parallel to the z-axis by a distance 'z' units. If z is positive, move upwards (in the positive z-direction); if z is negative, move downwards (in the negative z-direction).

The point where you finally arrive is the point P(x, y, z).

Octants and Signs of Coordinates

Similar to quadrants in 2D geometry, the three coordinate planes divide space into eight octants. The octants are distinguished by the signs of the coordinates (x, y, z).

The signs of the coordinates in each octant are as follows:

Octant	x	y	z
I	+	+	+
II	-	+	+
III	+	-	+
IV	-	-	+
V	+	+	-
VI	-	+	-
VII	+	-	-
VIII	-	-	-

Points lying on the coordinate axes or coordinate planes have one or two coordinates equal to zero, respectively. For example, a point on the x-axis will have coordinates (x, 0, 0), and a point on the xy-plane will have coordinates (x, y, 0).

Distance Between Two Points in Space

Let P1(x1, y1, z1) and P2(x2, y2, z2) be two points in three-dimensional space. We want to find the distance between these two points, denoted as P1P2.

We can use the distance formula, which is an extension of the Pythagorean theorem. Consider a rectangular parallelepiped (a box) with P1 and P2 as opposite vertices. The sides of this box are parallel to the coordinate axes.

The lengths of the sides of this box are |x2 - x1|, |y2 - y1|, and |z2 - z1|.

Using the Pythagorean theorem twice, we can derive the distance formula:

First, consider the diagonal of the base of the box in the xy-plane. Let this diagonal be 'd'. The square of this diagonal is given by:

d² = (x2 - x1)² + (y2 - y1)²

Now, consider the right-angled triangle formed by this diagonal 'd', the vertical edge of length |z2 - z1|, and the distance P1P2 as the hypotenuse.

By the Pythagorean theorem again:

(P1P2)² = d² + (z2 - z1)²

Substituting the expression for d²:

(P1P2)² = (x2 - x1)² + (y2 - y1)² + (z2 - z1)²

Therefore, the distance between points P1(x1, y1, z1) and P2(x2, y2, z2) is:

P1P2 = √[(x2 - x1)² + (y2 - y1)² + (z2 - z1)²]

Example 1: Calculating Distance

Find the distance between the points A(1, 2, 3) and B(4, 6, 8).

Here, (x1, y1, z1) = (1, 2, 3) and (x2, y2, z2) = (4, 6, 8).

Using the distance formula:

AB = √[(4 - 1)² + (6 - 2)² + (8 - 3)²]

AB = √[(3)² + (4)² + (5)²]

AB = √[9 + 16 + 25]

AB = √50

AB = √(25 × 2)

AB = 5√2 units.

Example 2: Distance from Origin

Find the distance of the point P(3, -4, 5) from the origin (0, 0, 0).

Using the distance formula with (x1, y1, z1) = (0, 0, 0) and (x2, y2, z2) = (3, -4, 5):

OP = √[(3 - 0)² + (-4 - 0)² + (5 - 0)²]

OP = √[(3)² + (-4)² + (5)²]

OP = √[9 + 16 + 25]

OP = √50

OP = 5√2 units.

Alternatively, the distance of a point (x, y, z) from the origin is simply √(x² + y² + z²).

Collinear Points

Three or more points are said to be collinear if they lie on the same straight line. In 3D geometry, we can check for collinearity using the distance formula.

Let A, B, and C be three points. They are collinear if the sum of the distances between two pairs of points is equal to the distance between the remaining pair. That is, if AB + BC = AC, or AC + CB = AB, or BA + AC = BC.

Example 3: Checking for Collinearity

Check if the points P(1, 2, 3), Q(2, 4, 6), and R(3, 6, 9) are collinear.

Calculate the distances between each pair of points:

PQ = √[(2 - 1)² + (4 - 2)² + (6 - 3)²] = √[1² + 2² + 3²] = √[1 + 4 + 9] = √14

QR = √[(3 - 2)² + (6 - 4)² + (9 - 6)²] = √[1² + 2² + 3²] = √[1 + 4 + 9] = √14

PR = √[(3 - 1)² + (6 - 2)² + (9 - 3)²] = √[2² + 4² + 6²] = √[4 + 16 + 36] = √56

Now, check the condition for collinearity:

PQ + QR = √14 + √14 = 2√14

We also know that √56 = √(4 × 14) = 2√14.

Since PQ + QR = PR (2√14 = 2√14), the points P, Q, and R are collinear.

Section Formula (Internal and External Division)

This formula helps find the coordinates of a point that divides the line segment joining two points in a given ratio.

Let P(x1, y1, z1) and Q(x2, y2, z2) be two points. Let R(x, y, z) be a point that divides the line segment PQ in the ratio m:n.

Internal Division:

If R divides PQ internally in the ratio m:n, then the coordinates of R are given by:

x = (m*x2 + n*x1) / (m + n)

y = (m*y2 + n*y1) / (m + n)

z = (m*z2 + n*z1) / (m + n)

External Division:

If R divides PQ externally in the ratio m:n, then the coordinates of R are given by:

x = (m*x2 - n*x1) / (m - n)

y = (m*y2 - n*y1) / (m - n)

z = (m*z2 - n*z1) / (m - n)

(Note: For external division, m must be greater than n, or the ratio is taken as m:(-n)).

Example 4: Section Formula (Internal)

Find the coordinates of the point R which divides the line segment joining P(2, 3, 4) and Q(5, 6, 7) internally in the ratio 1:2.

Here, (x1, y1, z1) = (2, 3, 4), (x2, y2, z2) = (5, 6, 7), m = 1, and n = 2.

x = (1*5 + 2*2) / (1 + 2) = (5 + 4) / 3 = 9 / 3 = 3

y = (1*6 + 2*3) / (1 + 2) = (6 + 6) / 3 = 12 / 3 = 4

z = (1*7 + 2*4) / (1 + 2) = (7 + 8) / 3 = 15 / 3 = 5

So, the coordinates of R are (3, 4, 5).

Example 5: Section Formula (External)

Find the coordinates of the point R which divides the line segment joining P(1, 2, 3) and Q(4, 5, 6) externally in the ratio 2:1.

Here, (x1, y1, z1) = (1, 2, 3), (x2, y2, z2) = (4, 5, 6), m = 2, and n = 1.

x = (2*4 - 1*1) / (2 - 1) = (8 - 1) / 1 = 7 / 1 = 7

y = (2*5 - 1*2) / (2 - 1) = (10 - 2) / 1 = 8 / 1 = 8

z = (2*6 - 1*3) / (2 - 1) = (12 - 3) / 1 = 9 / 1 = 9

So, the coordinates of R are (7, 8, 9).

Midpoint Formula

The midpoint formula is a special case of the section formula where the ratio m:n is 1:1.

If M(x, y, z) is the midpoint of the line segment joining P(x1, y1, z1) and Q(x2, y2, z2), then:

x = (x1 + x2) / 2

y = (y1 + y2) / 2

z = (z1 + z2) / 2

Example 6: Midpoint Formula

Find the midpoint M of the line segment joining A(3, 5, 7) and B(1, 9, 11).

x = (3 + 1) / 2 = 4 / 2 = 2

y = (5 + 9) / 2 = 14 / 2 = 7

z = (7 + 11) / 2 = 18 / 2 = 9

The midpoint M is (2, 7, 9).

Centroid of a Triangle

The centroid of a triangle is the point of intersection of its medians. It is also the center of mass of the triangle.

If the vertices of a triangle are A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3), then the coordinates of the centroid G are:

x = (x1 + x2 + x3) / 3

y = (y1 + y2 + y3) / 3

z = (z1 + z2 + z3) / 3

The centroid is essentially the average of the coordinates of the vertices.

Example 7: Centroid Calculation

Find the centroid of the triangle with vertices P(1, 2, 3), Q(4, 5, 6), and R(7, 8, 9).

x = (1 + 4 + 7) / 3 = 12 / 3 = 4

y = (2 + 5 + 8) / 3 = 15 / 3 = 5

z = (3 + 6 + 9) / 3 = 18 / 3 = 6

The centroid G is (4, 5, 6).

Summary of Key Concepts

The ability to represent points in 3D space and calculate distances between them is fundamental to understanding more complex 3D geometric concepts. The coordinate system with three axes (x, y, z) allows us to pinpoint any location. The distance formula, derived from the Pythagorean theorem, is essential for measuring lengths. The section formula (including midpoint) and centroid formula provide tools to find specific points related to line segments and triangles in 3D space.

Mastering these foundational concepts will pave the way for exploring vectors, lines, planes, and surfaces in three dimensions, which are critical topics in JEE Main Mathematics.