Equation of a Line in 3D

We all know the very popular equation of the straight line Y = m . X + C which a straight line in a plane. But here we are going to discuss the Equation of a Straight Line in 3-dimensional space. A Straight Line is uniquely characterized if it passes through the two unique points or it passes through a unique point in a definite direction. In Three Dimensional Geometry lines (straight lines) are usually represented in the two forms Cartesian Form and Vector form. Here we are going to discuss the two-point form of a straight line in 3-dimensions using both cartesian as well as vector form.

Equation of a Straight Line in Cartesian Form

For writing the equation of a straight line in the cartesian form we require the coordinates of a minimum of two points through which the straight line passes. Let’s say (x₁, y₁, z₁) and (x₂, y₂, z₂) are the position coordinates of the two fixed points in the 3-dimensional space through which the line passes.

Now to obtain the equation we have to follow these three 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). 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 straight line.

Example 1: 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

Example 2: If a straight line is passing through the two fixed points in the 3-dimensional whose position coordinates are A (2, -1, 3) and B (4, 2, 1) then its cartesian equation using the two-point form is given by

Solution:

l = (4 – 2), m = (2 – (-1)), n = (1 – 3)
l = 2, m = 3, n = -2
Choosing the point A (2, -1, 3)
The required equation of the line
L : (x – 2) / 2 = (y + 1) / 3 = (z – 3) / -2 or
L : (x – 2) / 2 = (y + 1) / 3 = (3 – z) / 2

Example 3: If a straight line is passing through the two fixed points in the 3-dimensional whose position coordinates are X (2, 3, 4) and Y (5, 3, 10) then its cartesian equation using the two-point form is given by

Solution:

l = (5 – 2), m = (3 – 3), n = (10 – 4)
l = 3, m = 0, n = 6
Choosing the point X (2, 3, 4)
The required equation of the line
L : (x – 2) / 3 = (y – 3) / 0 = (z – 4) / 6 or
L : (x – 2) / 1 = (y – 3) / 0 = (z – 4) / 2

Equation of a Straight Line in Vector Form

For writing the equation of a straight line in the vector form we require the position vectors of a minimum of two points through which the straight line passes. Let’s say ${\vec {a}}$ and ${\vec {n}}$ are the position vectors of the two fixed points in the 3-dimensional space through which the line passes.

Now to obtain the equation we have to follow these three steps:

Step 1: Find a vector parallel to the straight line by subtracting the corresponding position vectors of the two given points. ${\vec {p}}$ = ( ${\vec {b}}-{\vec {a}}$ ); Here ${\vec {p}}$ is the vector parallel to the straight line.
Step 2: Choose the position vector of either of the two given points say we choose ${\vec {a}}$ .
Step 3: Write the required equation of the straight line passing through the points whose position vectors are ${\vec {a}}$ and ${\vec {b}}$ . L : ${\vec {r}}$ = ${\vec {a}}$ + t . ${\vec {p}}$ or ${\vec {r}}$ = ${\vec {a}}$ + t . ( ${\vec {b}}-{\vec {a}}$ )

Where ${\vec {r}}$ is the position vector of any variable point lying on the straight line and t is the parameter whose value is used to locate any point on the line uniquely.

Example 1: If a straight line is passing through the two fixed points in the 3-dimensional whose position vectors are (2 i + 3 j + 5 k) and (4 i + 6 j + 12 k) then its Vector equation using the two-point form is given by

Solution:

${\vec {p}}$ = (4 i + 6 j + 12 k) – (2 i + 3 j + 5 k)
${\vec {p}}$ = (2 i + 3 j + 7 k) ; Here ${\vec {p}}$ is a vector parallel to the straight line
Choosing the position vector (2 i + 3 j + 5 k)
The required equation of the straight line
L : ${\vec {r}}$ = (2 i + 3 j + 5 k) + t . (2 i + 3 j + 7 k)

Example 2: If a straight line is passing through the two fixed points in the 3-dimensional space whose position coordinates are (3, 4, -7) and (1, -1, 6) then its vector equation using the two-point form is given by

Solution:

Position vectors of the given points will be (3 i + 4 j – 7 k) and (i – j + 6 k)
${\vec {p}}$ = (3 i + 4 j – 7 k) – (i – j + 6 k)
${\vec {p}}$ = (2 i + 5 j – 13 k) ; Here ${\vec {p}}$ is a vector parallel to the straight line
Choosing the position vector (i – j + 6 k)
The required equation of the straight line
L : ${\vec {r}}$ = (i – j + 6 k) + t . (2 i + 5 j – 13 k)

Example 3: If a straight line is passing through the two fixed points in the 3-dimensional whose position vectors are (5 i + 3 j + 7 k) and (2 i + j – 3 k) then its Vector equation using the two-point form is given by

Solution:

${\vec {p}}$ = (5 i + 3 j + 7 k) – (2 i + j – 3 k)
${\vec {p}}$ = (3 i + 2 j + 10 k) ; Here ${\vec {p}}$ is a vector parallel to the straight line
Choosing the position vector (2 i + j – 3 k)
The required equation of the straight line
L : ${\vec {r}}$ = (2 i + j – 3 k) + t . (2 i + 3 j + 7 k)

Last Updated : 11 Feb, 2021