Find tangents at a specified positions

Question

I have two similar problems that I have tried to solve for several hours now but I end up with wrong answers. So there is something I do not understand correctly.

Problem 1: The intersection between the ellipsoid and the plane $$2 x^2 + y^2 + z^2=4$$ $$x-y+2z=2$$ is an ellipsoid in the room. Calculate the tangent direction at the point $$(1,1,1)$$

Attempted solution 1 (wrong):

1) Check that the point satisfies both equations (ok) $2*1^2+1^2+1^2 = 4$ and $1-1+2*1=2$

2) To find the intersection setup both equations as: $2x^2+y^2+z^2-4=x-y+2z-2$

3) We can then write the equation as: $$g(\mathbf x)=2x^2-x+y^2+y+z^2-2z-2 = 0$$ $$g(\mathbf x)=2(x-\frac{1}{4})^2-2(\frac{1}{4})^2+(y+\frac{1}{2})^2-(\frac{1}{2})^2+(z-1)^2-1^2-2 = 0$$ $$g(\mathbf x)=2(x-\frac{1}{4})^2+(y+\frac{1}{2})^2+(z-1)^2 = \frac{19}{8}$$ 4) The normal of this new ellipsoid will be: $$\nabla g(\mathbf x) = (4x-1, 2y+1, 2z-2)$$ The tangent equation is: $$\nabla g(\mathbf a)\cdot(\mathbf x - \mathbf a)=0$$ 5) My "answer" $$\mathbf a = (1,1,1)$$ $$(3, 3, 0)\cdot(x-1, y-1, z-1) = 0$$ $$(x-1) + (y-1) = 0$$ But this is not a direction!

Correct answer to problem 1: $$(-1,1,1)$$

Problem 2: The intersection between the hyperbola and the plane $$x^2+y^2-z^2=1$$ $$x-y+z=3$$ goes along a curve $\gamma$. Calculate the equation for the tangent at the point $$(2,1,2)$$

Attmpted solution 2(wrong):

I use the same method as in problem 1 and I get a plane for the tangent. But it is a line in the answer. I do not understand how to get to this solution :-(

Correct answer to problem 2 $$(x,y,z)=(2,1,2) + t(1,4,3)$$

I take it I use the wrong method(s) to solve the problems?

Any can give me a hint of what is wrong with my approach here?

Answer 1

Probably not the solution you are expecting, but I'll post it anyway:

Problem 1: I note $f,g:\mathbb{R}^3\to\mathbb{R}$ the maps $f(x,y,z)=2x^2+y^2+z^2$ and $g(x,y,z)=x-y+2z$.

Since $df_{(x,y,z)}=(4x,2y,2z)$ (in matrix notation), only $0$ is a critical value of $f$, and then $4$ is a regular value and $Q=f^{-1}(4)$ is a regular surface with tangent plane

$$T_{(1,1,1)}Q=\ker(df_{(1,1,1)})=\{(u,v,w)\in\mathbb{R}^3\,;\,4u+2v+2w=0\}.$$

Since $dg_{(x,y,z)}=(1,-1,2)$, no point is critical value and then $2$ is a regular value and $P=g^{-1}(2)$ is a regular surface with tangent plane $$T_{(1,1,1)}P=\ker(dg_{(1,1,1)})=\{(u,v,w)\in\mathbb{R}^3\,;\,u-v+2w=0\}$$

(i.e. the tangent space of an affine space is the linear direction of this affine space).

Now, the intersection of $T_{(1,1,1)}P$ and $T_{(1,1,1)}Q$ is given by

$$\left\{\begin{array}{rcl} 4u+2v+2w&=&0\\ u-v+2w&=&0 \end{array}\right.\iff \left\{\begin{array}{rcl} u-v+2w&=&0\\ 6v-6w&=&0\\ \end{array}\right.\iff\left\{\begin{array}{rcl} u&=&-t\\ v&=&t\\ w&=&t \end{array}\right.,$$

so the tangent planes are transverse, i.e. $T_{(1,1,1)}P+T_{(1,1,1)}Q\overset{\ast}{=}\mathbb{R}^3$ (because of dimension reasons plus rank theorem:

$$\dim\left(T_{(1,1,1)}P+T_{(1,1,1)}Q\right)=\dim T_{(1,1,1)}P+\dim T_{(1,1,1)}Q-\dim(T_{(1,1,1)}P\cap T_{(1,1,1)}Q)=2+2-1=3$$

implying $\ast$),

implying that $P\cap Q$ is locally a curve near $(1,1,1)$ which tangent curve is given by

$$T_{(1,1,1)}P\cap T_{(1,1,1)}Q=\mathrm{span(-1,1,1)}.$$

Problem 2: You can do the same reasoning.

Answer 2

There’s no need to compute the intersection of the plane and ellipsoid directly. The direction of the tangent vector at any point on that ellipse is the same as the direction of the line of intersection of the plane $x-y+2z=2$ with the tangent plane to the ellipsoid at that point. This line is perpendicular to the normals of the two planes, so you can compute it by taking the cross product of the respective normals.

A normal to the tangent plane at the point $(x,y,z)$ is given by $\nabla(2x^2+y^2+z^2-4)=(4x,2y,2z)$, while a normal to the fixed plane can be read from its equation: $(1,-1,2)$. Plugging the coordinates of the point into $(1,-1,2)\times(4x,2y,2z)$ yields $(-6,6,6)$, which can be simplified to $(-1,1,1)$.