Straight Line in Space

A straight line in space can be defined as the intersection of two planes. Therefore it is determined by a system of equations of the form

\[\left\{ \begin{aligned} A_1x + B_1y + C_1z + D_1 = 0 \\ A_2x + B_2y + C_2z + D_2 = 0 \end{aligned} \right.\]

The intersection of planes is a straight line if and only if these planes are not parallel, which means that at least one of the determinants is not equal to zero:

\[\left| {\begin{array}{*{20}{c}} {B_1} & {C_1}\\ {B_2} & {C_2} \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {C_1} & {A_1}\\ {C_2} & {A_2} \end{array}} \right|^2 + \left| {\begin{array}{*{20}{c}} {A_1} & {B_1}\\ {A_2} & {B_2} \end{array}} \right|^2 \ne 0.\]

Canonical Equation of a Line in Space

Let a straight line be given and a non-zero vector s lying on the given line or parallel to it. Vector s is called the direction vector of this line. Let us derive the equation of a straight line passing through a given point P₁(x₁, y₁, z₁) and having a given direction vector s(a, b, c).

Let P(x, y, z) be an arbitrary point. It lies on the straight line if and only if the coordinates of the vector P₁P are proportional to the coordinates of the vector s, that is

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

This triple equation is called the canonical equation of the straight line in space or the equation in point-direction form.

Point direction form of the equation of a line — Figure 1.

Two-Point Form of the Equation of a Line

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

Equation of a Straight Line in Parametric Form

\[\left\{ \begin{aligned} x &= {x_1} + t\cos\alpha \\ y &= {y_1} + t\cos\beta \\ z &= {z_1} + t\cos\gamma \end{aligned} \right.\]

where the point P₁(x₁, y₁, z₁) lies on the line and cos α, cos β, cos γ are the direction cosines of the direction vector of the line, the parameter t is any real number.

Equation of a straight line in parametric form — Figure 3.

Solved Problems

Example 1.

Convert the equation of the following line to canonical form \[\left\{ \begin{aligned} x - 2y + 3z - 4 = 0 \\ 3x + 2y - 5z - 4 = 0 \end{aligned} \right.\]

Solution.

Let's find the coordinates of some point \(M_0\left({x_0,y_0,z_0}\right)\) belonging to the line. Let, for example, \(x_0 = 0.\) Find \(y_0, z_0\) from the system

\[\left\{ \begin{aligned} - 2y_0 + 3z_0 - 4 = 0 \\ 2y_0 - 5z_0 - 4 = 0 \end{aligned} \right..\]

Adding both equations we get

\[-2z_0 - 8 = 0, \Rightarrow z_0 = -4.\]

Substitute \(z_0\) into the first equation:

\[-2y_0 + 3 \cdot \left(-4\right) - 4 = 0, \Rightarrow -2y_0 - 16 = 0, \Rightarrow y_0 = -8.\]

Point \(M_0\left({0,-8,-4}\right)\) of the straight line is found.

Now we determine the direction vector of the line \(\mathbf{a}.\) This line is defined by the intersection of planes. Therefore it is perpendicular to each of the normal vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}.\) As \(\mathbf{a},\) you can take any vector perpendicular to vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}.\) For example, let's take their cross product:

\[{\mathbf{a} = \mathbf{n_1} \times \mathbf{n_2}}.\]

The coordinates of vectors \(\mathbf{n_1}\) and \(\mathbf{n_2}\) are known:

\[\mathbf{n_1}\left({1,-2,-3}\right),\;\;\mathbf{n_2}\left({3,2,-5}\right).\]

Therefore

\[{\mathbf{a} = \mathbf{n_1} \times \mathbf{n_2}} = \left| {\begin{array}{*{20}{r}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 1 & -2 & 3\\ 3 & 2 & -5 \end{array}} \right| = \left| {\begin{array}{*{20}{r}} -2 & 3\\ 2 & -5 \end{array}} \right|\mathbf{i} - \left| {\begin{array}{*{20}{r}} 1 & 3\\ 3 & -5 \end{array}} \right|\mathbf{j} + \left| {\begin{array}{*{20}{r}} 1 & -2\\ 3 & 2 \end{array}} \right|\mathbf{k} = \left({10-6}\right)\mathbf{i} - \left({-5-9}\right)\mathbf{j} + \left({2+6}\right)\mathbf{k} = 4\mathbf{i} + 14\mathbf{j} + 8\mathbf{k}.\]

The coefficients are all even and can be divided by 2. So the directing vector of the line has coordinates \({\mathbf{a}\left({2,7,4}\right)}.\) We obtain the following canonical equation of the straight line:

\[\frac{x}{2} = \frac{y + 8}{7} = \frac{z + 4}{4}.\]