Finding Tangent Plane and Normal Line Equations for a Given Surface

[ktobrien]

Homework Statement

Find equations of the following.

x²-2y²+z²+yz=7, (5,3,-3)
(a) the tangent plane
(b) the normal line to the given surface at the point

Homework Equations

I know it involves f_x, f_y, f_z

The Attempt at a Solution

I got 10x-15y-3z=7. Is this correct? Because its not true at the point (5,3,-3).

I got it by f(5,3,-3)+f_x(5,3,-3)(x-5)+f_y(5,3,-3)(y-3)+f_z(5,3,-3)(z+3)

As for the normal line I know the answers are
x=10t+5
y=-15+3
z=-3-3

 

[Mark44]

Homework Statement

Find equations of the following.

x²-2y²+z²+yz=7, (5,3,-3)
(a) the tangent plane
(b) the normal line to the given surface at the point

Homework Equations

I know it involves f_x, f_y, f_z

The Attempt at a Solution

I got 10x-15y-3z=7. Is this correct? Because its not true at the point (5,3,-3).

No. The point (5, 3 -3) has to satisfy both the equation of the plane and the equation of the surface.

I got it by f(5,3,-3)+f_x(5,3,-3)(x-5)+f_y(5,3,-3)(y-3)+f_z(5,3,-3)(z+3)

For one thing, this is not an equation, so there's no way to get an equation out of it. For another thing, if my memory is correct, the equation of the tangent plane is f_x(5, 3, -3)(x - 5) + f_y(5, 3, -3)(y - 3) + f_z(5, 3, -3)(z - (-3)) = 0.

You didn't show the partial derivatives that you calculated, so it might also be that you have an error in one or more of them.

As for the normal line I know the answers are
x=10t+5
y=-15+3
z=-3-3

 

[ktobrien]

Ive got it now. I used the linear approximation and set it = to 0. I just had to take out the f(5,3,-3). I figured it was something stupid like that. Thanks for the help though.