Prove Plane is perpendicular to a parameter line ?
2016-05-04 4:37 pm
Given : P : x - y - z + 4 = 0 parameter line D : x = 1 + t, y = 1 - t , z = 2 - t
回答 (2)
2016-05-04 5:16 pm
✔ 最佳答案
Let F(x,y,z)=x-y-z=-4 & the plane is called "Pi".grad(F)=i-j-k which is normal to the plane "Pi"
The given line L in Descartes' form is
(x-1)/1=(y-1)/(-1)=(z-2)/(-1)=t
=>its directional vector is A=i-j-k
Since grad(F)// A, "Pi" _|_ L.
2016-05-04 4:44 pm
The coefficients [1,-1,-1] give a normal vector to the plane.
We have to show a vector joining 2 points of the line goes in the same direction.
t=0 gives (x,y,z) = (1,1,2)
t=1 gives (x,y,z) = (2,0,1)
vector between the points is [2-1, 0-1, 1-2] = [1,-1,-1] which is in the same direction as [1,-1,-1] because in fact it is the same vector.
We have to show a vector joining 2 points of the line goes in the same direction.
t=0 gives (x,y,z) = (1,1,2)
t=1 gives (x,y,z) = (2,0,1)
vector between the points is [2-1, 0-1, 1-2] = [1,-1,-1] which is in the same direction as [1,-1,-1] because in fact it is the same vector.
收錄日期: 2021-05-01 20:41:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160504083729AA4bUyl