# Desperate for love and affection- converting a parametric equation to Cartesian

1. Sep 14, 2010

### jhg12345

1. The problem statement, all variables and given/known data
Give a Cartesian equation of the given hyperplane
The plane passing through (1,2,2) and orthogonal to the line x = (5,1,-1) + t(-1,1,-1)

2. Relevant equations

3. The attempt at a solution
So I've looked at this for a few hours and still can't figure out how to do it. It's supposed to be x(sub1) - x(sub2) + x(sub3) = 1, but I can't understand how one gets that solution. I know there should be a system of equations, with one of the equations being (1,-1,1) (which is orthogonal to the line in the question) but how about the other equations? Is the other hyperplane used for the system (5,1,-1) - (1,2,2) or something? Someone please help?!

2. Sep 14, 2010

### gabbagabbahey

Hint: If I gave you the normal to a plane and a point in the plane, could you give me the equation of the plane?

3. Sep 14, 2010

### jhg12345

Is the parametric equation the line x = (1 +t, 2-t, 2+t)? I simply plugged in the point and the normal of the vector from the question.
And I still can't figure out how to convert this to Cartesian form, but I might need to look at it some more.

4. Sep 14, 2010

### gabbagabbahey

No, a plane is not a line...

5. Sep 14, 2010

### jhg12345

Uhh, you know what I mean. Help please?

6. Sep 14, 2010

### hunt_mat

Do you know what vector describes the diection of the line you have?

7. Sep 14, 2010

### jhg12345

You take the vector and divide it by its distance. So in the case of (-1,1,-1), it would be -1/sqrt3, 1/sqrt3, -1/sqrt3

8. Sep 14, 2010

### hunt_mat

It doesn't have to be of unit length. Now what does it mean to be orthogonal? Say if I had a point (x,y,z) which was orthogonal to (-1,1,-1)?

9. Sep 14, 2010

### jhg12345

Orthogonal=perpendicular. So... (1,-1,1) obviously isn't orthogonal to my first line

10. Sep 14, 2010

### hunt_mat

Think in terms of dot products of vectors, what does it mean to be orthogonal?

11. Sep 14, 2010

### jhg12345

Oh yes the dot product has to be zero. But couldn't then the normal vector be any number of lines?

12. Sep 14, 2010

### gabbagabbahey

No, I don't. I asked if you could determine the equation of a plane, given the normal and a point in the plane. You spat out the equation of a line.

Don't worry about the line you are given yet. If I tell you that the normal to a plane is (a,b,c) and that the point $(x_0,y_0,z_0)$ lies in that plane, what is the equation of that plane?

13. Sep 14, 2010

### jhg12345

Lets see, (x,y,z)+t(a,b,c). So...

14. Sep 14, 2010

### gabbagabbahey

No, that's the equation of a line (Actually, it isn't even an equation! But assuming you meant $(x_0,y_0,z_0)+(a,b,c)t = 0$, its a line). Open up your textbook/notes and read the section on determining the equation of a plane (or Google it), and then try again.

15. Sep 14, 2010

### hunt_mat

You had the direction of te line correct, it was (-1,1,-1). There are an infitite number of lines which make up the plane which you're interested in.

16. Sep 14, 2010

### jhg12345

It's a(sub1)x(sub1)+a(sub2)x(sub2)+a(sub3)x(sub3) = c

17. Sep 14, 2010

### gabbagabbahey

Oh, and what are a(sub1), a(sub2), a(sub3) and "c" in terms of the vector and point I gave you?

18. Sep 14, 2010

### hunt_mat

If
$$(x_{1},x_{2},x_{3})$$
if perpendicular to (-1,1,-1) then?

19. Sep 14, 2010

### jhg12345

$ax_0+by_0+cz_0 = 0$

20. Sep 14, 2010

### gabbagabbahey

No.

Straight from wikipedia:

THIS is the equation you should have looked up.

Anyways, if I give you a line that is orthoganl to a plane, what can you say about the direction of the given line and the normal of the given plane?

Last edited by a moderator: May 4, 2017