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