Find the equation of the plane tangent to the surface at a given point 7xy + yz + 4xz - 48 = 0; (2,2,2)

Solution:

Given, the equation is 7xy + yz + 4xz - 48 = 0

We have to find the equation of the plane at a given point (2, 2, 2).

We have function,

f(x, y, z) = 7xy + yz + 4xz - 48

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

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

\(\frac{\partial f}{\partial x}(7xy+yz+4xz-48)=7y+4z\)

\(\frac{\partial f}{\partial y}(7xy+yz+4xz-48)=7x+z\)

\(\frac{\partial f}{\partial z}(7xy+yz+4xz-48)=y+4x\)

So, \(\bigtriangledown f(x, y, z)=7y+4z+7x+z+y+4x\\=7x+4x+7y+y+4z+z\\\bigtriangledown f(x, y, z)=11x\hat{i}+8y\hat{i}+5z\hat{k}\)

At the given point (2, 2, 2) the normal vector to the surface is given by

\(\bigtriangledown f(2, 2, 2)=11(2)\hat{i}+8(2)\hat{i}+5(2)\hat{k}\\=22\hat{i}+16\hat{i}+10\hat{k}\)

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

Where, \(\vec{r}=\begin{pmatrix} x\\ y\\z \end{pmatrix}.\vec{n}=\begin{pmatrix} 22\\16 \\10 \end{pmatrix}\) is the normal vector and a is any point in the given plane.

The equation of the plane is given by

\(\begin{pmatrix} x\\y \\z \end{pmatrix}.\begin{pmatrix} 22\\16 \\10 \end{pmatrix}=\begin{pmatrix} 2\\2 \\2 \end{pmatrix}.\begin{pmatrix} 22\\16 \\10 \end{pmatrix}\)

So, x(22) + y(16) + z(10) = 2(22) + 2(16) + 2(10)

22x + 16y + 10z = 44 + 32 + 20

22x + 16y + 10z = 96

22x + 16y + 10z - 96= 0

Therefore, the equation of the plane is 22x + 16y + 10z - 96= 0.

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Find the equation of the plane tangent to the surface at a given point 7xy + yz + 4xz - 48 = 0; (2,2,2)

Summary:

The equation of the plane tangent to the surface at a given point 7xy+yz+4xz-48=0; (2,2,2) is 22x + 16y + 10z - 96 = 0.