Найти грань и объем пирамиды в трехмерном пространстве

a = np.array([2, -2,0])
b = np.array([5, 3, 2])
c = np.array([-3,-2,0])
d = np.array([1, 2, 3])
 
vector_a = [b[0] - a[0], b[1] - a[1], b[2] - a[2]]
vector_b =[c[0] - a[0], c[1] - a[1], c[2] - a[2]]
vector_c =[d[0] - a[0], d[1] - a[1], d[2] - a[2]]
 
 
vectorniy_dobutok_b_c = np.cross(b,c)
mishanyi_dobutok_vectoriv = np.dot(a,vectorniy_dobutok_b_c)
 
print(mishanyi_dobutok_vectoriv )
obiom = mishanyi_dobutok_vectoriv / 6
print(obiom)

a = np.array([2, -2,0])
b = np.array([5, 3, 2])
c = np.array([-3,-2,0])
d = np.array([1, 2, 3])
 
vec_ab = b-a
vec_ac = c-a
vec_ad = d-a
 
# S_ABC = 1/2*|[AB,AC]|
# V_ABCD = 1/6 * |([AB,AC], AD)|
 
v1 = np.cross(vec_ab, vec_ac )
v2 = np.dot(v1, vec_ad)
 
print(f'Площадь грани ABC: {0.5 * np.linalg.norm(v1, ord = 2):.3f}')
print(f'Объем ABCD: {np.abs(v2)/6:.3f}')
 
# Площадь грани ABC: 13.463
# Объем ABCD: 5.833

import numpy as np
vector1 = 3
vector2 = 2
f = np.pi / 3
 
 
answer = (((vector1 ** 2) * 5) + (5 * vector2) * (vector1 * vector2 * np.cos(f)) + (vector2 ** 2))
#answer = np.dot( (5 * vector1 + vector2 ), (vector1 + vector2 ))
#answer = np.dot( ( vector1 + vector2 * 3 ), (3 * vector1 - 2 * vector2 ))
print(answer)
 
answer2 = np.sqrt((vector1 ** 2) + (vector2 ** 2) - ((2 * vector2 + vector1) * np.cos(f)))
print(answer2)

import numpy as np
 
def rotate_matrix(x):
    return np.array([[np.cos(x),-np.sin(x)],[np.sin(x),np.cos(x)]])
 
a = np.array([3,0])
phi = np.pi/3
b = np.matmul(rotate_matrix(phi),2*a/3)
 
answer = np.dot(5*a + b, a + b)
answer2 = np.linalg.norm(a + b)
 
print(answer)
print(answer2)