# Parametrization of straigth line in space

1. Apr 13, 2007

### kasse

How can I parametrize the straigth line C from (2,-1,3) to (4,2,-1)?

In the xy-plane I simply use the eq. y-y(0)=m(x-x(0)) to find the parametrization, but what should I do when we have 3 dimensions?

2. Apr 13, 2007

### Dick

A straight line from A to B can be parametrized by t*B+(1-t)*A where t runs from 0 to 1.

Last edited: Apr 13, 2007
3. Apr 13, 2007

### HallsofIvy

Staff Emeritus
There are two ways to approach this: If x is a linear function of the parameter, t, i.e. x= at+ b, when t= 0, x= 2, when t= 1, x= 4, what are a and b? If y is a linear function of the parametre, t, i.e. y= ct+ d, when t= 0, y= -1, when t= 1, y= 2, what are c and d? If z is a linear function of the parameter, t, i.e. z= et+ f, when t= 0, z= 3, when t= 1, z= -1, what are e and f?

Or,treat it as a vector equation: X(t)= <x(t),y(t),z(t)>= At+ B where A and B are vectors. When t= 0 X(0)= <2, -1, 3>= B. When t= 1, X(1)= <4,2,1>= A+ B and so A= <4, 2, 1>- B= <4,2,1>- <2,-1,3>= <2, 3, -2>: X(t)= <2, 3,-2>t+ <2, -1, 3>= <2t+2,3t-1,-2t+ 3>.

Notice that, in both of these methods, as well as Dicks method take t between 0 and 1. Since there are an infinite number of different parameterizations for a curve, you are free to do that.