Homework Help: Lines and Plains

1. May 4, 2010

spoc21

EDIT: Sorry I made a spelling error in the title: its supposed to be Planes
1. The problem statement, all variables and given/known data

Recall that there are three coordinates planes in 3-space. A line in R3 is parallel to xy-plane, but not to any of the axes. Explain what this tells you about parametric and symmetric equations in R3. Support your answer using examples.

2. Relevant equations

N/A

3. The attempt at a solution

I know that if the line is parallel to the xy plane, it will have the direction vector:
[x,y,0]

To find the parametric equation, we need to pick a fixed point on the line
(x1, y1, z1)

Parametric equation are:
(note: t is the scalar)

x =x1+xt
y =y1+yt
z =z1

I am not sure what I need to explain here, any help is much appreciated

Thanks!

2. May 4, 2010

Staff: Mentor

It's better to represent this as, for example, <A, B, 0>.
In your parametric equations above, to find x, you need to know x1 and x, and similarly for y. Instead of x and y on the right sides, use constants, such as A and B.
In the parametric equations, the only obvious thing I see is that the z coordinate is constant.

3. May 4, 2010

LCKurtz

It would be better to write [a,b ,0]. You use x and y for the coordinate variables.

Right there is why. The x in xt isn't the same as the x on the left. Write it as

x =x1+ at
y =y1+ bt
z =z1

You are on the right track. You have taken care of the condition that the line is parallel to the xy plane. It also says the line is not parallel to any coordinate axes. What additional facts about the parameterization does that give you?

 I see mark44 answered at the same time with similar comments

4. May 4, 2010

spoc21

Thanks Mark 44, and LCKurtz
Parametric:
x =x1+at
y =y1+bt
z =z1

Symmetric equation:
t = (x-x1)/a
t = (y - y1)/b
0 = z-z1

So basically, the z value is constant? I came to that conclusion too, but what about it not being parallel to any of the axis? I am fine with the former, but very confused with the latter..Could you please elaborate a little.
Thank you.

5. May 4, 2010

Staff: Mentor

If the line were parallel to one of the axes, only one equation would have a term with a t parameter. For example, if the line happened to be parallel to the x-axis, its parametric equations would be:
x = x1 + at
y = y1
z = z1

6. May 4, 2010

spoc21

so basically, my equations are correct, as they demonstrate that the line is not parallel to any other axis..and could you please explain what you think this result means..

Thanks!

7. May 4, 2010

LCKurtz

The condition that the line not be parallel to the x or y axis gives conditions that a and b must satisfy (or not satisfy). You need to qualify your answers with those conditions. What are they?

8. May 4, 2010

spoc21

Thank you,
ok, so the following are the parametric equations:
x = x1
y = y1
z = z1

Symmetric equations:

x-x1
y-y1
z-z1

Is this correct?, and what exactly does it mean in this context..

Thanks LCKurtz!

9. May 4, 2010

Staff: Mentor

No, this is just a single point. The parametric equations are as you had them in post 4.
No, these aren't even equations.

10. May 4, 2010

spoc21

Thanks Mark 44!
what would this result mean?, other than that the z is constant.. are there any connections between this, and the other two planes?

11. May 5, 2010

HallsofIvy

In x= x1+ at, y= y1+ bt, z= z1, saying that the line is NOT parallel to the x and y coordinate axes just says that neither a nor b is 0.

I notice that you still have not found the "Symmetric Equations". Just solve each of the parametric equations for the parameter, t, and set them equal.

In general if x= At+ p, y= Bt+ q, z= Ct+ r, then
$$t= \frac{x- p}{A}= \frac{y- q}{B}= \frac{z- r}{C}$$

If any one of those coefficients is 0, just do not include that fraction. Since, here, z is a constant, the "symmetric equations" will just set a fraction in x equal to a fraction in y.

12. May 5, 2010

spoc21

Thanks!
So, for the example, could I use the following:

Point A (1,-5,0)
Point B (3,-9,0)

Direction vector:

[3-1,-9-(-5),0]
= [2,-4,0]

So therefore the parametric equation would be:

[x,y,z] = (1,-5,0) + t[2,-4,0]

x = 1 + 2t
y = -5-4t
z = 0

Is my method correct?
Thank you

13. May 5, 2010

LCKurtz

HallsofIvy, you're obviously one of the old-timers around here, but I don't understand why you don't let the OP come to that conclusion either by himself or with leading questions.