Tangent Plane And Normal Vector.


by dcl
Tags: normal, plane, tangent, vector
dcl
dcl is offline
#1
Jun6-04, 01:06 AM
P: 55
I'm having trouble working out the tangent plane of an equation at a specified point (4,1,-2)
The equation being [tex]9x^2 - 4y^2 - 25z^2 = 40[/tex]

now
[tex]\nabla f = (18x, -8y, -50z)[/tex] yeh?
Just reading off this should give us the normal vector shouldn't it? (18,-8,-50)
and from that we can work out the equation of the plane.
18(x-4) - 8(y-1) -50(z-(-2)) = 0
Is this corrent or am I using a horribly flawed method?
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dcl
dcl is offline
#2
Jun6-04, 01:33 AM
P: 55
Think I've worked it out for myself.
Method was sorta wrong.
Once I have Grad F, all I need to do is sub in the values of the point and It will give me the normal vector and from that I can work out the equation.
I think thats right.


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