Cartesian Form of Equation of Line in 3D

The cartesian form of line is given by using the coordinates of two points located in space from which the line is passing. In this we will discuss two cases, when line passes through two points and when line passes through points and is parallel to a vector.

Case 1: 3D Equation of Line in Cartesian Form Passing Through Two Points

Let us assume we have two points A and B whose coordinates are given as A(x₁, y₁, z₁) and B(x₂, y₂, z₂).

Then the 3D equation of straight line in cartesian form is given as

[Tex]\bold{\frac{x – x_1} {x_2 – x_1} = \frac{y – y_1} {y_2 – y_1} = \frac{z -z_1} {z_2 – z_1}} [/Tex]

where, x, y and z are rectangular coordinates.

Derivation of Equation of Line Passing through two Points

We can derive the Cartesian form of 3D Equation of Straight Line by the use of following mentioned steps:

• Step 1: Find the DR’s (Direction Ratios) by taking the difference of the corresponding position coordinates of the two given points. l = (x₂ – x₁), m = (y₂ – y₁), n = (z₂ – z₁); Here l, m, n are the DR’s.
• Step 2: Choose either of the two given points say, we choose (x₁, y₁, z₁).
• Step 3: Write the required equation of the straight line passing through the points (x₁, y₁, z₁) and (x₂, y₂, z₂).
• Step 4: The 3D equation of straight line in cartesian form is given as L : (x – x₁)/l = (y – y₁)/m = (z – z₁)/n = (x – x₁)/(x₂ – x₁) = (y – y₁)/(y₂ – y₁) = (z – z₁)/(z₂ – z₁)

Where (x, y, z) are the position coordinates of any variable point lying on the straight line.

Example: If a straight line is passing through the two fixed points in the 3-dimensional whose position coordinates are P (2, 3, 5) and Q (4, 6, 12) then its cartesian equation using the two-point form is given by

Solution:

l = (4 – 2), m = (6 – 3), n = (12 – 5)

l = 2, m = 3, n = 7

Choosing the point P (2, 3, 5)

The required equation of the line

L: (x – 2) / 2 = (y – 3) /  3 = (z – 5) / 7

Case 2: 3D Equation of Line in Cartesian Passing through a Point and Parallel to a given Vector

Let us assume the line passes through a point P(x₁, y₁, z₁) and is parallel to a vector given as [Tex]\vec n = a\hat i + b\hat j + c\hat k [/Tex].

Then the equation of line is given as

[Tex]\bold{\frac{x – x_1} a = \frac{y – y_1} b = \frac{z -z_1} c} [/Tex]

where, x, y, z are rectangular coordinates and a, b, c are direction cosines.

Derivation of 3D Equation of Line in Cartesian Passing through a Point and Parallel to a given Vector

Let us assume we have a point P whose position vector is given as [Tex]\vec p[/Tex] from the origin. Let the line that passes through P is parallel to another vector [Tex]\vec n[/Tex]. Let us take a point R on the line that passes through P, then the position vector of R is given as [Tex]\vec r [/Tex].

Since, PR is parallel to [Tex]\vec n[/Tex] ⇒ [Tex]\overline {PR} = \lambda \vec n[/Tex]

Now if we move on the line PR then the coordinate of any point that lies on the line will have the coordinate in the form of (x₁ + λa), (y₁ + λb), (z₁ + λc), where λ is a parameter whose value ranges from -∞ to +∞ depending on the direction from P where we move.

Hence, the coordinates of the new point will be

x = x₁ + λa ⇒ λ = x – x₁/a

y = y₁ + λb ⇒ λ = y – y₁/b

z = z₁ + λc ⇒ λ = z – z₁/c

Comparing the above three equations we have the equation of line as

[Tex]\bold{\frac{x – x_1} a = \frac{y – y_1} b = \frac{z -z_1} c}[/Tex]

Example: Find the Equation of a Line passing through a point (2, 1, 3) and parallel to a vector 3i – 2j + k

Solution:

The Equation of line passing through a point and parallel to a vector is given as

(x – x₁)/a = (y – y₁)/b = (z – z₁)/c

From the question we have, x₁ = 2, y₁ = 1, z₁ = 3 and a = 3, b = -2 and c = k. Hence, the required equation of the line will be

⇒ (x – 2)/3 = (y – 1 )/-2 = (z – 3)/1

The equation of a line in a plane is given as y = mx + C where x and y are the coordinates of the plane, m is the slope of the line and C is the intercept. However, the construction of a line is not limited to a plane only.

We know that a line is a path between two points. These two points can be located anywhere whether they could be in a single plane or they could be in space. In the case of a plane, the location of the line is characterized by two coordinates arranged in an ordered pair given as (x, y) whereas in the case of space, the location of the point is characterized by three coordinates expressed as (x, y, z).

In this article, we will learn the different forms of equations of lines in 3D space.