In the three‐dimensional Cartesian coordinates, find the shortest distance of
the intersection, of x+2y+3z=6 and 3x+2y+z=6 to the origin (0,0,0)
. Also, what are the coordinates of the point that has this shortest
distance?
兩條平行線到原點的距離
2014-02-26 7:10 am
回答 (2)
2014-02-26 7:59 am
✔ 最佳答案
聯立x+2y+3z=6 , 3x+2y+z=6為空間兩面交線寫成參數式
兩式相減得 2x-2z=0 => x=z
令x=z=t，t為實數
代入x+2y+3z=6
=> t+2y+3t=6 =>y=3-2t
交線參數式
{ x=t
{ y=3-2t ，t為任意實數
{ z=t
上取任一點P(t,3-2t,t)與(0,0,0)距離d
d=√[t^2+(3-2t)^2+t^2]
=√6t^2-12t+9
=√6(t-1)^2+3
當t=1時，d有最小值√3 ---------Ans(1)
點P座標為(1,1,1) ---------Ans(2)
參考: Four-Year
2014-02-26 6:16 pm
問的是兩平面交線到原點距離, 標題卻是 "兩平行線到原點距離"...
收錄日期: 2021-05-04 01:55:53
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140225000015KK05860