How to find an equation of the plane that contains the line and is perpendicular to the plane
Favorites|Homepage
Subscriptions | sitemap
HOME > > How to find an equation of the plane that contains the line and is perpendicular to the plane

How to find an equation of the plane that contains the line and is perpendicular to the plane

[From: ] [author: ] [Date: 12-06-07] [Hit: ]
4,3) as a point on the line,......
line: (x+2)/3=(y-4)/2=3-z
plane: x-2y+z=5

-
The line is:

(x+2)/3 = (y-4)/2 = 3-z = t

Or in parametric form

x = 3t - 2, y = 2t + 4, z = (-1)t + 3

So the direction of the line is <3,2,-1>.

The normal direction to the plane is <1,-2,1> hence the cross product

<3,2,-1> x <1,-2,1> = <0,-4,-8>

is perpendicular to both. The plane must therefore have the form

(0)x - 4y - 8z = C

Setting t=0 in the equation for the line gives (-2,4,3) as a point on the line, hence

(0)(-2) - (4)(4) - (8)(3) = -40

So factoring out the common factor of four we get the plane

y + 2z = 10
1
keywords: plane,an,of,find,line,that,and,is,contains,How,perpendicular,to,equation,the,How to find an equation of the plane that contains the line and is perpendicular to the plane
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .