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
Aug23-08, 12:26 PM
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
Aug24-08, 05:54 AM
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 = [tex]\frac{z - a_z}{b_z}[tex]
Thanks
Defennder
Aug24-08, 07:36 AM
What's a_z and b_z?
HallsofIvy
Aug24-08, 09:42 AM
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.
Well your parametric equations looks ok, though more complicated than necessary. The Cartesian equations (what you name "symmetric") doesn't appear correct.
HallsofIvy
Aug25-08, 07:28 AM
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.