## Problem on the tangent plane of a surface

Find the tangent plane at $(1,1,1)$ to the surface $$x^2 + y^2 + z^2 + xy + xz = 5.$$
• ## Solution

Because we seek a plane, recall that
We know that the plane contains the point $\mathbf{x_0} = (1,1,1)$. Thus, all we need is a normal vector to the plane.
Because the plane is tangent to the given surface, the vector $\bfn$ will also be normal to the surface.
Recall that
We observe that the surface is given implicitly. That is, it is of the form $g(x,y,z) = c$, where $g(x,y,z) = x^2 + y^2 + z^2 + xy + xz$ and $c=1$.
The tangent plane is given by $$\left(\mathbf{x} - \langle 1,1,1 \rangle \right) \cdot \langle 4,3,3 \rangle = 0.$$
If $\bfx = (x,y,z)$, then this equation simplifies to \begin{align}
\bfx \cdot \langle 4,3,3 \rangle &= \langle 1,1,1 \rangle \cdot \langle 4,3,3 \rangle.\\
\end{align}
Hence, the tangent plane is $$4x + 3y + 3z = 10.$$