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

[jhg12345]

Homework Statement

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)

Homework Equations

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?!

 

[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?

 

[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.

 

[gabbagabbahey]

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.

No, a plane is not a line...

 

[jhg12345]

Uhh, you know what I mean. Help please?

 

[hunt_mat]

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

 

[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

 

[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)?

 

[jhg12345]

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

 

[hunt_mat]

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

 

[jhg12345]

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

 

[gabbagabbahey]

Uhh, you know what I mean. Help please?

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?

 

[jhg12345]

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

 

[gabbagabbahey]

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

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.

 

[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.

 

[jhg12345]

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

 

[gabbagabbahey]

Ok fail on the last reply; it's a(sub1)x(sub1)+a(sub2)x(sub2)+a(sub3)x(sub3) = c

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

 

[hunt_mat]

If
<br /> (x_{1},x_{2},x_{3})<br />
if perpendicular to (-1,1,-1) then?

 

[jhg12345]

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

<br /> ax_0+by_0+cz_0 = 0<br />

 

[gabbagabbahey]

<br /> ax_0+by_0+cz_0 = 0<br />

No.

Straight from wikipedia:

Definition[/PLAIN] with a point and a normal vector
In a three-dimensional space, another important way of defining a plane is by specifying a point and a normal vector to the plane.

Let \textbf{r}_0 be the position vector of some known point P_0 in the plane, and let \textbf{n} be a nonzero vector normal to the plane. The idea is that a point P with position vector \textbf{r} is in the plane if and only if the vector drawn from P_0 to P is perpendicular to \textbf{n}. Recalling that two vectors are perpendicular if and only if their dot product is zero, it follows that the desired plane can be expressed as the set of all points \textbf{r} such that

\textbf{n}\cdot(\textbf{r}-\textbf{r}_0) = 0

(The dot here means a dot product, not scalar multiplication.) Expanded this becomes

a(x-x_0)+b(y-y_0)+c(z-z_0) = 0

(Where \textbf{n}=(a,b,c) and \textbf{r}_0 = (x_0,y_0,z_0) )

which is the familiar equation for a plane.[2]

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?

 

[jhg12345]

No.

what can you say about the direction of the given line and the normal of the given plane?

They're the same!

 

[jhg12345]

Okay I think I figured it out thanks for the help gagga