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3D Coordinate Geometry - Equation of a Line

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• Sandeep Bhardwaj
• Mahindra Jain

Equation of a Line

Examples and problems.

To find the equation of a line in a two-dimensional plane, we need to know a point that the line passes through as well as the slope. Similarly, in three-dimensional space, we can obtain the equation of a line if we know a point that the line passes through as well as the direction vector , which designates the direction of the line. The formula is as follows:

The equation of a line with direction vector \(\vec{d}=(l,m,n)\) that passes through the point \((x_1,y_1,z_1)\) is given by the formula \[\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n},\] where \(l,m,\) and \(n\) are non-zero real numbers. \(_\square\)

Consider a line which passes through the point \(P=(x_1,y_1,z_1)\) and has direction vector \(\vec{d}=(l,m,n),\) where \(l,m,\) and \(n\) are non-zero real numbers. Let \(X=(x,y,z)\) be a random point on the line. Then the vector \(\vec{PX},\) which is the red arrow in the figure, will be parallel to \(\vec{d}.\) Hence we have \[\begin{align} \vec{PX}&=t\vec{d}\\ (x-x_1,y-y_1,z-z_1)&=t\cdot(l,m,n)\\ t&=\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}. \end{align}\] Therefore, any point \(X=(x,y,z)\) on the line will satisfy the equation \[\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}.\ _\square\]

The equation of a line with direction vector \(\vec{d}=(l,m,0)\) that passes through the point \((x_1,y_1,z_1)\) is given by the two formulas \[\frac{x-x_1}{l}=\frac{y-y_1}{m} \quad and \quad z=z_1,\] where \(l\) and \(m\) are non-zero real numbers.

The proof is very similar to the previous one.

Find the equation of a line with direction vector \(\vec{d}=(1,2,3)\) that passes through the point \(P=(-1,0,1).\) According to the formula above, the equation of the line is \[x+1=\frac{y}{2}=\frac{z-1}{3}.\ _\square\]

In similarity with a line on the coordinate plane, we can find the equation of a line in a three-dimensional space when given two different points on the line, since subtracting the position vectors of the two points will give the direction vector.

Find the equation of the line that passes through the points \(P=(3,-1,2)\) and \(Q=(-3,0,1).\) Subtracting the position vectors of the two points gives the direction vector, which is \[\vec{d}=\vec{PQ}=(-6,1,-1).\] We can either set \(P\) or \(Q\) as \((x_1,y_1,z_1).\) Using \(P\) would give \[\frac{x-3}{-6}=y+1=-(z-2),\qquad(1)\] and using \(Q\) would give \[\frac{x+3}{-6}=y=-(z-1).\qquad(2)\] Observe that adding 1 to all sides of (2) gives (1), which means that both equations are identical. \( _\square \)

What if a coordinate of the direction vector equals zero? Suppose the \(x\)-coordinate of the direction vector is zero. This indicates that all points on the line would have equal \(x\)-coordinates. Hence the equation for this case would look like

\[\begin{align} \frac{y-y_1}{m}&=\frac{z-z_1}{n}\\ x&=x_1. \end{align}\]

Similarly, in the case where two coordinates of the direction vector are zero (say, the \(x\)- and \(y\)-coordinates), the equation would look like

\[\begin{align} x&=x_1\\ y&=y_1. \end{align}\]

Find the equation of the line that passes through the two points \(P=(1,1,1)\) and \(Q=(-1,1,3).\) Subtracting the position vectors of the two points gives the direction vector, which is \[\vec{d}=\vec{PQ}=(-2,0,2).\] Since the \(y\)-coordinate of the direction vector is zero, the equation is \[\begin{align} \frac{x-1}{-2}&=\frac{z-1}{2}\\ y&=1\\ \\ &\text{or}\\ \\ \frac{x+1}{-2}&=\frac{z-3}{2}\\ y&=1.\ _\square \end{align}\]

The relationship between two different lines in a three-dimensional space is always one of the three: they can be parallel, skew, or intersecting at one point. If the direction vectors of the lines are parallel, then the lines are also parallel (provided that they are not identical). If the lines do not meet and their direction vectors are not parallel, then they are skew. If the lines meet and their direction vectors are not parallel, then the lines meet at a single point. As we can see, comparing the direction vectors usually gives useful information concerning two lines.

What is the relationship of the following two lines: \[\frac{x-2}{2}=y=\frac{z+1}{-3}\quad \text{and}\quad \frac{x+7}{-6}=-\frac{y}{3}=\frac{z+1}{9}?\] The direction vectors of the two lines are \(\vec{d_1}=(2,1,-3)\) and \(\vec{d_2}=(-6,-3,9).\) Since \(-3\vec{d_1}=\vec{d_2},\) the two direction vectors are parallel. This implies that the two lines are either identical or parallel. A point that the first line passes through is \((2,0,-1).\) Since this point does not satisfy the equation of the second line, the second line does not pass through this point. Therefore, the two lines are parallel. \(_\square\)

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3D Geometry

3D Geometry is used to represent a point, a line, or a plane with reference to the x-axis, y-axis, and z-axis respectively. The three-dimensional geometry has all the concepts similar to the two-dimensional coordinate geometry.

Let us check more about the concepts of 3D geometry, the representation of a point, line, plane, with the help of examples, FAQs.

What Is 3D Geometry?

The 3d geometry helps in the representation of a line or a plane in a three-dimensional plane, using the x-axis, y-axis, z-axis. The coordinates of any point in three-dimensional geometry have three coordinates, (x, y, z).

The three-dimensional cartesian coordinate system consists of three axes, the x-axis, the y-axis, and the z-axis, which are mutually perpendicular to each other and have the same units of length across all three axes. Similar to the two-dimensional coordinate system, here also the point of intersection of these three axes is the origin O, and these axes divide the space into eight octants. Any point in 3D Geometry is represented with the coordinates (x, y, z).

Further the coordinates of a points in the eight octants are (+x,+y,+z), (-x,+y,+z), (+x,+y,-z), (-x,+y,-z), (+x,-y,+z), (-x,-y,+z), (+x,-y,-z), (-x,-y,-z).

Notation of a point in a cartesian coordinate system is a way of presenting a point for easy understanding and calculations. The points in a cartesian coordinate system are written in parentheses, and separated by a comma. The examples of a point in a three-dimensional frame is (2, 5, 4). The origin is denoted by the O and the coordinates of a point is denoted by the point (x, y, z). Here the last alphabets of the alphabetical series are taken or the first alphabets of the word is taken to represent the coordinates of a point.

A coordinate is an address, which helps to locate a point in space. For a three-dimensional frame, the coordinates of a point is (x, y, z). Here let us take note of these three important terms.

• Abscissa: It is the x value in the point (x, y, z) and is the distance of this point along the x-axis, from the origin
• Ordinate: It is the y value in the point (x, y, z) and is the perpendicular distance of the point from the x-axis, and is parallel to the y-axis.
• Applicate: In a three-dimensional frame the point is (x, y, z), and the z -coordinate of the point and is referred to as applicate.

Three Dimensional Geometry - Important Concepts

The 3D geometry makes use of the three coordinates to represent a point. The important concepts with reference to three-dimensional geometry are direction ratio, direction cosine, distance formula, midpoint formula, and section formula. The following are the important concepts of 3D Geometry.

Direction Ratios

The point A,(a, b, c) is represented as a vector with the position vector as \(\vec OA = a\vec i + b\vec j + c\vec k \) and has the direction ratios a, b, c. This ratio represents the vector line with reference to the x-axis, y-axis, and z-axis respectively. Further, these direction ratios also help to derive the direction cosines.

Direction Cosine

Direction Cosine gives the relation of a vector or a line in a three-dimensional space, with each of the three axes. The direction cosine is the cosine of the angle subtended by this line with the x-axis, y-axis, and z-axis respectively. If the angles subtended by the line with the three axes are α, β, and γ, then the direction cosines are Cosα, Cosβ, Cosγ respectively. The direction cosines for a vector \(\overrightarrow A = a \hat i + b \hat j + c \ \hat k\) is Cosα = \(\frac{a}{\sqrt {a^2 + b^2 + c^2}}\), Cosβ = \(\frac{b}{\sqrt {a^2 + b^2 + c^2}}\), Cosγ = \(\frac{c}{\sqrt {a^2 + b^2 + c^2}}\). The direction cosines are also represented by l, m, n, and we can prove that l 2 + m 2 + n 2 = 1.

Distance Formula

The distance between two points \((x_1, y_1, z_1)\) and \(x_2, y_2, z_2) \) is the shortest distance, and is equal to the square root of the summation of the square of the difference of the x coordinates, the y-coordinates, and the z-coordinates of the two given points. The formula for the distance between two points is as follows.

D = \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\)

Mid-Point Formula

The formula to find the midpoint of the line joining the points \((x_1, y_1, z_1)\) and \(x_2, y_2, z_2) \) is a new point, whose abscissa is the average of the x values of the two given points, and the ordinate is the average of the y values of the two given points. The midpoint lies on the line joining the two points and is located exactly between the two points.

\((x, y) =\left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}, \dfrac{z_1 + z_2}{2}\right)\)

Section Formula

The section formula is useful to find the coordinates of a point that divides the line segment joining the points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) in the ratio \(m : n\). The point dividing the given two points lies on the line joining the two points and is available either between the two points or on the line, beyond the two points.

\((x, y) = \left(\dfrac{mx_2 + nx_1}{m + n}, \dfrac{my_2 + ny_1}{m + n}, \dfrac{mz_2 + nz_1}{m + n}\right) \)

3D Geometry Representation Of A Point, Line, Plane

The three dimensional geometry is used for the representation of a point, line, or a plane. Let us check the different forms of representation of a point, line, and plane in three-dimensional geometry.

Representation of a point in 3D Geometry

The point in a three-dimensional geometry can be represented either in cartesian form or a vector form. The two forms of representation of the point in a 3D geometry are as follows.

Cartesian Form: The cartesian form of representation of any point in 3D geometry uses three coordinates with reference to the x-axis, y-axis, and z-axis respectively. The coordinates of any point in a 3D geometry is (x, y, z). The x value of the point is called the abscissa, the y value is called the ordinate, and the z value is called the applicate.

Vector Form: The vector form of representation of a point P is a position vector OP, and is written as \(\vec OP = x \vec i + y \vec j + z \vec k\), where \(\vec i\), \(\vec j\), \(\vec k\) are the unit vectors along the x-axis, y-axis, and z-axis respectively.

Representation Of A Line in 3D Geometry

The equation of a line in a three-dimensional cartesian system can be computed from the following two methods. The two methods of finding the equation of a line are as follows.

• The equation of a line passing through a point 'a' and parallel to a given vector 'b' is as follows. r = a + λb
• The equation of a line passing through two given points, a and b, can be represented as r = a + λ(b - a)

Representation Of A Plane in 3D Geometry

The equation of a plane in a cartesian coordinate system can be computed through different methods based on the available inputs values about the plane. The following are the four different expressions for the equation of a plane.

• Normal Form: Equation of a plane at a perpendicular distance d from the origin and having a unit normal vector \(\hat n \) is \(\overrightarrow r. \hat n\) = d.
• Perpendicular to a given Line and through a Point: The equation of a plane perpendicular to a given vector \(\overrightarrow N \), and passing through a point \(\overrightarrow a\) is \((\overrightarrow r - \overrightarrow a). \overrightarrow N = 0\)
• Through three Non Collinear Lines: The equation of a plane passing through three non collinear points \(\overrightarrow a\), \(\overrightarrow b\), and \(\overrightarrow c\), is \((\overrightarrow r - \overrightarrow a)[(\overrightarrow b - \overrightarrow a) × (\overrightarrow c - \overrightarrow a)] = 0\).
• Intersection of Two Planes: The equation of a plane passing through the intersection of two planes \(\overrightarrow r .\hat n_1 = d_1\), and \(\overrightarrow r.\hat n_2 = d_2 \), is \(\overrightarrow r(\overrightarrow n_1 + λ \overrightarrow n_2) = d_1 + λd_2\).

Related Topics

The following topics help in a better understanding of 3D Geometry.

• Cartesian Coordinate System
• Polar Coordinates
• Coordinate Plane
• Coordinate Geometry
• Internal Division

Examples on 3D Geometry

Example 1: Find the direction ratios and direction cosines of a point (4, 5, -2) in 3D geometry.

The given point is (4, 5, -2) is represented as a vector \(\vec a = 4\vec i + 5\vec j - 2\vec k\)

Here we have |a| = \(\sqrt{4^2 + 5^2 + (-2)^2} = \sqrt{16 + 25 + 4} = \sqrt{45} = 3\sqrt 5 \)

Direction Ratios's: (a, b, c) = (4, 5, -2)

Direction Cosines: \(\frac{a}{\sqrt {a^2 + b^2 + c^2}, \frac{b}{\sqrt {a^2 + b^2 + c^2}, \frac{c}{\sqrt {a^2 + b^2 + c^2}\) = \(\left( \dfrac{4}{3\sqrt5}, \dfrac{5}{3\sqrt5}, \dfrac{-2}{3\sqrt5}\right)\).

Example 2: What is the equation of a line in three-dimensional geometry, passing through the points (1, 3, -2), and (-1, 4, 3)?

The given point are (1, 3, -2), and (-1, 4, 3).

The equation of a line passing through the two points is r = a + λ(b - a).

\(\vec r = (1\vec i + 3 \vec j -2 \vec k) + λ((-1\vec i + 4\vec j + 3\vec k) - (1\vec i + 3\vec j - 2 \vec k))\)

\(\vec r = (1\vec i + 3 \vec j -2 \vec k) + λ(-2\vec i + 1\vec j + 5\vec k)\)

\(x\vec i + y\vec j + z\vec k = (1 - 2λ)\vec i + (3 +λ)\vec j +( -2 + 5λ)\vec k\)

\((x - 1)\vec i + (y - 3)\vec j + (z + 2)\vec k = -2λ\vec i +λ\vec j + 5λ\vec k\)

\(\dfrac{(x - 1)}{-2} = \dfrac{( y - 3)}{1} = \dfrac{(z + 2)}{5}\)

Therefore, the equation of the line passing through the two point is \(\dfrac{(x - 1)}{-2} = \dfrac{( y - 3)}{1} = \dfrac{(z + 2)}{5}\).​​​​​​

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Practice Questions on 3D Geometry

Faqs on 3d geometry.

The 3d geometry helps in the representation of a line or a plane in a three-dimensional plane, using the x-axis, y-axis, z-axis. The coordinates of any point in three-dimensional geometry have three coordinates, (x, y, z). Similar to the two-dimensional coordinate system, here also the point of intersection of these three axes is the origin O, and these axes divide the space into eight octants. The x coordinate of a point is called abscissa, the y coordinate is called the ordinate, and the z coordinate is called applicate.

How Do You Represent A Point In 3D Geometry?

A point in a three-dimensional geometry can be represented either in cartesian form or a vector form. The two forms of representation of the point in a 3D geometry are as follows.

• The cartesian form of representation of any point in 3D geometry is (x, y, z) and is with reference to the x-axis, y-axis, and z-axis respectively. The x value of the point is called the abscissa, the y value is called the ordinate, and the z value is called the applicate.
• The vector form of representation of a point P is a position vector OP, and is written as \(\vec OP = x \vec i + y \vec j + z \vec k\), where \(\vec i\), \(\vec j\), \(\vec k\) are the unit vectors along the x-axis, y-axis, and z-axis respectively.

How Do You Represent A Line in 3D Geometry?

A line in a three-dimensional cartesian system can be computed from the following two equations. The two methods of finding the equation of a line are as follows.

How Do You Represent A Plane In 3D Geometry?

The equation of a plane in a cartesian coordinate system can be computed through four different methods.

• The equation of a plane perpendicular to a given vector \(\overrightarrow N \), and passing through a point \(\overrightarrow a\) is \((\overrightarrow r - \overrightarrow a). \overrightarrow N = 0\)
• The equation of a plane passing through three non collinear points \(\overrightarrow a\), \(\overrightarrow b\), and \(\overrightarrow c\), is \((\overrightarrow r - \overrightarrow a)[(\overrightarrow b - \overrightarrow a) × (\overrightarrow c - \overrightarrow a)] = 0\).
• The equation of a plane passing through the intersection of two planes \(\overrightarrow r .\hat n_1 = d_1\), and \(\overrightarrow r.\hat n_2 = d_2 \), is \(\overrightarrow r(\overrightarrow n_1 + λ \overrightarrow n_2) = d_1 + λd_2\).

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Equation of a Line in 3D

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.

Table of Content

What is Equation of a Line?

Equation of line in 3d, cartesian form of equation of line in 3d, vector form of equation of line in 3d, 3d lines formulas, solved examples on equation of a line in 3d.

The equation of a line is an algebraic way to express a line in terms of the coordinates of the points it joins. The equation of a line will always be a linear equation .

If we try to plot the points obtained from a linear equation it will be a straight line . The standard equation of a line is given as:

ax + by + c = 0

• a and b are Coefficients of x and y
• c is Constant Term

Other forms of the equation of line are mentioned below:

Now we will learn the equation of the line in 3D.

The equation of straight line in 3D requires two points which are located in space. The location of each point is given using three coordinates expressed as (x, y, z).

The 3D Equation of a Line is given in two formats, cartesian form and vector form . In this article we will learn the Equation of a line in 3D in both Cartesian and Vector Form and also learn to derive the equation. The different cases for equation of line are listed below:

• Line Passing through two points
• Line Passing through a given point and Parallel to a given Vector

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(x 1 , y 1 , z 1 ) and B(x 2 , y 2 , z 2 ).

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 = (x 2 – x 1 ), m = (y 2 – y 1 ), n = (z 2 – z 1 ); Here l, m, n are the DR’s.
• Step 2: Choose either of the two given points say, we choose (x 1 , y 1 , z 1 ).
• Step 3: Write the required equation of the straight line passing through the points (x 1 , y 1 , z 1 ) and (x 2 , y 2 , z 2 ).
• Step 4: The 3D equation of straight line in cartesian form is given as L : (x – x 1 )/l = (y – y 1 )/m = (z – z 1 )/n = (x – x 1 )/(x 2 – x 1 ) = (y – y 1 )/(y 2 – y 1 ) = (z – z 1 )/(z 2 – z 1 )

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

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(x 1 , y 1 , z 1 ) 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 (x 1 + λa), (y 1 + λb), (z 1 + 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 = x 1 + λa ⇒ λ = x – x 1 /a

y = y 1 + λb ⇒ λ = y – y 1 /b

z = z 1 + λc ⇒ λ = z – z 1 /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

The Equation of line passing through a point and parallel to a vector is given as (x – x 1 )/a = (y – y 1 )/b = (z – z 1 )/c From the question we have, x 1 = 2, y 1 = 1, z 1 = 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

Vector Form of the Equation of Line in 3D is given using a vector equation that involves the position vector of the points. In this heading, we will obtain the 3D Equation of the line in vector form for two cases.

Case 1: 3D Equation of Line Passing through Two Points in Vector Form

Let us assume we have two points A and B whose position vector is given as [Tex]\vec a[/Tex] and [Tex]\vec b[/Tex] .

Then the vector equation of the Line L is given as

[Tex]\vec l = \vec a + \lambda (\vec b – \vec a) [/Tex] where [Tex](\vec b – \vec a)[/Tex] is the distance between two points and λ is the parameter that lies on the line.

Derivation of 3D Equation of Line Passing through Two Points in Vector Form

Suppose we have two points A and B whose position vector is given as [Tex]\vec a[/Tex] and [Tex]\vec b[/Tex] . Now we know that a line is the distance between any two points. Hence, we need to subtract the two position vectors to obtain the distance.

⇒ [Tex]\vec d = \vec b – \vec a [/Tex]

Now we know that any point on this line will be given as the sum of position vector [Tex]\vec a \space or \space \vec b [/Tex] with the product of the parameter λ and the position vector of the distance between two points i.e. [Tex]\vec d [/Tex]

Hence, the equation of the line in the vector form will be [Tex]\vec l = \vec a + \lambda (\vec b – \vec a)[/Tex] or [Tex]\vec l = \vec b + \lambda (\vec a – \vec b)[/Tex]

Example: Find the vector equation of a line in 3D that passes through two points whose position vectors are given as 2i + j – k and 3i + 4j + k

Given that the two position vectors are given as 2i + j – k and 3i + 4j + k Distance d = (3i + 4j + k) – (2i + j -k) = i + 3j + 2k We know that equation of the line is given as [Tex]\vec l = \vec a + \lambda (\vec b – \vec a) [/Tex] Hence, the equation of the line will be [Tex]\vec l[/Tex] = 2i + j – k + λ(i + 3j + 2k)

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

Let’s say we have a point P whose position vector is given as [Tex]\vec p[/Tex] . Let this line be parallel to another line whose position vector is given as [Tex]\vec d [/Tex] .

Then the vector equation of the line ‘l’ is given as

[Tex]\vec l = \vec p + \lambda \vec d[/Tex] where λ is the parameter that lies on the line.

Derivation of the Vector Form of 3D Equation of Line Passing through a Point and Parallel to a Vector

Consider a point P whose position vector is given as [Tex]\vec p[/Tex] . Now let us assume this line is parallel to a vector [Tex]\vec d[/Tex] then, the equation of the line will be [Tex]\vec l = \lambda \vec d[/Tex] . Now since the line also passes through point P, then as we move away from Point P in either direction on the line then the position vector of the point will be in the form of [Tex]\vec p + \lambda \vec d [/Tex] . Hence, the equation of the line will be [Tex]\vec l = \vec p + \lambda \vec d[/Tex] where λ is the parameter that lies on the line.

Example: Find the Vector Form of the Equation of the line passing through the point (-1, 3, 2) and parallel to a vector 5i + 7j – 3k.

We know that the vector form of the equation of a line passing through a point and parallel to a vector is given as [Tex]\vec l = \vec p + \lambda \vec d[/Tex] Given that the point is (-1, 3, 2), hence the position vector of the point will be -i + 3j + 2k and the given vector is 5i + 7j – 3k. Therefore, the required equation of the line will be [Tex]\vec l [/Tex] = (-i + 3j + 2k) + λ(5i + 7j – 3k).

Similar Reads

• Equation of a Straight Line
• Tangent and Normal
• Slope of Line

Practice equations of line in 3D with these solved practice questions.

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

[Tex]{\vec {p}}[/Tex]  = (4 i + 6 j + 12 k ) – (2 i + 3 j + 5 k ) [Tex]{\vec {p}}[/Tex]  = (2 i + 3 j + 7 k ) ; Here  [Tex]{\vec {p}}[/Tex]  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 :  [Tex]{\vec {r}}[/Tex]  = (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

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

[Tex]{\vec {p}}[/Tex]  = (5 i + 3 j + 7 k) – (2 i + j – 3 k) [Tex]{\vec {p}}[/Tex]  = (3 i + 2 j + 10 k) ; Here  [Tex]{\vec {p}}[/Tex]  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:  [Tex]{\vec {r}}[/Tex]  = (2 i + j – 3 k) + t . (3 i + 2 j + 10k)

Example 4: 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

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 5: 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

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 Line in 3D – FAQs

What is equation of a line in 3d.

The Equation of a Line in 3D is given as (x – x 1 )/(x 2 – x 1 ) = (y – y 1 )/(y 2 – y 1 ) = (z – z 1 )/(z 2 – z 1 )

What is Cartesian Form of the Equation of a Line in 3D?

Cartesian Form of the Equation of Line in 3D is given for two cases Case 1: When the line passes through two points: [Tex]{\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] Case 2: When a line passes through one point and is parallel to a vector: [Tex]{\frac{x – x_1} a = \frac{y – y_1} b = \frac{z -z_1} c} [/Tex]

What is Vector Form of Equation of a Line in 3D?

Vector form of the equation of a line in 3D is given for two cases: Case 1: Line Passing through two Points: [Tex]\vec l = \vec a + \lambda (\vec b – \vec a) [/Tex] Case 2: Line Pass Through a Point and Parallel to a vector: [Tex]\vec l = \vec p + \lambda \vec d [/Tex]

What is Slope Point Equation of a Line?

Slope Point Equation of a line is given as y = mx + C where m is the slope

What is the Standard Equation of a Line?

Standard equation of a line is ax + by + c = 0

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5.3 Representations of a Line in Two and Three Dimensions

Two points P 1 and P 2 on a line, L, determine L.

L can be described parametrically as the set of points with coordinates those of P 1 + s * (P 2 - P 1 ) for some number s.

( P 2 - P 1 ) is a vector which points in the direction of L.

We do this out explicitly� L consists of the points obeying

x� = P 1x � + s * (P 2x - P 1x )

y� = P 1y � + s * (P 2y - P 1y )

and the equation for L is

(P 2y - P 1y ) x - (P 2x - P 1x )y = (P 2y - P 1y )P 1x - (P 2x - P 1x )P 1y

which when solved for y is

The constant C is called the y-intercept of the line L. It is the value of y on L where L meets the y axis.

In three dimensions, a line is determined by two equations. You can describe it as above parametrically (though now all points and vectors have three components), but you must find two vectors normal to (P 2 - P 1 ) to find equations that characterize it. You have infinite choice in doing so, but two convenient choices are

and you can require that the dot product of each of these with r (recall r = (x, y, z)) is what it is with P 1 . This gives you two equations which together determine the line.

To find a point on a line from equations you fix one coordinate arbitrarily and solve the two equations for the other two coordinates.

The following applet allows you two enter two arbitrary points. It then shows you the line in 3 space that your points determine, and the parametric representation of that line.

Exercises 5.1 Find two equations for the default line in this applet. Then choose two random points, and find two equations for the line they lie on.

3D Line Mathematics

Introduction

In my previous blog post “2D Line Mathematics Using Homogeneous Coordinates” , we have discussed the 2D line mathematics.

In this blog post, I would like to discuss the 3D line mathematics from a similar perspective.

3D Point Representations

Inhomogeneous coordinates.

The inhomogeneous coordinates for a 3D point are just ordinary two-value Cartesian coordinates.

$$ \mathbf{x} = (x, y, z) $$

Augmented Coordinates

The augmented coordinates for a 3D point are just the 3D inhomogeneous coordinates with an additional constant $1$.

$$ \bar{\mathbf{x}} = (x, y, z, 1) $$

Homogeneous Coordinates

The homogeneous coordinates are just the augmented coordinates scaled by some value $\tilde{w}$.

$$ \begin{align} \tilde{\mathbf{x}} &= \tilde{w} \bar{\mathbf{x}} \\ &= \tilde{w} (x, y, z, 1) \\ &= (\tilde{w}x, \tilde{w}y, \tilde{w}z, \tilde{w}) \\ &= (\tilde{x}, \tilde{y}, \tilde{z}, \tilde{w}) \\ \end{align} $$

where $\tilde{w} \in \mathbb{R}$.

When $\tilde{w} = 0$, $\tilde{\mathbf{x}}$ is called ideal point and do not have the corresponding inhomogeneous coordinates.

3D Line Representation

Given two points $(\mathbf{p}, \mathbf{q})$ represented using inhomogeneous coordinates, on a 3D line, by vector algebra, we could represent any point $\mathbf{r}$ on the 3D line as a linear combination of the two points.

$$ \begin{align} \mathbf{r} &= \mathbf{p} + \lambda (\mathbf{q} - \mathbf{p}) \\ &= (1 - \lambda)\mathbf{p} + \lambda \mathbf{q} \end{align} $$

Because any point on the 3D line can be represented using the expression above, the 3D line could also be represented using the exactly same expression.

Given one point $\mathbf{p}$ on a 3D line and its direction vector $\mathbf{d}$ represented using inhomogeneous coordinates, by vector algebra, we could represent any point $\mathbf{r}$ on the 3D line

$$ \begin{align} \mathbf{r} &= \mathbf{p} + \lambda \mathbf{d} \\ \end{align} $$

More concretely, $\mathbf{d}$ is just the $\mathbf{q} - \mathbf{p}$ in the previous expression using two points.

In terms of 3D line representation using homogeneous coordinates representation, we could derive it from the 3D line representation using inhomogeneous coordinates representation

$$ \begin{align} (\mathbf{r}, 1) &= \frac{1}{w_r} \tilde{\mathbf{r}} \\ &= \Big((1 - \lambda)\mathbf{p} + \lambda \mathbf{q}, 1\Big) \\ &= \Big((1 - \lambda) \frac{1}{w_p} \tilde{\mathbf{p}} [:2] + \lambda \frac{1}{w_q} \tilde{\mathbf{q}} [:2], 1\Big) \\ &= (1 - \lambda) \frac{1}{w_p} \tilde{\mathbf{p}} + \lambda \frac{1}{w_q} \tilde{\mathbf{q}} \\ \end{align} $$

$$ \begin{align} \tilde{\mathbf{r}} &= (1 - \lambda) \frac{w_r}{w_p} \tilde{\mathbf{p}} + \lambda \frac{w_r}{w_q} \tilde{\mathbf{q}} \\ \end{align} $$

• Degrees of Freedom

It is not difficult to find that in order to determine a 3D line using the above two-point representations, we need 6 parameters. However, in fact, 3D line only has 4 degrees of freedom (DOF), and we only need 4 parameters to determine a 3D line explicitly.

Any 3D line is tangent to a unique sphere centered at the origin with a specific direction. The sphere could be determined by the radius $r$. The point $\mathbf{m}$ where the 3D line and the sphere meet could be determined by two additional angles $(\theta, \phi)$, $\mathbf{m} = (r \cos \theta \cos \phi, r \sin \theta \cos \phi, r \sin \phi)$. Finally, because the plane where $\mathbf{m}$ sits and is tangent to the sphere is unique, the orientation of the 3D line will be determined by one additional angle $\omega$. Taken together, a 3D line only has 4 degree of freedom, and a 3D is determined uniquely by 4 parameters.

However, it does not necessary mean the 6-DOF 3D line representation is inferior to the 4-DOF 3D line representation. It really depends on the problem we are going to solve. In many scenarios, the 6-DOF 3D line representation is very convenient for problem solving.

Another way to think about the 3D line DOF is using the two-point 6-DOF representations and the translational invariance. Because

$$ \begin{align} \mathbf{r} &= (r_x, r_y, r_z) \\ &= \mathbf{p} + \lambda \mathbf{d} \\ &= (p_x, p_y, p_z) + \lambda (d_x, d_y, d_z) \\ \end{align} $$

We have the following equation.

$$ \lambda = \frac{r_x - p_x}{d_x} = \frac{r_y - p_y}{d_y} = \frac{r_z - p_z}{d_z} $$

Because the translational invariance, the value of $(d_x, d_y, d_z)$ cannot change freely. If $d_x$ is determined, $d_y$ and $d_z$ are determined as well. For example, suppose $\mathbf{p} = (p_x, p_y, p_z) = (1, 2, 3)$, and we find a point $\mathbf{r}$ on the 3D line $\mathbf{r} = (r_x, r_y, r_z) = (5, 4, 9)$. So the equation becomes

$$ \begin{align} \lambda &= \frac{r_x - p_x}{d_x} = \frac{r_y - p_y}{d_y} = \frac{r_z - p_z}{d_z} \\ &= \frac{5 - 1}{d_x} = \frac{4 - 2}{d_y} = \frac{9 - 3}{d_z} \\ &= \frac{4}{d_x} = \frac{2}{d_y} = \frac{6}{d_z} \\ \end{align} $$

We are allowed to choose $d_x$ freely, say $d_x = 2$. But once $d_x$ is determined, we cannot choose $d_y$ and $d_z$ freely. In this case, when $d_x = 2$, we must have $d_y = 1$ and $d_z = 3$. Therefore, we only have 1 DOF for $(d_x, d_y, d_z)$, and the DOF for a 3D line is $3 + 1 = 4$.

• 2D Line Mathematics Using Homogeneous Coordinates
• 4-DOF of a 3D Line - Stack Exchange

https://leimao.github.io/blog/3D-Line-Mathematics/

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• 1 Introduction
• 2.1 Inhomogeneous Coordinates
• 2.2 Augmented Coordinates
• 2.3 Homogeneous Coordinates
• 3 3D Line Representation
• 4 Degrees of Freedom
• 5 References

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Equation of a line passing through two points in 3d

This online calculator finds equation of a line in parametrical and symmetrical forms given coordinates of two points on the line

You can use this calculator to solve the problems where you need to find the line equation that passes through the two points with given coordinates. Enter coordinates of the first and second points, and the calculator shows both parametric and symmetric line equations. As usual, you can find the theory and formulas below the calculator.

First point

Second point, finding equation of a line in 3d.

A point and a directional vector determine a line in 3D. You can find the directional vector by subtracting the second point's coordinates from the first point's coordinates.

From this, we can get the parametric equations of the line.

If we solve each of the parametric equations for t and then set them equal, we will get symmetric equations of the line.

Similar calculators

• • Equation of a line given two points
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• Organic Chemistry

Organic Compounds: Structural Representation

What is an organic compound.

The covalently bonded carbon-containing compounds called organic compounds are substantial for life on earth.

The specific branch of science that deals with organic compounds are called organic chemistry. Organic chemistry encompasses every characteristic feature of an organic compound, its properties, and behaviour under different conditions. Here we will discuss the structural representation of organic molecules.

Table of Contents

• Structural Representation of Organic Compounds
• Complete Structural Formula
• Condensed Structural Formula
• Bond Line Structural Formula
• 3-D Representation of Organic Compounds

Structural Representation of Organic Compounds:

Though an organic compound has only one chemical formula, structurally it can be depicted in numerous ways. The three ways by which a compound can be represented are; 1. Complete Structural Formula 2. Condensed Structural Formula 3. Bond line Structural Formula

Complete Structural Formula:

The Lewis dot structure is considered the complete structural formula. In Lewis’s structure, the covalent bonds in the compound are denoted by a dash (―). This helps to emphasize the number of bonds formed by the atoms present in the molecule. Every single bond, a double bond, and a triple bond are represented by one dash, double dash, and triple dash respectively. It illustrates every single bond formed between every atom in the compound, thus called the complete structural formula.

For example, the complete structural formula for ethane, ethene and ethyne is shown as below:

Condensed Structural Formula:

In condensed structural formulas, the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens.

Bond Line Structural Formula:

A bond line structural formula is another way of structural representation of organic compounds.  Here, every bond is represented as a line in a zigzag manner. If not specified, every terminal is assumed to be a methyl (-CH 3 ) group.

3-D Representation of Organic Compounds:

The organic compounds which were represented using structural formulas in the previous section are three-dimensional compounds. In order to draw the 3-D structure of an organic compound, we can use wedge-dash representation.

In wedge-dash representation, the bond that protrudes out of the plane of paper towards the viewer is denoted by a solid wedge while that projects away from the viewer or into the plane of the paper is denoted by a dashed wedge and the bond in the plane of the paper is represented by a line. For example, the wedge-dash representation for methane molecule is shown as below:

To learn more about the structural representation of organic compounds, download BYJU’S -The Learning App.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Representation of a line in 3D projective space

From Hartley and Zisserman Multiple View Geomerty p. 69, regarding representations of a line in $\mathbb{P^3}$:

"Suppose $\mathbf{A}, \mathbf{B}$ are two (non-coincident) space points. Then the line jointing these points is represented by the span of the row space of the 2x4 matrix $W$ composed of $\mathbf{A^T}$ and $\mathbf{B^T}$ as rows:

$$ W^T = \begin{bmatrix} \mathbf{A^T}\\ \mathbf{B^T}\\ \end{bmatrix} $$

Then: The span of $W^T$ is the pencil of points $\lambda\mathbf{A} + \mu\mathbf{B}$ on the line.

Can someone help me understand why a line seems to be a two-dimensional space with two free parameters $\lambda$ and $\mu$? I assume it is related to the fact that each point is in its homogeneous representation.

• projective-geometry
• projective-space

• 2 $\begingroup$ Exactly, it's homogeneous coordinates. Just like a projective point seems to be a one-dimensional space with one free parameter. $\endgroup$ –  Angina Seng Commented Jun 1, 2017 at 20:38
• $\begingroup$ Just as a point in the projective space $\mathbb{RP}^3$ corresponds to a line through the origin in $\mathbb R^4$, a line in $\mathbb{RP}^3$ corresponds to a plane through the origin in $\mathbb R^4$. $\endgroup$ –  amd Commented Jun 1, 2017 at 22:46

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