Problem on the tangent plane of a surface
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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
A plane is specified by a point and a normal vector
To specify a plane, we need a point $\bfx_0$ and a normal vector $\bfn$.
The plane going through $\bfx_0$ with normal vector $\bfn$ is given by $$(\mathbf{x} - \mathbf{x_0}) \cdot \mathbf{n} = 0$$
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
Normal vector to an implicitly defined surface
Gradient is perpendicular to level curves
Level curves and surfaces
To find a normal vector to a surface, view that surface as a level set of some function $g(x,y,z)$.
A normal vector to the implicitly defined surface $g(x,y,z) = c$ is $\nabla g(x,y,z)$.
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 $c=1$.
Definition of the gradient
A normal vector to the surface is given by by the gradient \begin{align}\nabla g(x,y,z) &= \langle 2x + y + z, 2y + x, 2z + x \rangle, \\
\nabla g(1,1,1) &= \langle 4, 3, 3 \rangle.
\end{align}
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 $$4x + 3y + 3z = 10.$$
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