The cone $z=\sqrt{x^2+y^2}$ and the plane $z=1+y$
I parameterized the plane and put it into vector form:
$t=1+y \to y=t-1$
$z= 1+t-1 \to z = t$
$y= t-1$, $z=t$
Since I'm finding the intersection, I can just plug the parameters from the plane into the cone and get x:
$t = \sqrt{x^2+(t-1)^2} \to t^2 = x^2 + t^2 – 2t +1 \to x^2 = -2t+1 \to x = \sqrt{-2t+1}$
for the vector equation I have:
$\Big(\sqrt{-2t+1}i+t-1j+tk\Big)$
Best Answer
I would approach it this way.
$1+y = \sqrt{x^2 + y^2}\\ 1+2y + y^2 = x^2+ y^2\\ 1+2y = x^2\\ y = \frac 12 x^2 - \frac 12$
now I would say: $x = t\\ y = \frac 12 t^2 - \frac 12\\ z = \frac 12 t^2 + \frac 12$