# parametrization

• December 10th 2009, 02:02 AM
dizizviet
parametrization
I'm not sure when my instructor covered this but i don't even know what the question is asking and what to even do...
Find a parametrization for the line segment beginning at P1 and ending at P2

• December 10th 2009, 03:40 AM
HallsofIvy
Quote:

Originally Posted by dizizviet
I'm not sure when my instructor covered this but i don't even know what the question is asking and what to even do...
Find a parametrization for the line segment beginning at P1 and ending at P2

Parametric equations for a straight line would be of the form x= at+ b, y= ct+ d, z= et+ f with t a variable.

x goes from 1 to 0. Can you find x= at+ b so that when t=0, x= 1 and when t= 1, x= 0?

y is always -2. Can you find y= ct+ d so that when t= 0, y= -2 and when t= 1, y= -2? (Yes, that is really easy!)

z goes from 2 to 1/4. Can you find z= et+ f so that when t= 0, z= 2 and when t= 1, z= 1/4?

(I just chose t= 0 and 1 at the two points to make it easy. There are many different parametric equations for a single line.)
• December 12th 2009, 03:22 PM
dizizviet
this may sound stupid but when i plug in those variables i what do i do with the a or b variable?
• December 13th 2009, 05:58 AM
Plato
Here is the answer: $\left( {1 - t, - 2,2 - \frac{7}{4}t} \right)$

Now you tell us why that is the answer.
• December 13th 2009, 05:31 PM
dizizviet
i did the work..
P1: (1,-2,2) P2: (0,-2,1/4)
(1,-2,2)-(0,-2,1/4) = (1,0,7/4)
(0,-2,1/4) + t(1,0,7/4) = (t,-2, 1/4 + 7/4t)
But when i did this in reverse
(0,-2,1/4) - (1,-2,2) = (-1,0,-7/4)
(1,-2,2) + t(-1,0,-7/4) = (1-t,-2,2-7/4t)
I got the answer but i was wondering why i had to put P2 first instead of P1... and how is the top equations that HallsofIvy gave to me doesn't seem to be used while i was working on this. Or was it being used without me even realizing it?
And where is the rules at? i can't seem to find it....