Create a circle perpendicular to a vector in a 3D coordinate system

Question

I have been working on this problem for a few days already, but could not figure it out completely.

Let's assume that I have two points, pointO (0,0,0) and pointA (1,1,1). Then we can calculate the vector of OA = <1,1,1>

I have a circle on the X-Y plane and the coordinates I derived is from X_circle = radius * cos(theta) Y_circle = radius * cos(theta) Z_circle = 0

Based on these circle coordinates (x,y,z), I want to project/translate it so that the circle is perpendicular to the vector OA.

I have followed this as my reference to this problem. However, my Z-coordinates are not translated. Where did I miss? Any feedback would be greatly appreciated.

I also provide my code for this problem. Please refer below.

def vector(point0, point1):
    return np.array((point1[0]-point0[0], point1[1]-point0[1], point1[2]-point0[2]))

def magnitude(vec):
    return np.array(math.sqrt((vec[0]**2+vec[1]**2+vec[2]**2)))

def unit_vector(vec, mag):
    return np.array(vec/mag)

# convert points into list of coordinates x, y, z
def conv_point_list(point):  
    x_list.append(point[0])
    y_list.append(point[1])
    z_list.append(point[2])

point0 = [0,0,0]
point1 = [1,1,1]

vec = vector(point0, point1)
print(vec)
mag = magnitude(vec)
print(mag)
uvec = unit_vector(vec, mag)
print(uvec)

# put to a list
x_list = []
y_list = []
z_list = []
conv_point_list(point0)
conv_point_list(point1)

# Plot a circle on a 2D plane
angle = np.linspace( 0 , 2 * np.pi , 150 ) 
radius = 0.4
# circle data points [r*cos(theta), r*sin(theta), 0]
x_circle = radius * np.cos( angle )  # r*cos(theta) 
y_circle = radius * np.sin( angle )  # r*sin(theta) 
z_circle = np.zeros(150)

# Translation
x_circle_t = np.multiply(RxRy[0], x_circle)
y_circle_t = np.multiply(RxRy[1], y_circle)
z_circle_t = np.multiply(RxRy[2], z_circle)

# Import libraries
# https://pythonguides.com/matplotlib-3d-scatter/
from mpl_toolkits import mplot3d
import numpy as np
import matplotlib.pyplot as plt
import plotly.graph_objects as go

# Create Figure
fig = plt.figure(figsize = (10, 7))
ax = plt.axes(projection ="3d")

# Create Plot
ax.scatter3D(x_list, y_list, z_list, marker='o', s=20, label='points', c='r')
ax.plot(x_list, y_list, z_list) # vector

# Plot a circle on a 2D plane
ax.plot(x_circle, y_circle, z_circle, c='green', label='circumference')
ax.plot(x_circle_t, y_circle_t, z_circle_t, c='orange', label='circumference_T')

# Parameters
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')

# Add legend
ax.legend(loc=1)

# Show plot
plt.show()

Results

Answer 1

If $\omega = \frac{OA}{\|OA\|}$ is a unitary vector and $v_0$ is a vector orthogonal to $\omega$, you can parametrize the circle centered on $(0,0,0)$ passing through $v_0$ in a plane orthogonal to $\omega$ by the formula \begin{equation} v(t) = \cos(t) v_0 + \sin(t) \omega\times v_0 \end{equation}

In your case, if $r$ is the radius of the circle, one can choose $v_0 = \frac{r}{\sqrt{2}}(1, -1, 0)^T$ and we have $\omega = \frac{1}{\sqrt{3}}(1, 1, 1)^T$. It follows that \begin{equation} v(t) = \frac{r}{\sqrt{6}}\begin{bmatrix} \sqrt{3}\cos(t)+ \sin(t)\\ -\sqrt{3}\cos(t)+ \sin(t)\\ -2\sin(t) \end{bmatrix} \end{equation} It is easy to check that $\|v(t)\| = r$ and $x(t)+ y(t)+z(t) = 0$.