Find intersection of the two surfaces $x^2-y^2-z^2=1$ and $x+y=1$ Find intersection of this two surfaces
$$x^2-y^2-z^2=1$$
 and
$$x+y=1.$$
I know that the first is hyperboloid of two sheet and the second is plane, 
but how can i find the intersection? Is it possible do calculus in one variable? How?

On the surface of the hyperboloid, given any $z$ the quantities $x$ and $y$ are related by $x^2 - y^2 = 1 + z^2$. On the plane $x$ and $y$ are related by $x+y=1$, so on the intersection of the surfaces they satisfy $$x-y = (x-y)(x+y) = 1 + z^2.$$
Add the equations $x-y = 1+z^2$ and $x+y=1$ to get $$x = 1 + \frac{z^2}{2}$$ and thus $$y = - \frac{z^2}{2}.$$ Since $z$ is arbitrary you can use it to parameterize the curve: $$ \left( 1 + \frac{t^2}{2}, - \frac{t^2} 2, t \right), \quad -\infty < t < \infty.$$
