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Coordinate Axes and Coordinate Planes in 3D space

Points, Lines and Planes

Last Updated : 09 Jun, 2025

Points, Lines, and Planes are basic terms used in Geometry that have a specific meaning and are used to define the basis of geometry.

We define a point as a location in 3-D or 2-D space that is represented using coordinates.
We define a line as a geometrical figure that is extended in both directions to infinity.
Similarly, a plane is defined as the collection of all such lines, i.e., it is a 3-D space on which the line passes.

In this article, we will learn about Points, Lines, and Planes in detail, including their solved examples and problems based on them.

Table of Content

Points, Lines, and Planes in Geometry

In basic geometry, fundamental concepts like points, lines, and planes form the foundation upon which more complex geometric ideas are built. Points are precise locations in space, devoid of size or dimension, represented simply by dots.

Lines are infinite paths stretching in two opposite directions, composed of an unending series of points. They are defined by any two distinct points lying on them. Beyond lines, planes extend infinitely in all directions, forming flat, two-dimensional surfaces.

A plane can be defined by any three non-collinear points or by a line and a point not on the line. These basic elements serve as the starting point for exploring the intricacies of geometric shapes, relationships, and constructions.

What is a Point?

A Point in geometry is defined as a location in the space that is uniquely determined by an ordered triplet (x, y, z) where x, y, & z are the distances of the point from the X-axis, Y-axis, and Z-axis respectively in the 3-Dimensions and is defined by ordered pair (x, y) in the 2-Dimensions where, x and y are the distances of the point from the X-axis, and Y-axis, respectively. It is represented using the dot and is named using the capital English alphabet. The figure added below shows a point P in 3-D at a distance of x, y, and z from the X-axis, Y-axis, and Z-axis, respectively.

Point

Collinear and Non-Collinear Points

When 3 or more points are present on the straight line then such types of points as known as Collinear pointsand if these points do not present on the same line, then such types of points are known as non-collinear points.

Collinear and Non-Collinear Points

Coplanar and Non-Coplanar Points

When the group of points is present on the same plane then such types of points are known as coplanar points and if these points do not present on the same plane, then such types of points are known as non-coplanar points.

Coplanar and Non-Coplanar Points

What is a Line?

A Line in three-dimensional geometry is defined as a set of points in 3D that extends infinitely in both directions It is the smallest distance between any two points either in 2-D or 3-D space. We represent a line with L and in 3-D space, a line is given using the equation,

L: (x - x₁) / l = (y - y₁) / m = (z - z₁) / n
where
(x, y, z) are the position coordinates of any variable point lying on the line
(x₁, y₁, z₁) are the position coordinates of a point P lying on the line
l, m, & n are the direction ratios of the line.

In 3D we can also form a line by the intersection of two non-parallel planes.

Line — Points, Lines, and Planes

Line Segment

A line segment is defined as the finite length of the line that is used to join two points in 2-D and 3-D. It is the shortest distance between two points. A line segment between two points A and B is denoted as, AB

Line Segment

A line has infinite length whereas a line segment is a part of a line and has finite length.

Mid-Point

Midpoint is defined as the point on the line segment which divides the line segment into two equal parts. Suppose we have two points A and B and the line segment joining these two points is AB and not the point P on the line is called the midpoint if it breaks the line into two equal parts such that,

AP = PB

Thus, P is called the midpoint of line segment AB. The image added below shows the line segment AB with P as the midpoint.

Mid-Point

Rays

A ray is defined as a line that has a fixed end point in one direction but can be extended to infinity in the other direction. It is of infinite length. We define the ray joining points O and A and extending to infinity towards A as

Rays

Intersecting and Parallel lines

In 2-D any two lines can either meet at some point or they never meet at some point. The lines that meet at some point are called intersecting lines. The distance between the intersecting line keeps on decreasing as we move toward the point of intersection, and at the point of intersection of these lines, the distance between them becomes zero. When two lines intersect an angle is formed between them.

Two lines that never meet each other in 2-D planes are called parallel lines. For parallel lines, the distance between them is always constant.

The images below show Intersecting and Parallel lines.

Intersecting and Parallel lines

Perpendicular Lines

Intersecting lines that intersect at right angles are called perpendicular lines. The angle between these perpendicular lines is always the right angle or 90 degrees.

The perpendicular lines are shown in the image added below:

Perpendicular Lines

What is a Plane?

A Plane in three-dimensional (3D) geometry is a surface such that the line segment joining any two points lies completely on it. It is the collection of all the points and can be extended infinitely in any of the two dimensions.

The general form of a plane in 3D is a first-degree equation in x, y, z i.e. We represent a plane in 3-D as,

(ax + by + cz + d = 0)
where
(x, y, z) represents the coordinates of a variable point on the plane.

A plane has only two dimensions length and breadth and it can be infinitely stretched in these two dimensions.

Plane

➣ Read More, Cartesian Plane

Solid

A solid is a 3-D concept we also called, space. We defined the solid as the extended plane that has three dimensions length breadth and height. A solid can be extended infinitely to incorporate all the space in 3-D.

Vector Form of Equation of Plane in Normal Form

The vector form of the equation of a plane in normal form is given by:

\pi : \vec{r} \cdot \hat{n} = d
where:
π represents a plane in 3D space.({\vec {r}} . {\vec{n}} = d)
{\vec{r}} vector is the position vector of a general point lying on the plane,
n̂ is the unit vector normal to the plane
d is the distance of a plane from the origin

Note: The vector equation of the plane in the form ({\vec {r}} . {\vec{n}} = d) is said to be in the normal form only when {\vec {n}} is a unit vector normal to the plane and d is the distance of the plane from the origin. If {\vec {n}} is not a unit vector then we have to divide the above equation by |{\vec{n}}| on both sides in order to convert it into the normal form. \vec{r} \cdot \left( \frac{\vec{n}}{|\vec{n}|} \right) = \frac{d}{|\vec{n}|} \quad \text{or} \quad \vec{r} \cdot \hat{n} = \frac{d}{|\vec{n}|}. This can be understood with the help of the example discussed below,

Example: The vector equation of the plane in 3D space which is at a distance of 8 units from the origin and normal to the vector (2 i+ j + 2 k) is given by?

Solution:

d = 8 and {\vec {n}} = (2 i+ j + 2 k)
n̂ = (2 i+ j + 2 k) / √(2² + 1² + 2²)
n̂ = (2 i+ j + 2 k) / √9
n̂ = (2/3) i+ (1/3) j + (2/3) k
Hence the required vector equation of the plane in normal form is
{\vec {r}} . ((2/3) i+ (1/3) j + (2/3) k) = 8 which can be simplified as {\vec {r}} . (2 i+ 1 j + 2 k) = 24

Cartesian Form of Equation of a Plane in Normal Form

The cartesian form of the equation of a plane in normal form is given by:

π: lx + my + nz = p
where:
π again represents a plane in 3D space
l, m, n are the DC's i.e. direction cosines of the normal to the plane always satisfies this condition (l² + m² + n² = 1)
p is the distance of the plane from the origin

Note: Any cartesian equation of the plane in the form (ax + by + cz + d = 0) is said to be in the normal form only when a, b, c are the direction cosines of the normal to the plane and |d| is the distance of the plane from the origin.

If a, b, and c are not the direction cosines of the normal to the plane then we have to follow these steps:

Step 1: Keep the terms of x, y, and z on the LHS and take the constant term d on the RHS.
Step 2: If the constant term on the RHS is negative then make it positive by multiplying with (-1) on both sides of the equation.
Step 3: Divide term on the both sides of the equation by √(a²+ b² + c²).

After applying these steps the coefficients of x, y, and z on the LHS will become the direction cosines of the normal to the plane, and the constant term on the RHS will become the distance of the plane from the origin.

Example: A plane in the 3D space is represented by (2x + y + 2z - 24 = 0) then the cartesian equation of this plane in the normal form is given by.

Solution:

Given equation of plane,
2x + y + 2z = 24
Dividing both sides of the above equation by √(2² + 1² + 2²) = √9 = 3
The given equation of plane in cartesian form is,
(2/3) x + (1/3) y + (2/3) z = 8
Here,
l = 2/3 , m = 1/3 , n = 2/3 are the direction cosines
p = 8 is the distance from the origin.

Distance of a Point from a Plane in Cartesian Form

The distance of a point P(x_o, y_o, z_o) from a plane π:(a x + b y + c z +d = 0) in the cartesian form is defined as the length (L) of the perpendicular drawn from that point to the plane. That is calculated using the formula,

L = |a x_o + b y_o + c z_o + d| / √(a² + b² + c²)

Now let's look at the example to know more.

Example: Find the distance of the point (2, 1, 0) from the plane (2 x + y + 2 z + 5 = 0).

Solution:

x_o = 2, y_o = 1, z_o = 0
a = 2, b = 1, c = 2, d = 5
L = |(2 × 2) + (1 × 1) + (0 × 2) + 5| / √(2², 1², 2²)
⇒ L = 10 / √9
⇒ L = 10/3
Thus, the required distance is 10/3 units.

Distance of a Point from a Plane in Vector Form

The distance of a point P having position vector {\vec {a}} from a plane π: {\vec {r}}.{\vec {n}} = d in vector form is defined as the length (L) of the perpendicular drawn from that point to the plane.

L = \frac{| \vec{a} \cdot \vec{n} - d |}{| \vec{n} |}

Example: The distance of a point with position vector (2 i + j + 0 k) from the plane {\vec {r}} . (2 i + j + 2 k) = 5 is given by?

Solution:

{\vec {a}}     = 2 i + j + 0 k
{\vec {n}}     = 2 i + j + 2 k
|{\vec {n}}    | = √(2², 1², 2²) = √9 = 3
d = 5 (given)
{\vec {a}}.{\vec {n}}     = (2 × 2) + (1 × 1) + (0 × 2) = 5
⇒ L = |5 - 5| / 3
⇒ L = 0
Thus, the required distnce is 0 units.

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Points, Lines, and Planes Solved Examples

Example 1:The vector equation of the plane in 3D space which is at a distance of 5 units from the origin and normal to the vector (4 i+ 3 k) is given by?

Solution:

d = 5 and {\vec {n}}     = (4 i+ 3 k)
n̂ = (4 i + 3 k) / √(4² + 0² + 3²)
⇒ n̂ = (4 i+ 3 k) / √(25)
⇒ n̂ = (4/5) i+ (3/5) k
Hence the required vector equation of the plane in normal form is,
{\vec {r}}     . ((4/5) i+ (3/5) k = 5
⇒ {\vec {r}}     . (4 i+ 3 k) = 25

Example 2:Find the distance of the point (0, 1, 0) from the plane (3 y + 4 z = 7).

Solution:

Given Point,
(x_o,y_o,z_o) = (0, 1, 0)
x_o = 0, y_o = 1, z_o = 0
Equation of the plane,
3 y + 4 z = 7
Comapring with ax + by +cz + d = 0
a = 0, b = 3, c = 4, d = -7
Distanc of a point form a plane is given using the formula,
L = |a x_o + b y_o + c z_o + d| / √(a² + b² + c²)
⇒ L = |0 + (3 × 1) + (4 × 0) - 7| / √(0², 3², 4²)
⇒ L = |3 - 7| / √(25)
⇒ L = 4/5
Thus, the required distance is 4/5 units

Example 3:Find the distance of a point (5, 3, 0) from the plane\vec{r} \cdot (4 \hat{i} + 3 \hat{j}) = 8

Solution:

{\vec {a}}     = 5 i +3 j + 0 k
{\vec {n}}     = 4 i +3 j + 0 k
|{\vec {n}}    | = √(4²+ 3²+ 0²)
= √(25) = 5
d = 8 (given)
{\vec {a}}     . {\vec {n}}     = (5 × 4) + (3 × 3) + (0 × 0)
= 29
⇒ L = |29 - 8| / 5
⇒ L = 21/5
Thus, the required distance is 21/5 units

Example 4: Find the distance between points, (2, 4) and (5, 6).

Solution:

Given point, (2, 4) and (5, 6)
Comapring with (x₁, y₁) and (x₂, y₂)
(x₁, y₁) = (2, 4) and (x₂, y₂) = (5, 6)
Using the distance formula,
d = √[(x₂- x₁)² + (y₂ - y₁)²]
⇒ d = √[(5-2)² + (6-4)²]
⇒ d = √13
Thus, the distance between two points is √13 units.

Practice Questions onPoints, Lines and Planes

Identify which of the following sets of points are collinear: A(2, 3), B(5, 7), C(8, 11).
Determine if the points D(4, 6), E(4, 6), and F(4, 8) lie on the same line.
Given three non-collinear points, how many distinct lines can be drawn through them?
Determine whether the points G(2, 3, 1), H(4, 1, 3), and I(6, 5, 2) lie on the same plane.
Find the equation of the line passing through the points P(2, 1) and Q(4, 5) in slope-intercept form.
Determine the distance between the points R(3, 2, 4) and S(7, 5, 8).

Coordinate Axes and Coordinate Planes in 3D space

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