Lines and Planes in Space (3D)

Nerd Cafe

PART 1: Vectors in 3D

A. Position Vector

A position vector points from the origin to a point in space. If a point P(x, y, z), then the position vector is:

$$\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$$

B. Vector Between Two Points

If A(x₁,y₁,z₁) and B(x₂,y₂,z₂):

$$\overrightarrow{AB}=(x_{b}-x_{a},y_{b}-y_{a},z_{b}-z_{a})$$

C. Python Example:

import numpy as np

A = np.array([1, 2, 3])
B = np.array([4, 0, 5])

AB = B - A
print("Vector AB:", AB)

D. Output:

Vector AB: [ 3 -2  2]

PART 2: Line in Space

A. Vector Equation of a Line

A line passing through point

$$\overrightarrow{r_{0}}=(x_{0},y_{0},z_{0})$$

with direction vector

$$\overrightarrow{v}=(a,b,c)$$

is:

$$\overrightarrow{r(t)}=\overrightarrow{r_{0}}+t\overrightarrow{v}$$

This gives:

$$\begin{matrix} x=x_{0}+at \\ y=y_{0}+bt \\ z=z_{0}+ct \end{matrix}$$

B. Python Example:

import matplotlib.pyplot as plt

# Line: point + t * direction
r0 = np.array([1, 2, 1])
v = np.array([2, 1, 3])
t = np.linspace(-5, 5, 100)

line = r0[:, None] + v[:, None] * t

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.plot(line[0], line[1], line[2], label='Line in 3D')
ax.scatter(*r0, color='red', label='Point r0')
ax.quiver(*r0, *v, length=1, color='green', label='Direction vector')
ax.legend()
plt.show()

C. Output:

PART 3: Plane in Space

A. General Equation of a Plane

Given a point

$$P_{0}=(x_{0},y_{0},z_{0})$$

and normal vector

$$\overrightarrow{n}=(a,b,c)$$

then:

$$a(x-x_{0})+b(y-y_{0})+c(z-z_{0})=0$$

Which simplifies to:

$$ax+by+cz=d$$

where:

$$d=ax_{0}+by_{0}+cz_{0}$$

B. Python Example:

from sympy import symbols, Eq, solve

x, y, z = symbols('x y z')
point = np.array([1, 2, 3])
normal = np.array([2, -1, 4])

# Plane equation: a(x-x0) + b(y-y0) + c(z-z0) = 0
a, b, c = normal
x0, y0, z0 = point

plane_eq = Eq(a*(x - x0) + b*(y - y0) + c*(z - z0), 0)
print("Plane Equation:", plane_eq)

C. Output:

Plane Equation: Eq(2*x - y + 4*z - 12, 0)

PART 4: Angle Between Two Lines

Given direction vectors v₁,v₂​​:

$$cos\theta=\frac{\overrightarrow{v_{1}}.\overrightarrow{v_{2}}}{|\overrightarrow{v_{1}}||\overrightarrow{v_{2}}|}$$

A. Python Example:

v1 = np.array([1, 2, 3])
v2 = np.array([4, 5, 6])

cos_theta = np.dot(v1, v2) / (np.linalg.norm(v1) * np.linalg.norm(v2))
theta_rad = np.arccos(cos_theta)
theta_deg = np.degrees(theta_rad)

print("Angle between lines (degrees):", theta_deg)

B. Output:

Angle between lines (degrees): 12.933154491899135

PART 5: Intersection of Line and Plane

To find intersection:

1. Substitute parametric equations of the line into the plane equation.

2. Solve for t.

3. Plug t back into the line to find the point.

A. Python Example:

# Line: r(t) = [1 + 2t, 2 + t, 3 + 4t]
x, y, z, t = symbols('x y z t')
line = [1 + 2*t, 2 + t, 3 + 4*t]

# Plane: 2x - y + 4z - 10 = 0
plane_eq = Eq(2*line[0] - line[1] + 4*line[2] - 10, 0)
t_val = solve(plane_eq, t)[0]

intersection = [coord.subs(t, t_val) for coord in line]
print("Intersection point:", intersection)

B. Output:

Intersection point: [15/19, 36/19, 49/19]

PART 6: Distance from a Point to a Plane

Given point

$$P=(x_{1},y_{1},z_{1})$$

and plane

$$ax+by+cz+d=0$$

so:

$$Distance=\frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$

A. Python Example:

point = np.array([2, 3, 4])
normal = np.array([2, -1, 4])
d = -10  # from plane equation: 2x - y + 4z - 10 = 0 → d = -10

numerator = abs(np.dot(normal, point) + d)
denominator = np.linalg.norm(normal)

distance = numerator / denominator
print("Distance from point to plane:", distance)

B. Output:

Distance from point to plane: 1.5275252316519468

PART 7: Distance Between Parallel Planes

If both planes have the same normal vector:

$$ax+by+cz+d_{1}=0\;\;and\;\;ax+by+cz+d_{2}=0$$

so:

$$Distance=\frac{|d_{2}-d_{1}|}{\sqrt{a^{2}+b^{2}+c^{2}}}$$

PART 8: Angle Between Line and Plane

Given a line with direction d and plane normal n:

$$sin\theta=\frac{|\overrightarrow{d}.\overrightarrow{n}|}{|\overrightarrow{d}||\overrightarrow{n}|}$$

A. Python Example:

d = np.array([1, 1, 0])  # direction vector of line
n = np.array([0, 0, 1])  # normal vector of plane

sin_theta = abs(np.dot(d, n)) / (np.linalg.norm(d) * np.linalg.norm(n))
theta_rad = np.arcsin(sin_theta)
theta_deg = np.degrees(theta_rad)

print("Angle between line and plane (degrees):", theta_deg)

B. Output:

Angle between line and plane (degrees): 0.0

PART 9: Line of Intersection of Two Planes

If two planes intersect, the line lies in both. To find it:

1. Take both plane equations.

2. Solve the system symbolically.

A. Python Example:

x, y, z = symbols('x y z')

# Two planes:
eq1 = Eq(2*x + y - z, 5)
eq2 = Eq(x - y + 2*z, 3)

sol = solve([eq1, eq2], (x, y), dict=True)
print("Intersection in parametric form:")
print(sol)

B. Output:

Intersection in parametric form:
[{x: 8/3 - z/3, y: 5*z/3 - 1/3}]

Keywords

lines in space, planes in space, vector equations, parametric equations, dot product, cross product, normal vector, point-plane distance, projection onto plane, 3D geometry, vector algebra, line-plane intersection, plane equation, line direction vector, coplanar vectors, distance from point to line, shortest distance, scalar projection, geometric visualization, vector mathematics, nerd cafe