1. ## Intersection of surfaces

The surfaces $z=\sqrt{x^2+y^2}$ and $x^2+y^2+z^2=1, z\geq$ intersect along a horizontal circle. Find an equation describing this circle. The angle between their normals at the intersection points never changes. What is this angle?

And the hint is an intersection line of two surfaces, $f(x,y,z)=0$ and $g(x,y,z)=0$, is obtained as a solution of two simultaneous equations $f(x,y,z)=0$, $g(x,y,z)=0$, but I have no idea how to solve that

2. Originally Posted by chella182
The surfaces $z=\sqrt{x^2+y^2}$ and $x^2+y^2+z^2=1, z\geq$ intersect along a horizontal circle. Find an equation describing this circle. The angle between their normals at the intersection points never changes. What is this angle?

And the hint is an intersection line of two surfaces, $f(x,y,z)=0$ and $g(x,y,z)=0$, is obtained as a solution of two simultaneous equations $f(x,y,z)=0$, $g(x,y,z)=0$, but I have no idea how to solve that

$x^2+y^2-z^2=0$ and the 2nd equation is
$x^2+y^2+z^2=1$

Subtract columnwise. You'll get $2z^2=1~\implies~z=\frac12 \sqrt{2}$

2. Plug in this value into the first equation. You'll get:

$x^2+y^2=\frac12$

3. This is the equation of a circle. Include the value of z you'll get a circle parallel to the x-y-plane in a distance of z from this plane.

3. ## Re: Intersection of surfaces

Hi! I know that this thread was started years ago but I have exactly the same question, weirdly enough, and thanks earboth because your explanation was really easy to understand BUT I was wondering about the part where you need to find the angle between their normals at intersection point....

Now I might be completely wrong but I got that the normals would be:
$n_1=2(x,y,-z)$ and
$n_2=2(x,y,z)$

which are in opposite $z$ directions so I would think that the angle would be $pi$ radians.

Am I completely wrong, and if so then how is it meant to be calculated?

(Also sorry, I can't seem to figure out how to do the Pi sign!!)

4. ## Re: Intersection of surfaces

Two vectors with opposite z components but non-zero x and y components are not opposite. for example, the vector, (1, 0, 1) and (1, 0, -1) are perpendicular (their dot product is 0) and so have angle between them $\pi/2$, not $\pi$. Opposite vectors, with angle $\pi$ have all three components opposite.

Remember that the dot product of two vectors, $\vec{u}$ and $\vec{v}$, can be written as $\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)$ where $\theta$ is the angle between the two vectors. In particular, 2<x, y, -z> and 2<x, y, z> both have length $2\sqrt{x^2+ y^2+ z^2}= 2$ because the points satisfy $x^2+ y^2+ z^2= 1$. That is, $\cos(\theta)= 2(x^2+ y^2- z^2)/2= 0$ because points on this curve also satisfy $x^2+ y^2- z^2= 0$. That is, the vectors do NOT have angle $\pi$ radians, they have angle $\pi/2$.

You can get $\pi$ by using \pi in LaTeX: [ tex ]\pi[ /tex ] without the spaces.

You are also wrong about the normal to the circle, n1. That circle, as said before, is of the form $x^2+ y^2= 2$ with $z= \sqrt{2}/2$. Since the z component is constant, the circle lies in a plane parallel to the xy-plane. The center of the circle is at $(0, 0, \sqrt{2}/2)$ and any point on the circle is of the form $(x, y, \sqrt{2}/2)$. A normal to the circle is $(x, y, 0)$, not (x, y, -z). You are correct that a normal to the sphere with center at (0, 0, 0) is parallel to a radius of that sphere and so is (x, y, z). The vector $(x, y, 0)$ has length $\sqrt{x^2+ y^2}= \sqrt{1/2}= \sqrt{2}/2$ still. The dot product of the vectors (x, y, 0) and (x, y, z) is $x^2+ y^2= 1/2$.

That is, the correct angle between the vectors is given by $cos(\theta)= \frac{1/2}{(1)(\sqrt{2}/2}= \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2}$ so that $\theta= \pi/4$.

5. ## Re: Intersection of surfaces

Thank you so much!!!