Find tangent plane to surface x^2 + 2xy - y^2 + z^2 = 7 at

[Tulip007]

Find the tangent plane to the surface, x^2 + 2xy - y^2 + z^2 = 7, at the point (1, -1, 3). Using the tangent plane as an approximation to the surface, find the approximate height of the surface above the (x,y) plane at the point (1.1, -1.2). How does this result compare to the true height?

This is what I have done so far, although I do not know if it is completely right:
gradient F = 2i - 2j =2k
gradient F (1,-1,3) = 2 (1)i + 2 (-1)j + 2 (3)k
= 2 (x -1) - 2 (y +1) + 6 (z - 3) = 0
= 2x - 2 - 2y - 2 + 6z - 18 = 0
= 2x - 2y + 6z = 18 + 2 + 2
= 2x - 2y + 6z = 22

After this I am stuck, I have no idea what the next process is.

 

[Dr. Flim-Flam]

Re: Find the tangent plane of the surface

Part A: F(x,y,z) = x^2 + 2xy - y^2 + z^2 - 7 = 0

Fsubx(x,y,z) = 2x+2y, Fsuby(x,y,z) = 2x-2y, Fsubz 9x,y,z) = 2z

Fsubx(1,-1,3) = 0, Fsuby(1,-1,3) = 4, Fsubz(1,-1,3) = 6

Hence, 0(x-1)+4(y+1)+6(z-3) = 0, 2y+3z = 7. Check: 2(-1) + 3(3) = 7.

Don't understand Part B.