Find the point of intersection of the line and surface

[steller]

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:

8 + 25t + 19t^2

I dont know if this is the right approach or what to do next.
Thank you

 

[HallsofIvy]

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:

8 + 25t + 19t^2

I dont know if this is the right approach or what to do next.
Thank you

What happened to the "="?

What you should have is 8+ 25t+ 19t^2= 0. Solve that equation for t, the put that value into the parametric equations for the line to get the coordinates of the point.

 

[tkhunny]

Well, it used to be an equation and now it isn't. You should fix that.

Find the value of t and you will have it.

 

[steller]

What happened to the "="?

What you should have is 8+ 25t+ 19t^2= 0. Solve that equation for t, the put that value into the parametric equations for the line to get the coordinates of the point.

This may sound dumb but i am at a loss on how to solve for t. I moves the 8 to the right hand of the equation.
But i can only pull out one t from the equation.

t(19t + 25) = -8

 

[DrPhil]

This may sound dumb but i am at a loss on how to solve for t. I moves the 8 to the right hand of the equation.
But i can only pull out one t from the equation.

t(19t + 25) = -8

It is a quadratic equation, which may have 0, 1, or 2 solutions - that is, the line may miss, kiss, or intersect twice. BUT you have an error in the linear term .. the -t of the left should be ADDED on the right:

19 t^2 + 27 t + 8 = 0

If you don't see how to factor that right off, calculate the "discriminant" of the quadratic, (b^2 - 4ac). If that comes out to be a perfect square, then the expression does factor - but at that point I would just go ahead with the quadratic formula.