Question

Question:

Find the equation of the tangent plane and normal to the surface

2𝑥𝑧² − 3𝑥𝑦 − 4𝑥 = 7 at point

( 1, −1,2 )

Answer 1

Solution;

Now, as gradient of 𝑓 is normal at a point to the level surface, we have;

$\nabla f=\tilde{i}\frac{\delta f}{\delta x}+\tilde{j}\frac{\delta f}{\delta y}+\tilde{k}\frac{\delta f}{\delta z}$

Since , $f=2xz^2-3xy-4x-7$

We have;

$\frac{\delta f}{\delta x}=2z^2-3y-4$

$\frac{\delta f}{\delta y}=-3x$

$\frac{\delta f}{\delta z}=4xz$

Therefore;

$\nabla f=\tilde{i}(2z^2-3y-4)+\tilde{j}(-3x)+\tilde{k}(4xz)$

At point (1,-1,2),we have the gradient as;

$\nabla f=\tilde{i}(2(2^2)-3(-1)-4)+\tilde{j}(-3×1)+\tilde {k}(4×1×2)$

$\nabla f=7\tilde i-3\tilde j +8\tilde k$ Is the tangent.

If the vector position at point (1,-1,2) is $\vec r$ then we have;

$\vec r=\tilde i-\tilde j+2\tilde k$

Let Q be another current point on the level surface through P(1,-1,2) with position vector $\vec R$ .

Then ;

$\vec {PQ}$ =$\vec R-\vec r$ =$(X-1)\tilde i +(Y+1)\tilde j+(Z-2)\tilde k$

Since the tangent $\nabla f$ and $\vec {PQ}$ are perpendicular vectors ,their dot product is 0,we have

$[(X-1)\tilde i+(Y+1)\tilde I+(Z-2)\tilde k].[7\tilde i -3\tilde j+8\tilde k]=0$

Hence;

7(X-1)-3(Y+1)+8(Z-2)=0

7X-3Y+8Z=26

If the position vector of the current point Q on the normal is $\vec R$ and (X,Y,Z) are its coordinates,then the equation of the normal at point P(x,y,z) to a level surface is given by

$\frac{X-x}{\frac{\delta f}{\delta x}}=\frac{Y-y}{\frac{\delta f}{\delta y}}=\frac{Z-z}{\frac{\delta f}{\delta z}}$

Here,point (x,y,z) is (1,-1,2) and $\frac{\delta f}{\delta x}=7$ ,$\frac{\delta f}{\delta y}=-3$ and $\frac{\delta f}{\delta z}=8$

Therefore,the equation of the normal at the point (1,-1,2) to a given level surface is

$\frac{X-1}{7}=\frac{Y+1}{-3}=\frac{Z-2}{8}$