Solve for Vector Intersections: r=<t,t2,-3t> & 2x-y+z=-2 | Check My Work

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Homework Help Overview

The problem involves finding the intersection points between a parametric curve defined by the vector function r = and a plane described by the equation 2x - y + z = -2.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the process of substituting the parametric equations into the plane equation and question the correctness of the quadratic formed. There is confusion regarding the signs and coefficients in the quadratic equation derived from the intersection condition.

Discussion Status

Participants are actively engaging with the problem, identifying errors in the setup and discussing the implications of those errors. Some guidance has been offered regarding the correct substitution of variables into the plane equation, and there is acknowledgment of misunderstandings in the approach taken.

Contextual Notes

There is mention of a misunderstanding regarding the use of the components of the vector function in relation to the plane equation, which has led to incorrect conclusions about the intersection points.

reddawg
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Homework Statement



Please check my work for the following problem:

Find the point(s) where the curve r = <t,t2,-3t> intersects the plane 2x-y+z=-2.



2. The attempt at a solution

t + t2 -3t = -2

(t-2)(t+0) = -2

t=0 and t=-2

plugging those values in yields: r(0) = <0,0,0>

r(-2) = <-2,4,6>

Is that what the problem is asking for?
 
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reddawg said:

Homework Statement



Please check my work for the following problem:

Find the point(s) where the curve r = <t,t2,-3t> intersects the plane 2x-y+z=-2.



2. The attempt at a solution

t + t2 -3t = -2

(t-2)(t+0) = -2

t=0 and t=-2

plugging those values in yields: r(0) = <0,0,0>

r(-2) = <-2,4,6>

Is that what the problem is asking for?

Check your answers. <0,0,0> is not on the plane. Why? (Hint: you solved the wrong quadratic - check it).
 
I see that <0,0,0> is not on the plane however I'm not sure what other quadratic I could solve.
 
reddawg said:
I see that <0,0,0> is not on the plane however I'm not sure what other quadratic I could solve.

There's a wrong sign and a coefficient mistake in your quadratic. It's just an error. There's nothing wrong with your procedure Just fix it.
 
Last edited:
I honestly can't see the mistake all I did was match the signs with the given equations.
 
reddawg said:
I honestly can't see the mistake all I did was match the signs with the given equations.

Then you are doing something strange. You have 2x-y+z=(-2). x=t, y=t^2 and z=(-3t). If I put those values of x, y and z into the plane equation, I get 2t-t^2-3t=(-2). Why didn't you? What did you do?
 
Oh I see, I misunderstood the process. The values of x, y, z go into the plane equation.

I was using the components of r which is where I got positive t^2.

Thank you.
 
reddawg said:
Oh I see, I misunderstood the process. The values of x, y, z go into the plane equation.

I was using the components of r which is where I got positive t^2.

Thank you.

You're welcome. I still don't see quite what you did, but that's ok. Just don't do it again.
 

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