Find the point of intersection of the line and surface

Question

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}

\begin{array}{*{20}c} \frac{{ - 27 \pm 11 }}{{38}} \end{array}

and

\begin{array}{*{20}c}\frac{{ - 27 + 11 }}{{38}} and \frac{{ - 27 - 11 }}{{38}} \end{array}

\begin{array}{*{20}c} t= \frac{{ 8 }}{{19}} \end{array} and \begin{array}{*{20}c} t= -1 \end{array}

Then just plug these $t$ values into the parametric equations to get the points?

$(1) \ x = 1+t$ \begin{array}{*{20}c} x= 1 + \frac{{ 8 }}{{19}} \end{array}

$(2) \ y = 2+3t$ \begin{array}{*{20}c} x= 2 + 3 \frac{{ 8 }}{{19}} \end{array}

$(3) \ z = 1-t$ \begin{array}{*{20}c} x= 1 - \frac{{ 8 }}{{19}} \end{array}

Do the same for $t= -1$?

for the points $( 27/19, 26/19, 11/19)$ and $( 0, -1,2)$

Is this correct?

Answer 1

Your method is entirely correct, but as noted in the comment above, you may pick up unwanted solutions. Your point $(0, -1, 2)$ is on the surface, but this is not the case for $(27/19, 26/19, 11/19)$. (This point doesn't satisfy the equation of the surface. So if you double check your computations, and it results in the same coordinates, you can "throw it out," and keep $(0, -1, 2)$.

ADDED The root $t =\dfrac{8}{19}$ should be $t = \left(-\dfrac{8}{19}\right)$. Try computing the coordinates with the negative root $t = -\dfrac{8}{19}$, and see what this brings. Perhaps the point with the modified coordinates will satisfy the equation of the surface!

For $t = \left(-\dfrac{8}{19}\right)$, I get the point $\left(\dfrac{11}{19}, \dfrac{14}{19}, \dfrac{27}{19}\right)$. We need to simply check if it satisfies the equation of the surface. And so it does! (See another confirmation, below, thanks to @RecklessReckoner).

Therefore, we have two solutions!