# Homework Help: Rearranging parametric equations

1. Nov 24, 2014

### Evo8

1. The problem statement, all variables and given/known data
I am given a line that passes through the points (x0, y0, z0) and (x,y,z) and a plane in 3D space being defined by these three nonlinear points (x1, y1, z1), (x2, y2, z2), (x3, y3, z3).

Im directed to use cramers rule to find the intersection of the line and plane. Applying cramer's rule should be fairly straight forward once I get to the equations I need.
2. Relevant equations

x = x0 + (x - x0)t

x= x1 + (x2 - x1)u + (x3 - x1)v

3. The attempt at a solution

What I think I need is a system of 3 equations to apply cramers rule and find the intersection point.

My three variables are t,u and v. so I think I would need 3 equations with 3 variables each. Im not sure what to do with the terms that have no unknown variable but just a value or "known" variable.

Any hints towards the right direction?

Thank you!

2. Nov 24, 2014

### Evo8

After looking at this a bit further perhaps im making it more complicated then it really is.

I can simply rearrange to the following:

x-x0 = (x - x0)t + 0(u) + 0(v)

x-x1 = 0(t) + (x2-x1)u +(x3-x1)v

If this is the case then I need to see how to apply cramers rule to a 2x3 matrix....if possible

3. Nov 24, 2014

### Staff: Mentor

You can't use Cramer's rule here as it uses a determinant, and you need a square matrix for a determinant to be defined.

4. Nov 24, 2014

### Evo8

Thats what I found as well so i feel I am doing something wrong to get the equations needed....Can I create another equation with substitution? sounds weird to say so im going to guess not but I've been playing with the equations for a little while now...

5. Nov 24, 2014

### LCKurtz

Are you using something like Maple for this? A general answer will be long enough and with enough subscripted variables that the chances of writing it correctly, or even reading it to see if it is correct, will be very small. Of course, if you are given specific points, it's much simpler.

6. Nov 24, 2014

### Evo8

I wasnt planning on using maple or Matlab. But I do have some online tools and my TI-89 if the determinants or any of those operaitons become too much to do by hand.

However all of the information thats given to me is in the first post. Im still tinkering with this one...

7. Nov 24, 2014

### Ray Vickson

You get three equations: one by equating the x-coordinates of the line and the plane, one by equating the y-coordinates, and one more by equating the z-coordinates. So, you have three equations, and three variables t, u and v.

8. Nov 24, 2014

### Evo8

as simple as it may be im not sure how to go about equating the 3 equations. Maybe I'm not as familiar with parametric equations as I should be.

For the line i have the coordinates:
x0, y0, z0
x, y, z

For the plane I have:
x1, y1, z1
x2, y2, z2
x3, y3, z3

I see that one of the given equations has the coordinates from the line and the other has only coordinates from the plane. The intersection point (x,y,z) I would assume is within both equations. I'm still lost on how to equate these 3 equations...

9. Nov 24, 2014

### Staff: Mentor

You don't "equate" equations.
What is the parametric equation of the line?
Can you write parametric equations for the plane? You need a normal to the plane, which you can get by 1) forming a vector from P1 to P2 and a vector from P1 to P3, and then 2) calculating the cross product of the two vectors.

Once you have a normal take an arbitrary point in the plane P(x, y, z) and one of the given points and form that vector. The dot product of that vector and the normal should be zero, and that will give you your plane equation.

Then you can equate coordinates as Ray described.

10. Nov 24, 2014

### Evo8

Ill have to go back through and re-read your method here but the two parametric equations are given in the first post.
x=x0 + (x` - x0)t <-- I think this one is for the line

x= x1 + (x2 - x1)u + (x3-x1)v <---- this one is for the plane

Now I want to add that there should be arrows above all of those terms indicating a vecor except for the t,u and v terms.

11. Nov 25, 2014

### Evo8

So I got some sleep and took another look at what you wrote. Its makes sense to me up to a certain point.

Here is what I've done so far:
$$P_1=\left( \begin{array}{ccc} x_1 \\ y_1 \\ z_1 \end{array} \right) P_2=\left( \begin{array}{ccc} x_2 \\ y_2 \\ z_2 \end{array} \right) P_1=\left( \begin{array}{ccc} x_3 \\ y_3 \\ z_3 \end{array} \right)$$

$$P_1P_2=\left( \begin{array}{ccc} x_2-x_1 \\ y_2-y_1 \\ z_2-z_1 \end{array} \right) P_1P_3=\left( \begin{array}{ccc} x_3-x_1 \\ y_3-y_1 \\ z_3-z_1 \end{array} \right) P_1P_2xP_1P_3=\left( \begin{array}{ccc} (y_2-y_1)(z_3-z_1)-(z_2-z_1)(y_3-y_1) \\ (z_2-z_1)(x_3-x_1)-(x_2-x_1)(z_3-z_1) \\ (x_2-x_1)(y_3-y_1)-(y_2-y_1)(x_3-x_1) \end{array} \right)$$

This is where I'm not 100% sure what you mean.

Ok an arbitrary point $$P<x,y,z>$$ and one of the given points from the vector lets choose $$<x_1,y_1,z_1>$$ Then you say the dot product of "that" vector and the normal vector should be zero. Do I take a cross product of the first two points (the arbitrary one and the selected point form the plane and THEN do the dot product of the result with the normal vector?

Lets say this is correct and i get zero. This is just a check right? Now I need to "equate coordinates" from the equation that I get from the dot product? Im not sure I fully follow here. I know I might be asking you to break it down to a really low level here but I just dont see it...

Thanks again for your help on this.

12. Nov 25, 2014

### Ray Vickson

I cannot figure out what you are doing, but I do think you are over-complicating the problem. Using notation
$$\vec{x} = \langle{x,y,z}\rangle,\\ \vec{x}_0 = \langle{x_0,y_0,z_0}\rangle, \:\vec{x}_1 = \langle{x_1,y_1,z_1}\rangle\\ \vec{p} = \langle{x'-x_0,y'-y_0,z'-z_0}\rangle\\ \vec{u} = \langle{x_2-x_1,y_2-u_1,z_2-z_1}\rangle \\ \vec{v} = \langle{x_3-x_1,y_3-y_1,z_3-z_1}\rangle$$
we have:
$$\vec{x} = \vec{x}_0 + r \vec{p} \;\; \text{(equation of line)}\\ \vec{x} =\vec{x}_1 + s \vec{u} + t \vec{v} \;\; \text{(equation of plane)}$$
You need the same $\vec{x}$ on both the line and the plane. That will give you three equations in the three unknowns $r,s,t$.

13. Nov 25, 2014

### Evo8

I'm much more confused now...

Im not sure if your using different variables for the unknowns but it seems to change through out and is different then my original problem. In addition the two parametric equations are different then the given ones.

Where did r and s come from? The original unknowns are t,u and v. Even if the r is supposed to be t the parametric eq for the line is
$$\vec{x} = \vec{x_0}+(\vec{x^\prime}-\vec{x_0})t$$
and $$\vec{x}=\vec{x_1}+(\vec{x_2}-\vec{x_1})u+(\vec{x_3}-\vec{x_1})v$$

However the equations probably aren't even that critical for me right now. I just don't get the concept of what to do with these. You comment of "you need the same $$\vec{x}$$ on both the line and the plane" I dont really get what that means. I do understand that the $$\vec{x}$$ is the point where both the plane and line will meet and that is what I need but how to apply? Im lost and its frustrating because this seem to have been meant to be an easier problem...

14. Nov 25, 2014

### Ray Vickson

It doesn't matter what we call the parameters; you can re-name them to anything you want. I preferred to use the symbols u and v to denote vectors, so I needed other names for the scalars. However, go back to whatever notation makes you most comfortable.

Of course the equation for the line is what you wrote, but in case you missed it, I set $\vec{p} = \langle{x'-x_0,y'-y_0,z'-z_0}\rangle$, so my equation is EXACTLY the same as yours---just a lot easier to write.

You say don't see what "you need the same $\vec{x}$ on both the line and the plane" means. Are you not trying to find the point $\vec{x}$ where the line and the plane meet? Is that not just a single point that is the same whether you regard it as being on the line or on the plane? (That is, assuming that the line does not lie completely in the plane.)