Curve being tangent to a surface Verify that the curve $\displaystyle x^{2}-y^{2}+z^{2}=1, xy+xz = 2$ is tangent to the surface $xyz-x^{2}-6y+6=0$ at the point (1, 1, 1). 

I have found the gradient vector for the surface, which would be: 
$\displaystyle\frac{\partial F}{\partial x}=yz-2x=-1$
$\displaystyle\frac{\partial F}{\partial y}=xz-6=-5$
$\displaystyle\frac{\partial F}{\partial z}=xy=1$
However, I'm not sure on how to proceed on the other question (it is a curve, but they are giving us two different equations). How can i work around this problem? I have tried finding the gradient vector for the curve, but i'm not sure if it will work as it is a curve (and the gradient vectors results actually differ for both surfaces).
 A: You should prove that the tangent vector of the curve at $(1,1,1)$ is contained in the tangent plane of the surface, or is orthogonal to the normal vector to the surface at $(1,1,1)$ which is given by the gradient. You calculated the normal vector which is $(-1,-5,1)$. 
Now you need to calculate the tangent vector to the curve in $(1,1,1)$. The curve is given by the intersection of two surfaces, so the tangent vector would be orthogonal to both normal vectors in $(1,1,1)$ of the two surfaces. 

$(S_1):  x^2-y^2+z^2=1$ has normal vector $(2x,-2y,2z)$ which evaluated in your point gives $v_2=(2,-2,2)$.
$(S_2): xy+xz=2$ has normal vector $(y+z,x,x)$ which evaluated in your point gives $v_3=(2,1,1)$.
Solve now the system given by $v\cdot v_2=0$ and $v\cdot v_3=0$ to get $v_1-v_2+v_3=0$ and $2v_1+v_2+v_3=0$. Eliminate $v_1$ to get $3v_2-v_3=0$. So if $v_2=\alpha$ we have $v_3=3\alpha$ and $v_1=-2\alpha$. So $v=\alpha(-2,1,3)$.
The vector $v$ is the tangent vector to the curve at $(1,1,1)$, and it should be orthogonal to $(-1,-5,1)$. Indeed if we calculate the scalar product we get $(-2,1,3)(-1,-5,1)=2-5+3=0$. 
