# Homework Help: Vector Equation in Plane Space

1. Apr 2, 2006

### Hollysmoke

I know it's not a difficult question, but I don't really know how to solve the question, and the textbook doesn't help all that much:

Determine a vector equation for a plane containing two parallel lines-
r=(0,1,3) + t(-6,-3,6)
r=(-4,5,-4)+ s(4,2,-4)

I used (0,1,3) as the position vector but where do I go from there?

2. Apr 2, 2006

### dav2008

One of the ways to find the equation of a plane is to first find two vectors that are in that plane.

Can you find those two vectors from your two lines?

3. Apr 2, 2006

### Hollysmoke

You mean directional vectors, right?

4. Apr 2, 2006

### dav2008

Yes.

I guess in this case the two lines are actually parallel so the two directional vectors are linerly dependent so you'll need to find another vector that would lie in the plane.

Basically in the end it will be just like the equation of a line but instead of only one parameter the plane will have two parameters since it's 2-dimensional.

For a line:
$$\vec{r} = \vec{u} + s\vec{v}$$
and similarly for a plane:
$$\vec{r} = \vec{u} + s\vec{v} + t\vec{w}$$

Of course you'll also need a point on a plane along with the two vectors to define it.

Edit: http://en.wikipedia.org/wiki/Plane_(mathematics) If you scroll down to "Planes embedded in R3 you can read up on that to get some more details regarding other situations dealing with planes should they arise.

Last edited: Apr 2, 2006
5. Apr 2, 2006

### Hollysmoke

I'm not getting the right answer and I think I'm starting it wrong...

6. Apr 2, 2006

### dav2008

Can you show what you are doing?

7. Apr 2, 2006

### Hollysmoke

r=sa+tb

I got as far as r=(0,1,3) + s(2,1,-2) but I don't know how to figure out the directional vector for b

8. Apr 2, 2006

### dav2008

You just need two independent vectors that are on the plane. You already found one of them by using the movement vector for the lines. The second one can be found by finding two points $$\vec{a}$$ and $$\vec{b}$$ on the plane and finding the vector from one to the other, $$\vec{ab}$$

9. Apr 2, 2006

### Hollysmoke

So I should find the vector between (2,1,-2) and (-6,-3,6)?

10. Apr 2, 2006

### dav2008

Pick a point on line 1 and another point on line 2 and find the vector from one to the other.

(2,1,-2) and (-6,-3,6) are not points on the line; they're the movement vectors of the lines.

11. Apr 2, 2006

### Hollysmoke

OH! OMG I GET IT!

So I can use the Position vectors, and I'll get (4,-4,7)!

0-(-4)=4
1-5=-4
3-(-4)=7

12. Apr 2, 2006

### Hollysmoke

Thank you! 8D