Find the equation for the tangent plane to the surface x^2yz+3y^2=2xz^2-8z at the point (1,2,-1)?
I will start with the final form of equation you must arrive at. Take a look at this equation, get used to it and then we will understand together what we must do to reach to this form:
`N(r - r_0) = 0`
If you are confused, do not worry at all. You will understand right away what is the meaning of this equation. As you can see, N and `(r - r_0)` represent two vectors, N is the normal vector to the plane `x^2yz+3y^2=2xz^2-8z` and `r - r_0` is the vector that lies in the tangent plane. Therefore, the product `N(r - r_0) = 0` must be the dot product of the vectors.
Why the dot product is equal to 0? Well, the answer is simple. The dot product is 0 because the normal to the surface is normal to the tangent plane, too.
All you need to do is to start looking for the vector equation of normal N.
N = `grad` `F|_(1,2,-1) = F_x|_(1,2,-1) + F_y|_(1,2,-1) + F_z|_(1,2,-1)`
`N = (2xyz - 2z^2)|_(1,2,-1)*i + (x^2z+6y)|_(1,2,-1)*j + (xy-4xz+8)|_(1,2,-1)*k`
Just put 1,2,-1 instead of x,y,z and start calculate partial derivatives:
`N = -6i+11j+14k`
Now, it is the time to calculate the equations of vectors `r_0` and r:
`r_0 = i+2j-k`
`r = x i + yj + zk`
Calculate `r-r_0` subtracting the coefficients of i,j,k;
`r-r_0 = (x-1)i+(y-2)j+(z+1)k`
Now it's time for the dot product:
`N*(r-r_0) = 0`
Multiplicate coefficients of i,j,k of vectors N and `r-r_0` :
-`6(x-1) + 11(y-2) + 14(z+1 ) = 0`
The answer to this problem: `6x-11y-14z+2=0`
If you have any questions regarding the explanation, please feel free to ask what you did not quite understood.