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.
Im completely lost. My book has not example of this and I dont remember seeing a problem like this.
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.
Im completely lost. My book has not example of this and I dont remember seeing a problem like this.
Hello, kenshinofkin!
I agree with Mathstud28 . . . The problem has no solution.
Find the coordinates of the point(s) of intersection of the line: .$\displaystyle \begin{array}{ccc}x &= &1 - t \\ y &= &2 + 3t \\ z &=& 1 - t\end{array}$
. . and the surface: .$\displaystyle z \:= \:x^2 + 2y^2$
The usual way is to substitute the parametric equations into the surface's equation.
Then we have: .$\displaystyle z \:=\:x^2 + 2y^2$
. . . . . . . . . $\displaystyle 1-t \;=\;(1-t)^2 + 2(2+3t)^2$
But this simplifies to: .$\displaystyle 19t^2 + 23t + 8 \:=\:0$ . . . which has no real solutions.
. . The line does not intersect the elliptic paraboloid.