# Vectors: Please check my work

1. Feb 4, 2014

### reddawg

1. The problem statement, all variables and given/known data

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?

2. Feb 4, 2014

### Dick

Check your answers. <0,0,0> is not on the plane. Why? (Hint: you solved the wrong quadratic - check it).

3. Feb 4, 2014

### reddawg

I see that <0,0,0> is not on the plane however I'm not sure what other quadratic I could solve.

4. Feb 4, 2014

### Dick

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: Feb 4, 2014
5. Feb 4, 2014

### reddawg

I honestly can't see the mistake all I did was match the signs with the given equations.

6. Feb 4, 2014

### Dick

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???

7. Feb 4, 2014

### reddawg

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.

8. Feb 4, 2014

### Dick

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