How to Find a Line Parallel to Two Planes?

Finding a line parallel to two planes involves understanding both the geometric and algebraic properties of planes and lines in three-dimensional space.

Step-by-Step Process

  1. Identify the Normal Vectors
    Each plane in 3D space can be represented by an equation of the form:
    $Ax + By + Cz = D$
    The coefficients $A$, $B$, and $C$ form the normal vector $textbf{n} = (A, B, C)$ of the plane. Let’s say we have two planes with normal vectors $textbf{n}_1 = (A_1, B_1, C_1)$ and $textbf{n}_2 = (A_2, B_2, C_2)$

  1. Find a Direction Vector Parallel to Both Planes
    A line parallel to both planes must be perpendicular to both normal vectors. To find such a direction vector, we can use the cross product of the two normal vectors:
    $textbf{d} = textbf{n}_1 times textbf{n}_2$
    The cross product of two vectors $textbf{a} = (a_1, a_2, a_3)$ and $textbf{b} = (b_1, b_2, b_3)$ is given by:
    $textbf{a} times textbf{b} = (a_2b_3 – a_3b_2, a_3b_1 – a_1b_3, a_1b_2 – a_2b_1)$

  1. Define the Line Equation
    Once we have the direction vector $textbf{d}$, we need a point through which the line passes. Any point will work, but often the origin or a point of intersection of the planes (if they intersect) is used. The parametric form of the line is:
    $textbf{r}(t) = textbf{r}_0 + ttextbf{d}$
    where $textbf{r}_0$ is a point on the line, $textbf{d}$ is the direction vector, and $t$ is a parameter.

Example

Consider two planes with equations:

  1. $2x + 3y – z = 5$
  2. $x – y + 4z = 2$

  1. Normal Vectors
    The normal vectors are:
    $textbf{n}_1 = (2, 3, -1)$
    $textbf{n}_2 = (1, -1, 4)$

  1. Cross Product
    The direction vector $textbf{d} = textbf{n}_1 times textbf{n}_2$ is:
    $textbf{d} = (3*4 – (-1)(-1), (-1)(1) – 2*4, 2*(-1) – 3*1)$
    $textbf{d} = (12 – 1, -1 – 8, -2 – 3)$
    $textbf{d} = (11, -9, -5)$

  1. Line Equation
    Using the origin (0, 0, 0) as a point on the line, the parametric equation is:
    $textbf{r}(t) = (0, 0, 0) + t(11, -9, -5)$
    $textbf{r}(t) = (11t, -9t, -5t)$

Conclusion

By following these steps, you can find a line parallel to two planes. This line will have a direction vector that is the cross product of the normal vectors of the planes, ensuring it is parallel to both planes.

Citations

  1. 1. Khan Academy – Parallel and Perpendicular Planes
  2. 2. MathWorld – Plane
  3. 3. Paul’s Online Math Notes – Equations of Planes

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