SOLUTION: Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1) Would really appreciate if someone could show me with steps and

Algebra ->  Vectors -> SOLUTION: Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1) Would really appreciate if someone could show me with steps and       Log On


   



Question 205178: Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1)
Would really appreciate if someone could show me with steps and notes.
Thanks, -Nick

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
Hi all, can anyone please show me how to; Find the equation of the line that contains points (2,1,-3) and (-1,4,1)
Would really appreciate if someone could show me with steps and notes.
Thanks,

A line parallel to the vector v = ‹a,b,c› and passing through the
point P(x%5B1%5D,y%5B1%5D,z%5B1%5D) is represented by the
parametric equations

x=x%5B1%5D%2Bat, y=y%5B1%5D%2Bbt, z=z%5B1%5D%2Bct

or as the symmetric equations:

%28x-x%5B1%5D%29%2Fa=%28y-y%5B1%5D%29%2Fb=%28z-z%5B1%5D%29%2Fc

if none of a,b, or c are 0.

Begin by using the points P(2,1,-3) and Q(-1,4,1) 
to find a direction vector for the line passing through 
P and Q, given by
    __
v = PQ = ‹-1-(2),4-1,1-(-3)› = ‹-3,3,4›

So we substitute in 

x=x%5B1%5D%2Bat, y=y%5B1%5D%2Bbt, z=z%5B1%5D%2Bct

with ‹a,b,c› = ‹-3,3,4›  and the point P(x%5B1%5D,y%5B1%5D,z%5B1%5D) = P(2,1,-3)

x=2-3t, y=1%2B3t, z=-3%2B4t

That's the parametric equations for the line.

If you want the symmetric equation of the line, we
substitute in

%28x-x%5B1%5D%29%2Fa=%28y-y%5B1%5D%29%2Fb=%28z-z%5B1%5D%29%2Fc

%28x-2%29%2F%28-3%29=%28y-1%29%2F3=%28z-%28-3%29%29%2F4

%28x-2%29%2F%28-3%29=%28y-1%29%2F3=%28z%2B3%29%2F4

Edwin