2D 座標架
Author: 藍邦偉  老師

$ O $

$ 向量減法的幾何意義 $

$A$

$B$

$C$

$Step=%v$

$C_2$

$B_3$

$ 欲求\overrightarrow{AB}-\overrightarrow{AC} $

$ 決定A,B,C\;三點的位置 $

$ 作-\overrightarrow{AC} $

$ 將\overrightarrow{AB}沿\overrightarrow{AC_2}平移至C_2 $

$ 作\overrightarrow{AB_3}=\overrightarrow{AB}+\overrightarrow{AC_2}$
$\;\;\;\;\;\;\;\;\;\;\;\;=\overrightarrow{AB}+(-\overrightarrow{AC} )$

$ 將\overrightarrow{AB_3} 沿\overrightarrow{AC}平移到C$
$則：\;\overrightarrow{CB}=\overrightarrow{AB}-\overrightarrow{AC}$

u = Vector((x(Corner[1]), 0), (x(Corner[3]), 0))
v = Vector((0, y(Corner[1])), (0, y(Corner[3])))
O = (0, 0)
Q_2: 0.7 * Corner[1] + 0.3 * Corner[2]
Q_3: (x(Q_2), y(Corner[4]))
poly1, q_1, q_2, q_4, q_3 = Polygon(Corner[1], Q_2, Q_3, Corner[4])
A = (3.1835626052361174, -1.1423240780026416)
B = (4.310350250332438, 2.047057213559152)
C = (6.85039697571906, -0.3402042322206336)
s = 0 (slider from 0 to 5, step 1)
t1 = 0 (slider from 0 to 1, step 0.01)
t2 = 0 (slider from 0 to 1, step 0.01)
t3 = 0 (slider from 0 to 1, step 0.01)
t4 = 0 (slider from 0 to 1, step 0.01)
B_1: A + t1 * Vector[A, B]
C_1: A + t1 * Vector[A, C]
u_1 = Vector(A, B_1)
v_1 = Vector(A, C_1)
C_2: A - t2 * Vector[A, C]
u_2 = Vector(A, C_2)
t5 = 0 (slider from 0 to 1, step 0.01)
A_3: A - t3 * Vector[A, C]
B_3: B - t3 * Vector[A, C]
b = Vector(A_3, B_3)
A_4: A + t4 * Vector[A, B_3]
c = Vector(A, A_4)
A_5: A + t5 * Vector[A, C]
B_5: A_4 + t5 * Vector[A, C]
u_4 = Vector(A_5, B_5)
poly2, a, d, e, f = Polygon(A, B, B_3, A_3)
ss = 4
poly3, a_4, a_1, a_5, b_5 = Polygon(A_4, A, A_5, B_5)
D = (-6.38458333024281, 5.007261406326085)