# Determining whether three points lie on a straight line in three dimension

1. Aug 26, 2008

### k1point618

1. The problem statement, all variables and given/known data
Determine whether the points lie on straight line
A(2, 4, 2) B(3, 7, -2) C(1, 3, 3)

2. Relevant equations

3. The attempt at a solution
I've looked up at the equation for lines in three dimension, and it appears to be
x=x_0+at
y=y_0+bt
z=z_0+ct

i tried to take the x y z for A and B and try to solve for a, b, c. Then if the same a, b, c work for BC, then ABC is on a line. That is my thought, but i can't manage to do the first part. I don't know how to use the information given and the equations to start with...

2. Aug 26, 2008

### Dick

You don't have to work all that hard to get the equation for the line. In vector form the equation is [x,y,z]=A+(B-A)*t. Do you see why that gives you [x,y,z]=A at t=0 and [x,y,z]=B at t=1?? Can you translate that into equations for x, y and z?

3. Aug 26, 2008

### k1point618

This part i understand.
But i'm still not sure about how to translate that in to equations for x, y and z.

4. Aug 26, 2008

### Dick

B-A=[1,3,-4], right? So you have [x,y,z]=[2,4,2]+[1,3,-4]*t. I read off x=2+t. I just equated the first component of the two sides. What do you get for y and z?

5. Aug 26, 2008

### k1point618

i see...

so y= 4+3t and z=2-4t?
and from here, i can use the x, y, z equation for points BC to see if it's a line?

6. Aug 26, 2008

### Dick

Nah, just see if C is on the line, you don't need another set of equations. If there is a t that solves all three, then it's on the line. If not, not.

7. Aug 27, 2008

### k1point618

Thank you very much. I've got it