Answer:
A.[tex]-x+2y+z=2-\pi[/tex]
B.[tex]x=\frac{-t}{\sqrt{6}}+1+\pi, \ y=\frac{2t}{\sqrt{6}}+1, \ z=\frac{t}{\sqrt{6}}+1[/tex]
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):
[tex]f(x,y,z)= 12arctan(yz)-3x+3z[/tex]:
To proceed to find the tangent plane at (1+π,1,1), we use the following equation for the tangent plane:
[tex]\nabla{f(x_{0},y_{0},z_{0})*(x-x_{0},y_{0},z-z_{0})=0[/tex]
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:
[tex]\nabla{f(x,y,z)}=(\frac{\delta{f}}{\delta{x}},\frac{\delta{f}}{\delta{y}},\frac{\delta{f}}{\delta{x}})[/tex]
Now we differentiate f with respect to x,y and z to find those coordinates:
[tex]\nabla{f(x,y,z)}=(-3,\frac{12z}{1+(yz)^{2}},3)[/tex]
[tex]\nabla{f(1+\pi,1,1)}=(-3,\frac{12}{2},3)=(-3,6,3)\\[/tex]
We are ready to use the equation for the tangent plane
[tex](-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[/tex]
The tangent plane has an equation [tex]-x+2y+z=2-\pi[/tex], 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:
[tex]\|v\|=\sqrt{(-1)^{2}+2^{2}+1^{2}}=\sqrt{6}\\v_{1}=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})[/tex]
And because that line has to pass through (1+π,1,1) we conclude the vector equation for this line is the following:
[tex]\overrightarrow{V}(t)=(\frac{-1}{\sqrt{6}},\frac{2}{\sqrt{6}},\frac{1}{\sqrt{6}})t+(1+\pi,1,1)[/tex]
and from this equation:
[tex]x=\frac{-t}{\sqrt{6}}+1+\pi\\y=\frac{2t}{\sqrt{6}}+1\\z=\frac{t}{\sqrt{6}}+1[/tex]
