Determine all points at which the surfaces share the same tangent line.

Question

Determine all points at which the surfaces $x^2+y^2+z^2=3$ and $x^3+y^3+z^3 =3$ share the same tangent line.

I know how to get the same tangent line for the curves, but I'm not sure how to go about this for the surfaces!

Answer 1

In surface case: one just get a bit little analogy one do on two dimension.

For the first surface: $f(x,y,z)= x^2+y^2+z^2-3$,

Take partial derivative: $\frac {\partial f}{\partial x} = 2x$,$\frac {\partial f}{\partial y} = 2y$, $\frac {\partial f}{\partial z} = 2z$

So the gradient for the point on the surface is: $(2x,2y,2z)$.

For the second surface, its gradient is: $(3x^2,3y^2,3z^3)$

Since gradient along every point on the surface is perpendicular to the tangent surface where point localizes,1

So sharing a same space means sharing same normalized gradient at a specific point, so let $ (x/3,y/3,z/3) = \frac{1}{\sqrt{4x^2+4y^2+4z^2}} (2x,2y,2z) = \frac{1}{\sqrt{9x^6+9y^6+9z^6}}(3x^2,3y^2,3z^3)$ and get a system of equations:

$$x^6-9x^2+y^6+z^6 =0 \tag{1}$$ $$x^6+y^6-9y^2+z^6 =0 \tag{2}$$ $$x^6+y^6+z^6-9z^2 =0 \tag{3}$$

From $(2)-(3) \implies z = y \text{ or } z = -y$

Same way From $(1)-(2), \implies y = -x \text{ or } y = x $.

So there are four possibility (i.e. $(x,x,x)$,$(x,-x,x)$,$(x,x,-x)$,$(x,-x,-x)$ ) put it into $x^3+y^3+z^3-3=0$ and solve it.

EDIT: a intuitive explanation about why statement1 holds in two dimension:

for example $x^2+y^2 =1$, its gradient is $(2x,2y)$. After normalized, it become $(x,y)$.

See this picture:

It is apparent that gradient perpendicular to tagent line where the point lies.