Find an equation of the tangent plane to the given surface at the specified point. x²+ y²+ z²= 0; (0,0,3)

Solution:

Given, the equation is x² + y² + z² = 0

We have to find the equation of the plane at a given point (0, 0, 3).

We have function,

f(x, y, z) = x² + y² + z²

In order to find the normal at any particular point point in vector space, we use gradient operator,

\(\bigtriangledown f(x, y, z)=\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k}\)

\(\frac{\partial f}{\partial x}(x² + y² + z²)=2x\)

\(\frac{\partial f}{\partial y}(x² + y² + z²)=2y\)

\(\frac{\partial f}{\partial z}(x² + y² + z²)=2z\)

So, \(\bigtriangledown f(x, y, z)=2x+2y+2z\\\bigtriangledown f(x, y, z)=2x\hat{i}+2y\hat{i}+2z\hat{k}\)

At the given point (0, 0, 3) the normal vector to the surface is given by

\(\bigtriangledown f(0, 0, 3)=2(0)\hat{i}+2(0)\hat{i}+2(3)\hat{k}\\=6\hat{k}\)

The general form of vector equation is \(\vec{r}.\vec{n}=\vec{a}.\vec{n}\)

Where, \(\vec{r}=\begin{pmatrix} x\\ y\\z \end{pmatrix}.\vec{n}=\begin{pmatrix} 0\\0 \\6 \end{pmatrix}\) is the normal vector and a is any point in the given plane.

The equation of the plane is given by

\(\begin{pmatrix} x\\y \\z \end{pmatrix}.\begin{pmatrix} 0\\0 \\6 \end{pmatrix}=\begin{pmatrix} 0\\0 \\3 \end{pmatrix}.\begin{pmatrix} 0\\0 \\6 \end{pmatrix}\)

So, x(0) + y(0) + z(6) = 0(0) + 0(0) + 3(6)

6z = 18

z = 18/6

z = 3

Therefore, the equation of the tangent plane is z = 3