# Thread: Find the point of intersection of the line and surface

1. ## Find the point of intersection of the line and surface

I have an odd problem with no solution. I am completely lost on how to solve this.

Problem:

Find the coordinates of the point(s) of intersection of the line $x = 1+t, y = 2+3t, z = 1-t$
and the surface $z = x^2 +2y^2$
Attempt:

(1) $x = 1+t$
(2) $y = 2+3t$
(3) $z = 1-t$

Subbing in (1), (2), (3) into the surface i have

$1-t = (1+t)^2 +2(2+3t)^2$

Solving i get:

$19t^2+27t+8=0$

At this point i need to solve for t

It appears to be a quadratic equation.

$\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}$

$\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {(27)^2 - 4(19)(8)} }}{{2(19)}}} \end{array}$

$\begin{array}{*{20}c} {t = \frac{{ - 27 \pm \sqrt {121} }}{{38}}} \end{array}$

$\frac{{ - 27 \pm 11 }}{{38}}$

$\frac{{ - 27 + 11 }}{{38}}$ and $\frac{{ - 27 - 11 }}{{38}}$

$t= \frac{{ - 8 }}{{19}}$ and $t = -1$

Then just plug these t values into the parametric equations to get the points?
Did i do this right?

right