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
$$(x_{1},x_{2},x_{3})$$
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?

$ax_0+by_0+cz_0 = 0$

 

[gabbagabbahey]

$ax_0+by_0+cz_0 = 0$

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