equations for the line of intersection of two planes

heyhi

2007-10-25 7:51 pm

Find the equation for the line of intersection L between the planes x-y-z = 7 and x-2y+z=1.

[Hint:
we can sort out things to get the symmetric equations for L
(x-13)/3 = (y-6)/2 = (z-0)/1 ,to find the line of intersection is to find a direction vector "v" and a position vector "a" for this line. ]

But i don't really know how to find a and v, can anyone help?

更新1:

the symmetric equations for L, that's from the hint (x-13)/3 = (y-6)/2 = (z-0)/1 what i need is to present the equation in the form of "vector r = a+tv ", can u help me on that?

回答 (1)

myisland8132

2007-10-25 9:25 pm

✔ 最佳答案

上面個位理解錯了個Hint啦
因為由(x-13)/3 = (y-6)/2 = (z-0)/1﹐都可以直接看出a=(13,6,0)和v=(3,2,1)
個Hint只不過是說要設法由條件L is the line of intersection between the planes x-y-z = 7 and x-2y+z=1.中設法找出a和v罷了。[相反來說﹐若果你已經找到條直線﹐你都解決了條問題啦﹐根本上不用找a和v]
首先你要找方向比
設直線L的方向比是[a:b:c]﹐則
因直線L垂直於平面 x-y-z = 7,x-2y+z=1.
所以
a-b-c=0 [因為與平面的法線向量垂直]
a-2b+c=0
得b+c-2b+c=0;b=2c,a=3c
直線L的方向比是[a:b:c]=[3:2:1]
接著要找L上的一點P
因該點必在兩個平面上
x-y-z = 7 ...(1)
x-2y+z=1...(2)
(1)-(2):y-2z=6
令z=k,則y=6+2k,x=7+y+z=3k+13
所以可以設k=0
x=13,y=6,z=0 為直線L上之一點
現在a=(13,6,0);v=[3:2:1]
由對稱式(symmetric equations) 可得L的方程為
(x-13)/3 = (y-6)/2 = (z-0)/1

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