Intersection of a plane with cylinder 
The intersection of the plane $x+y+z=1$ with the cylinder $x^{2}+y^{2}=1$  is an ellipse. What are its major and minor axes? 

I don't know how to proceed.
 A: Hint Substitution shows that the reflections $(x, y, z) \mapsto (-x, y, z)$ and $(x, y, z) \mapsto (x, -y, z)$ fix the plane and cylinder and thus the ellipse and thus the major and minor axes.
A: Using cylindrical coordinates,
$$z = g (\theta) := 1 - \cos(\theta) - \sin(\theta)$$
Differentiating $g$ and finding where the derivative vanishes, we obtain the trigonometric equation
$$\sin (\theta) = \cos (\theta)$$
which has two solutions and maps to two points on the ellipse. Finding the middle point between these two points, we find the center of the ellipse, which is $(0, 0, 1)$. The squared Euclidean distance between a point on the ellipse and the center of the ellipse is given by
$$f (\theta) := 1 + \left( g (\theta) - 1 \right)^2 = 1 + \left( \cos(\theta) + \sin(\theta) \right)^2 = 2 + 2 \cos(\theta) \sin (\theta) = 2 + \sin (2\theta)$$
Differentiating $f$ and finding where the derivative vanishes, we obtain four possible values of $\theta$, two of which we had already found by determining where the derivative of $g$ vanishes. Thus, we have a total of $4$ critical points on the ellipse:

Computing the distances between the appropriate pairs of points, we find the desired lengths.
A: An intuition-based solution:
From symmetry of $x + y + z = 1$ , it has equal intercepts on $x$ and $y$ axes and equally inclined to them.
$\Rightarrow$ Considering it as 2-D function, $ z = f(x, y) = 1 - ( x + y )$, it's gradient is $\nabla f = \frac{df}{dx} \hat{i} + \frac{df}{dy} \hat{j} = - \hat{i} - \hat{j} $ i.e. $|| \ y = x$
Also, Cylinder $x^2 + y^2 = 1$ is radially symmetrical around origin
$\Rightarrow$ For it's intercept on cutting plane, $x + y + z = 1$,


*

*It will be ellipse [ $\because$ conic section for cylinder ]

*it will be symmetric around origin i.e. center of ellipse is origin

*it will stretch the farthest along the gradient of the plane  


$\Rightarrow$ Major Axis of ellipse of intersection $|| \ y = x$ [ $\because$ point farthest from center $\in$ Major Axis ]
$\Rightarrow$ Major Axis of ellipse of intersection lies along $y = x$ [ $\because$ center $\in$ Major Axis ]
$\Rightarrow$ Minor Axis of ellipse of intersection lies along $y = -x$ [ $\because$ Major Axis $\perp$ Minor Axis, center $\in$ Minor Axis ]  
Hence


*

*Major Axis: $y = x, \ x + y + z = 1$ or $(\lambda, \lambda, 1 - 2\lambda), \lambda \in \mathbb{R}$

*Minor Axis: $y = -x, \ x + y + z = 1$ or $(\lambda, -\lambda, 1), \lambda \in \mathbb{R}$
A: The given plane $E_0:\ x+y+z=1$ cuts the plane $E_1:\ z=1$ in the line $g:\ y=-x$. And that plane $E_1$ cuts the given cylinder $C:\ x^2+y^2=1$ in the circle $c:\ x^2+y^2=1\ \&\ z=1$. Therefore the minor axis of that ellipsis in question surely has the same radius $b=r=1$.
As the slope of the plane $E_0$ wrt. plane $E_1$ is just $m=\tan(\alpha)=1$ (obtained in the plane $E_2$ perpendicular to $g$), i.e. $\alpha=45°$, we clearly get the size of the major axis by means of Pythagoras as $a^2=r^2+z^2=1+1=2$.
Therefore, wrt. an own 2D coordsystem of $E_0$, you could write your searched for ellipsis as $$e:\ \frac{X^2}2+\frac{Y^2}1=1$$
Here $Y$ is to be taken along $g$ and $X$ is along the intersection line $h=E_0\cap E_2$.
--- rk
