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

In summary, the homework asks for a point where the curve r=<t,t2,-3t> intersects the plane 2x-y+z=-2. When solving for t and t2, the student got t=0 and t=-2 due to a wrong sign and coefficient mistake.
  • #1
reddawg
46
0

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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  • #2
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).
 
  • #3
I see that <0,0,0> is not on the plane however I'm not sure what other quadratic I could solve.
 
  • #4
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:
  • #5
I honestly can't see the mistake all I did was match the signs with the given equations.
 
  • #6
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?
 
  • #7
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
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.
 

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

1. What is a vector?

A vector is a mathematical object that has both magnitude (size) and direction. It is often represented by an arrow pointing in the direction of the vector with its length representing the magnitude.

2. How are vectors represented?

Vectors can be represented using a variety of notations, including as a column or row of numbers, or using a geometric form such as an arrow or line segment. In mathematics and physics, vectors are commonly denoted with a boldface letter or an arrow above the letter.

3. How do you add or subtract vectors?

To add or subtract vectors, you must first ensure that they are in the same dimension (e.g. both 2D or both 3D). Then, you can simply add or subtract each component of the vectors together. This can be done visually using the parallelogram method or mathematically using the component method.

4. Can vectors be multiplied?

Yes, vectors can be multiplied by a scalar (a single number) or by another vector. Scalar multiplication results in a vector with the same direction as the original vector, but with a different magnitude. Vector multiplication, also known as the dot product or cross product, results in a scalar or vector, respectively.

5. What are some real-world applications of vectors?

Vectors are used in many fields, including physics, engineering, and computer graphics. Some common applications include calculating velocity and acceleration in physics, representing forces in engineering, and creating 3D graphics in computer programs.

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