Write parametric and symmetric equations for the z-axis.

[lamerali]

Write parametric and symmetric equations for the z-axis.

I'm not sure i am on the right track; here is my attempt to an answer.

[0, 0, z] where z can equal any number.

a = [0, 0, 1]
b = [0, 0, z]

Parametric equations

x = 0
y = 0
z = 1 + tz

Symmetric equations
\frac{x - 0}{0} = \frac{y - 0}{0} = \frac{z - 1}{z}
0 = 0 = -1

I'm not certain that my attempt to this question is correct; if anyone can guide me in the right direction if needed i would be grateful!
Thanks :)

 

[Defennder]

The parametric equations cannot contain the terms x,y,z on the RHS. So z = 1 + tz is not correct. Think in terms of vectors. We want to obtain the equation of a line for the z-axis in vector form: OP + t\vec{v} where v is the direction of the line and OP any point on the line.

For the 2nd one, take notes of the values of x,y along the z-axis.

 

[lamerali]

so would the parametric equation for z = a_z + tb_z

and for the second one since the values of x and y are always equal to zero on the z-axis; would the symmetric equations be:

x = y = z
0 = 0 = \frac{z - a_z}{b_z}<br /> <br /> Thanks

 

[Defennder]

What's a_z and b_z?

 

[HallsofIvy]

Write parametric and symmetric equations for the z-axis.

I'm not sure i am on the right track; here is my attempt to an answer.

[0, 0, z] where z can equal any number.

a = [0, 0, 1]
b = [0, 0, z]

Two points on the z axis are [0, 0, 1] and [0, 0, 2] but certainly not "[0, 0, z]" because z is not a specific number.

Parametric equations

x = 0
y = 0
z = 1 + tz

Symmetric equations
\frac{x - 0}{0} = \frac{y - 0}{0} = \frac{z - 1}{z}
0 = 0 = -1

I'm not certain that my attempt to this question is correct; if anyone can guide me in the right direction if needed i would be grateful!
Thanks :)

 

[lamerali]

ok so using the points [0,0,1] and [0,0,2] on the z-axis will i get the parametric and symmetric equations as follows:

parametric equations
x = 0
y = 0
z = 1 + 2t

and the symmetric equations

\frac{x - 0}{0} = \frac{y - 0}{0} = \frac{z - 1}{2}
0 = 0 = \frac{z - 1}{2}

is this anywhere near correct?
Thanks

 

[Defennder]

Well your parametric equations looks ok, though more complicated than necessary. The Cartesian equations (what you name "symmetric") doesn't appear correct.

 

[HallsofIvy]

ok so using the points [0,0,1] and [0,0,2] on the z-axis will i get the parametric and symmetric equations as follows:

parametric equations
x = 0
y = 0
z = 1 + 2t

x= 0, y= 0, z= t describes exactly the same line.

and the symmetric equations

\frac{x - 0}{0} = \frac{y - 0}{0} = \frac{z - 1}{2}
0 = 0 = \frac{z - 1}{2}

Well, first, (x-0)/0 and (y- 0)/0 are NOT equal to 0! Multiply the entire set of equations by 0.

is this anywhere near correct?
Thanks

the "symmetric equations" describing the z axis are x= y= 0.

 

[lamerali]

Great i think i got it!
THANKS! :D

 

[ugeous]

Im struggling with this problem as well. I got parametric no problem, but I don't quite understand how to put the symmetric equation on paper. Also, I thought that symmetric equation only exists when the denominator does not equal to zero. How exactly would I write out the equation. Thanks

 

[Defennder]

The denominator of what, to be exact? There shouldn't be any denominator equating to 0.

 

[HallsofIvy]

Im struggling with this problem as well. I got parametric no problem, but I don't quite understand how to put the symmetric equation on paper. Also, I thought that symmetric equation only exists when the denominator does not equal to zero. How exactly would I write out the equation. Thanks

If in setting up the formulas "blindly", you get
\frac{x- 0}{0}= \frac{y- 0}{0}= \frac{z-1}{2}
Then, because you can't have a "0" denominator, as Defennder said, you must have x-0= 0 and y- 0= 0. Of course, that gives you nothing for the "(z-1)/2" to be equal to so the equations are simply x= y= 0.

 

[ugeous]

Okay I see. Just to make it clear for me, say you had an example where you symmetric equation was (x-1)/2=(y-2)/3=(z-4)/0 would that mean the final equation is (x-1)/2=(y-2)/3=z?

Thanks again!

 

[HallsofIvy]

No, why would you think so? That contradicts what Defennder and I just said! Because you cannot divide by 0, you must have z-4= 0 or z= 4. The equations for the line are (x-1)/2= (y-2)/3 and z= 4. Taking the joint value of (x-1)/2 and (y-2)/3 as t, that would correspond to parametric equations x= 1+ 2t, y= 2+ 3t, z= 4.

 

[ugeous]

Indeed it does! haha. I wrote that comment around 1am..just about time my brains stopped working haha. I understand what you mean. Thanks for you help guys! I can see clearly now!