Question

Help with finding equation of tangent plane and parametric equations of normal line to surface given by x^2+y^2+z^2-2xy+4xz-x+y=12 at P(1,0,2). 

Answer 1

The given equation can be written in the form \(F(x, y, z)=0,\) where \(F(x, y, z)=x^{2}+y^{2}+z^{2}-2 x y+4 x z-x+y-12\). The partial derivatives of \(F\) are

$$ \begin{aligned} F_{x}(x, y, z) &=\frac{\delta\left(x^{2}+y^{2}+z^{2}-2 x y+4 x z-x+y-12\right)}{\delta x} \\ &=2 x-2 y+4 z-1 \\ F_{y}(x, y, z) &=\frac{\delta\left(x^{2}+y^{2}+z^{2}-2 x y+4 x z-x+y-12\right)}{\delta y} \\ &=2 y-2 x+1 \end{aligned} $$

And

$$ \begin{aligned} F_{z}(x, y, z) &=\frac{\delta\left(x^{2}+y^{2}+z^{2}-2 x y+4 x z-x+y-12\right)}{\delta z} \\ &=2 z+4 x \end{aligned} $$

In particular at point (1,0,2) it becomes,

$$ \begin{aligned} F_{x}(1,0,2) &=2(1)-2(0)+4(2)-1 \\ &=2+8-1 \\ &=9 \\ F_{y}(1,0,2) &=2(0)-2(1)+1 \\ &=0-2+1 \\ &=-1 \end{aligned} $$

And

$$ \begin{aligned} F_{z}(1,0,2) &=2(2)+4(1) \\ &=4+4 \\ &=8 \end{aligned} $$

Next using the equation of tangent plane

$$ F_{x}(a, b, c)(x-a)+F_{y}(a, b, c)(y-b)+F_{z}(a, b, c)(z-c)=0 $$

We find that an equation of the tangent plane to the \(F\) at (1,0,2) is

$$ \begin{array}{r} 9(x-1)-1(y-0)+8(z-2)=0 \\ 9 x-9-y+8 z-16=0 \\ 9 x-y+8 z=25 \end{array} $$

Therefore equation of the tangent plane will be given by equation \(9 x-y+8 z=25\)

Next using the equation of normal line (in symmetric form)

\(\frac{x-a}{F_{x}(a, b, c)}=\frac{y-b}{F_{y}(a, b, c)}=\frac{z-c}{F_{z}(a, b, c)}\)

We obtain the following parametric equation of normal line to the \(F\) at (1,0,2)

\(\frac{x-1}{9}=\frac{y-0}{-1}=\frac{z-2}{8}\)

\(\frac{x-1}{9}=-y=\frac{z-2}{8}\)

Therefore equation of the normal line will be given by equation \(\frac{x-1}{9}=-y=\frac{z-2}{8}\)