求3維拋物線由起點 [1,0,0]經頂點[1,1,1] 落至[0,1,0] 的向量公式 最好是像t.A這種形式(t為0到1的純量)?

設 A[1,0,0] 經頂點 B[1,1,1] 落至 C[0,1,0]

其中 AB=BC=CA=√2

若把A當二維原點(0,0),AC為x軸, AC中點D為(√2/2,0), C為(√2,0), B(√2/2, √6/2)

x座標可表為 √2 t (0<=t<=1)

y座標可表為 - √6 (√2 t - √2/2)^2 + √6/2

= - 2√6 (t-1/2)^2 + √6/2

= √6 [ - 2 (t-1/2)^2 + 1/2]

A原點座標 (0,0) 改為 (1,0,0) , x 軸方向向量 (1, 0) 改為 (-1/√2,1/√2,0)

y 軸方向向量 (0,1) 改為 (1/√6, 1/√6, 2/√6)

所以拋物線參數座標為

x = 1 - t + [ - 2 (t-1/2)^2 + 1/2] = - 2 (t-1/2)^2 - t + 3/2

y = t + [ - 2 (t-1/2)^2 + 1/2] = - 2 (t-1/2)^2 + t + 1/2

z = 2 [ - 2 (t-1/2)^2 + 1/2] = - 4 (t-1/2)^2 +1

Ans: [- 2 (t-1/2)^2 - t + 3/2, - 2 (t-1/2)^2 + t + 1/2, - 4 (t-1/2)^2 +1]