# Thread: Intersection of line and surface

1. ## Intersection of line and surface

Find the coordinates of the points where the line

$\displaystyle x = t, y = 1 + t, z = 5t$

intersects the surface

$\displaystyle z = x^2 + y^2$

2. Hello, looseenz2!

I have a strange answer . . . Did I mess up somewhere?

Find the coordinates of the points where the line: .$\displaystyle \begin{Bmatrix}x &=& t \\ y &=& 1 + t \\ z &=& 5t\end{Bmatrix}$ .[1]
. . intersects the surface: .$\displaystyle z \:=\: x^2 + y^2$ .[2]

Substitute [1] into [2]: .$\displaystyle 5t \:=\:t^2 + (1+t)^2 \quad\Rightarrow\quad 3t^2 - 2t - 1 \:=\:0$

. . . . . . . . . . . . . . . . $\displaystyle (t-1)(3t+1) \:=\:0 \quad\Rightarrow\quad t \:=\:1,\:-\tfrac{1}{3}$

$\displaystyle \text{The points of intersection are: }\;(1,2,5),\;{\color{red}\rlap{//////////}}\left(\text{-}\tfrac{1}{3},\:\tfrac{2}{3},\:\text{-}\tfrac{5}{3}\right)$

The second point is extraneous. .
(I don't know why it appeared.)

$\displaystyle z \,=\,x^2+y^2$ is a paraboloid completely above the $\displaystyle xy$-plane.

There can be no intersection below the $\displaystyle xy$-plane.