# Tangent Plane And Normal Vector.

1. Jun 6, 2004

### dcl

I'm having trouble working out the tangent plane of an equation at a specified point (4,1,-2)
The equation being $$9x^2 - 4y^2 - 25z^2 = 40$$

now
$$\nabla f = (18x, -8y, -50z)$$ 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?

2. Jun 6, 2004

### dcl

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.