# Thread: Find the tangent plane to the surface

1. ## Find the tangent plane to the surface

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?

2. Originally Posted by njr008
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?
The Cartesian equation of the plane is ax + by + cz = d.

ai + bj + ck is a normal to the plane. If you know a normal, you know the values of a, b and c. To get the value of d, substitute the known point (1, -1, 3) into the Cartesian equation:

a(1) + b(-1) + c(3) = d.

Oh yes ...... the values of a, b and c are got from the normal. If the surface has the equation x = f(x, y), then a normal is:

$\displaystyle \frac{\partial f}{\partial x} i + \frac{\partial f}{\partial y} j - k$.

Substitute x = 1 and y = -1 to get a normal at (1, -1, 3).