Answer:
A.![-x+2y+z=2-\pi](https://tex.z-dn.net/?f=-x%2B2y%2Bz%3D2-%5Cpi)
B.![x=\frac{-t}{\sqrt{6}}+1+\pi, \ y=\frac{2t}{\sqrt{6}}+1, \ z=\frac{t}{\sqrt{6}}+1](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-t%7D%7B%5Csqrt%7B6%7D%7D%2B1%2B%5Cpi%2C%20%5C%20y%3D%5Cfrac%7B2t%7D%7B%5Csqrt%7B6%7D%7D%2B1%2C%20%5C%20z%3D%5Cfrac%7Bt%7D%7B%5Csqrt%7B6%7D%7D%2B1)
Step-by-step explanation:
A. At first, it´s useful to move everything to one side and name it as a function f(x,y,z):
:
To proceed to find the tangent plane at (1+π,1,1), we use the following equation for the tangent plane:
![\nabla{f(x_{0},y_{0},z_{0})*(x-x_{0},y_{0},z-z_{0})=0](https://tex.z-dn.net/?f=%5Cnabla%7Bf%28x_%7B0%7D%2Cy_%7B0%7D%2Cz_%7B0%7D%29%2A%28x-x_%7B0%7D%2Cy_%7B0%7D%2Cz-z_%7B0%7D%29%3D0)
Where (x₀,y₀,z₀) is the specified point where we want the tangent plane to connect. Now we need to find the gradient vector of f:
![\nabla{f(x,y,z)}=(\frac{\delta{f}}{\delta{x}},\frac{\delta{f}}{\delta{y}},\frac{\delta{f}}{\delta{x}})](https://tex.z-dn.net/?f=%5Cnabla%7Bf%28x%2Cy%2Cz%29%7D%3D%28%5Cfrac%7B%5Cdelta%7Bf%7D%7D%7B%5Cdelta%7Bx%7D%7D%2C%5Cfrac%7B%5Cdelta%7Bf%7D%7D%7B%5Cdelta%7By%7D%7D%2C%5Cfrac%7B%5Cdelta%7Bf%7D%7D%7B%5Cdelta%7Bx%7D%7D%29)
Now we differentiate f with respect to x,y and z to find those coordinates:
![\nabla{f(x,y,z)}=(-3,\frac{12z}{1+(yz)^{2}},3)](https://tex.z-dn.net/?f=%5Cnabla%7Bf%28x%2Cy%2Cz%29%7D%3D%28-3%2C%5Cfrac%7B12z%7D%7B1%2B%28yz%29%5E%7B2%7D%7D%2C3%29)
![\nabla{f(1+\pi,1,1)}=(-3,\frac{12}{2},3)=(-3,6,3)\\](https://tex.z-dn.net/?f=%5Cnabla%7Bf%281%2B%5Cpi%2C1%2C1%29%7D%3D%28-3%2C%5Cfrac%7B12%7D%7B2%7D%2C3%29%3D%28-3%2C6%2C3%29%5C%5C)
We are ready to use the equation for the tangent plane
![(-3,6,3)*(x-1-\pi,y-1,z-1) = 0\\3+3\pi-3x+6y-6+3z-3=0\\-3x+6y+3z=6-3\pi\\-x+2y+z=2-\pi](https://tex.z-dn.net/?f=%28-3%2C6%2C3%29%2A%28x-1-%5Cpi%2Cy-1%2Cz-1%29%20%3D%200%5C%5C3%2B3%5Cpi-3x%2B6y-6%2B3z-3%3D0%5C%5C-3x%2B6y%2B3z%3D6-3%5Cpi%5C%5C-x%2B2y%2Bz%3D2-%5Cpi)
The tangent plane has an equation
, and the orthogonal vector to this plane is one made of the coefficients of the plane, a normal vector for this plane is (-1,2,1).
To find a normal line to this surface in (1+π,1,1) we find a normal line to the plane, and because we know that (-1,2,1) is a normal vector, then the line has to have the same direction, so we normalize that vector to get the direction:
And because that line has to pass through (1+π,1,1) we conclude the vector equation for this line is the following:
![\overrightarrow{V}(t)=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})t+(1+\pi,1,1)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BV%7D%28t%29%3D%28%5Cfrac%7B-1%7D%7B%5Csqrt%7B6%7D%7D%2C%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%7D%2C%5Cfrac%7B1%7D%7B%5Csqrt%7B6%7D%7D%29t%2B%281%2B%5Cpi%2C1%2C1%29)
and from this equation:
![x=\frac{-t}{\sqrt{6}}+1+\pi\\y=\frac{2t}{\sqrt{6}}+1\\z=\frac{t}{\sqrt{6}}+1](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-t%7D%7B%5Csqrt%7B6%7D%7D%2B1%2B%5Cpi%5C%5Cy%3D%5Cfrac%7B2t%7D%7B%5Csqrt%7B6%7D%7D%2B1%5C%5Cz%3D%5Cfrac%7Bt%7D%7B%5Csqrt%7B6%7D%7D%2B1)