What is a Straight Line in 3D Space?

A straight line in 3D space is one of the fundamental concepts in geometry. Unlike a line in 2D space that only requires two coordinates (x and y), a line in 3D space requires three coordinates (x, y, and z) to define its position. Let’s delve into the specifics of how a straight line is represented, its equations, and its properties.

Parametric Equations of a Line

In 3D space, a straight line can be represented using parametric equations. These equations express the coordinates of any point on the line as functions of a single parameter, usually denoted as $t$

General Form

The general form of the parametric equations of a line is:

$begin{cases}
x = x_0 + at \
y = y_0 + bt \
z = z_0 + ct
end{cases}$

Here, $(x_0, y_0, z_0)$ is a point on the line, and $(a, b, c)$ is the direction vector of the line. The parameter $t$ varies over all real numbers.

Example

Consider a point $(1, 2, 3)$ and a direction vector $(4, 5, 6)$. The parametric equations of the line passing through this point and in the direction of this vector are:

$begin{cases}
x = 1 + 4t \
y = 2 + 5t \
z = 3 + 6t
end{cases}$

As $t$ changes, these equations provide the coordinates of different points on the line.

Symmetric Equations of a Line

Another way to represent a line in 3D space is using symmetric equations. These equations eliminate the parameter $t$ and directly relate the coordinates $x$, $y$, and $z$

General Form

The symmetric form of the equations of a line is:

$frac{x – x_0}{a} = frac{y – y_0}{b} = frac{z – z_0}{c}$

Example

Using the same point $(1, 2, 3)$ and direction vector $(4, 5, 6)$, the symmetric equations of the line are:

$frac{x – 1}{4} = frac{y – 2}{5} = frac{z – 3}{6}$

Vector Form of a Line

A line in 3D space can also be represented in vector form. This form uses vectors to express the position of any point on the line.

General Form

The vector form of a line is:

$mathbf{r} = mathbf{r_0} + tmathbf{d}$

Here, $mathbf{r}$ is the position vector of any point on the line, $mathbf{r_0}$ is the position vector of a specific point on the line, and $mathbf{d}$ is the direction vector.

Example

Using the point $(1, 2, 3)$ and direction vector $(4, 5, 6)$, the vector form of the line is:

$mathbf{r} = begin{pmatrix} 1 \ 2 \ 3 end{pmatrix} + t begin{pmatrix} 4 \ 5 \ 6 end{pmatrix}$

Intersection of Two Lines

Finding the intersection of two lines in 3D space can be more complex than in 2D space. Two lines may intersect, be skew (not intersecting and not parallel), or be parallel.

Method to Find Intersection

To find the intersection, set the parametric equations of the two lines equal to each other and solve for the parameters.

Example

Consider two lines with parametric equations:

Line 1:

$begin{cases}
x = 1 + 4t \
y = 2 + 5t \
z = 3 + 6t
end{cases}$

Line 2:

$begin{cases}
x = 2 + 7s \
y = 3 + 8s \
z = 4 + 9s
end{cases}$

Set the equations equal to each other:

$begin{cases}
1 + 4t = 2 + 7s \
2 + 5t = 3 + 8s \
3 + 6t = 4 + 9s
end{cases}$

Solve these equations to find $t$ and $s$. If a solution exists, the lines intersect at that point.

Distance Between Two Skew Lines

To find the distance between two skew lines, use the following formula:

$D = frac{|mathbf{d_1} times mathbf{d_2} cdot (mathbf{r_2} – mathbf{r_1})|}{|mathbf{d_1} times mathbf{d_2}|}$

Here, $mathbf{d_1}$ and $mathbf{d_2}$ are the direction vectors of the lines, and $mathbf{r_1}$ and $mathbf{r_2}$ are points on the lines.

Conclusion

Understanding straight lines in 3D space involves grasping their parametric, symmetric, and vector forms. These representations help solve various geometric problems, such as finding intersections and distances between lines. Mastering these concepts is crucial for advanced studies in mathematics, physics, and engineering.

Citations

  1. 1. Khan Academy – Parametric Equations
  2. 2. MIT OpenCourseWare – 3D Geometry
  3. 3. Paul’s Online Math Notes – 3D Space

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ