從Q(-1 , 2, 3)和2-x=y=-1/2z找垂直座標

Question

從Q(-1 , 2, 3)和 2-x = y = -1/2z 找垂直座標 (就是它們的兩線垂直的交差座標)怎麼找?教我

打錯了
是 2 -x = y -3 = -1/2 z (負一半 z)

by the way,
knowlegde, can you do it again?
plrease.. tthank you..

YES! 果然市天使!

Answer 1

請問是z是在分子還是分母呀? -z/2 還是 -1/(2z)?

2005-08-18 10:14:58 補充：
令 2-x = y-3 = -z/2 = t (參數式)
x= 2 - t, y = t+3, z = -2t

點到線距離
sqrt((2-t+1)^2 + (t+3-2)^2 + (-2t-3)^2)
= sqrt(6t^2 + 8t + 19)
= sqrt(6 (t+2/3)^2 + 19- 8/3)

求最小距離,則 t = -2/3
因此垂直座標為 ( 8/3, 7/3, 2/3)

2005-08-19 08:31:27 補充：
對厚,z=-2t, 所以應該是 4/3, 感謝knowledge大大

Answer 2

The line : vector n = (1,1,2)
and, the any point on line can discribe 2-x=y=-1/2z=t
so
x=2-t
y=t
z=-2t
so, Qp=(2-t,t,-2t)
and, vector QQp*n=-1
so
(2-t)+(t)+(-2t)*2=-1
and, t=3/4
so, Qp = (5/4,3/4,-3/2)

2005-08-18 10:06:10 補充：
-1/(2z)<---it is not line, so i think it must be -z/2

2005-08-18 23:37:58 補充：
The line
(x-2)/(-1) = y-3 = z/(-2)
so, vector n = (-1,1,-2)
and, the any point on line can discribe 2-x=y-3=-1/2z=t
so
x=2-t
y=t+3
z=-2t
vector QQp=(3-t,t+1,-2t-3)
and, vector (QQp*n) = 0
so
(3-t)*(-1)+(t+1)+(-2t-3)*(-2)=0
and, t = -2/3
so, Qp = (8/3,7/3,4/3)