
#1
May410, 11:19 AM

P: 87

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 3space. A line in R3 is parallel to xyplane, 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 (x_{1}, y_{1}, z_{1}) Parametric equation are: (note: t is the scalar) x =x_{1}+xt y =y_{1}+yt z =z_{1} I am not sure what I need to explain here, any help is much appreciated Thanks! 



#2
May410, 01:18 PM

Mentor
P: 20,937

Where are your symmetric equations? 



#3
May410, 01:24 PM

HW Helper
Thanks
PF Gold
P: 7,176

x =x_{1}+ at y =y_{1}+ bt z =z_{1} 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? [edit] I see mark44 answered at the same time with similar comments 



#4
May410, 01:33 PM

P: 87

Lines and Plains
Thanks Mark 44, and LCKurtz
Parametric: x =x_{1}+at y =y_{1}+bt z =z_{1} Symmetric equation: t = (xx_{1})/a t = (y  y_{1})/b 0 = zz_{1} 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
May410, 01:37 PM

Mentor
P: 20,937

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 xaxis, its parametric equations would be:
x = x_{1} + at y = y_{1} z = z_{1} 



#6
May410, 02:08 PM

P: 87

Thanks! 



#7
May410, 03:16 PM

HW Helper
Thanks
PF Gold
P: 7,176

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
May410, 03:27 PM

P: 87

ok, so the following are the parametric equations: x = x_{1} y = y_{1} z = z_{1} Symmetric equations: xx_{1} yy_{1} zz_{1} Is this correct?, and what exactly does it mean in this context.. Thanks LCKurtz! 



#9
May410, 04:43 PM

Mentor
P: 20,937





#10
May410, 11:07 PM

P: 87

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
May510, 06:47 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,879

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 [tex]t= \frac{x p}{A}= \frac{y q}{B}= \frac{z r}{C}[/tex] 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
May510, 01:12 PM

P: 87

Thanks!
So, for the example, could I use the following: Point A (1,5,0) Point B (3,9,0) Direction vector: [31,9(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 = 54t z = 0 Is my method correct? Thank you 


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